borderline of finite with infinite Chapt13.40085 Maxwell Equations placing demands on mathematics #631 New Physics #751 ATOM TOTALITY 5th ed

[Archimedes Plutonium]

The number pi as 10^603 as a whole number and the border between
finite and infinite is this:


31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000


And for short, let me call the number simply as 3x10^603.

In year 2011, I had a computer check to see if that number had a cube
root
that was a whole number. The computer said it had no even whole number
cube root. If it had, and if it were a even whole number cube root for
the first
time in the digits of pi would have been significant to say the
least.

An even whole number cube root, if it existed, would be Nx10^201 with
its last digit "0".

Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

[hanson]

"Archimedes Plutonium" <plutonium....@gmail.com> wrote:
The number pi as 10^603 as a whole number and

___ the border between finite and infinite___ is this:

31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000
>

And for short, let me call the number simply as 3x10^603, which is
Ludwig Poehl's alias Archie-pooh's own personal Idiot Quotient,
wherein, the whole entire Universe is just one big atom

where dots of the electron-dot-cloud are galaxies

-- Archimedes Plutonium, <http://www.iw.net/~a_plutonium>
>
hanson wrote:
You are serious, Archie, aren't you?! ... Kudos!, but most of
all, thanks for the laughs!... ahahahAHAHAHA.. ROTFLMAO
ahahaha... ahahahanson

[Archimedes Plutonium]

On Jun 15, 1:18 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> The number pi as 10^603 as a whole number and the border between
> finite and infinite is this:
>
> 31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
> 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
> 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
> 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
> 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
> 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
> 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
> 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
> 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
> 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000
>
> And for short, let me call the number simply as 3x10^603.
>
> In year 2011, I had a computer check to see if that number had a cube
> root
> that was a whole number. The computer said it had no even whole number
> cube root. If it had, and if it were a even whole number cube root for
> the first
> time in the digits of pi would have been significant to say the
> least.
>
> An even whole number cube root, if it existed, would be Nx10^201 with
> its last digit "0".
>

If you are not good at math, you probably are wondering what I am
doing here,
asking about cube roots.

If you look at the number "e" it is a constant of 2.71828.. and I
could have written
it out to 603 digits and removed the decimal point and asked if it is
cube-root whole number anywhere along its string of digits? 271828..
and the answer is yes because the
first two digits 27 is 3^3.

The same sort of question I am asking about pi, if perchance the first
time that pi is
evenly cube rooted is this number of pi:

31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000

And where is pi evenly cube rooted for the first time in its digit
string? Perhaps not
at 10^603 and somewhere else.

[Archimedes Plutonium]

On Jun 15, 5:23 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Jun 15, 1:18 pm, Archimedes Plutonium

(snipped in parts)

>
> If you look at the number "e" it is a constant of 2.71828.. and I
> could have written
> it out to 603 digits and removed the decimal point and asked if it is
> cube-root whole number anywhere along its string of digits? 271828..
> and the answer is yes because the
> first two digits 27 is 3^3.
>
> The same sort of question I am asking about pi, if perchance the first
> time that pi is
> evenly cube rooted is this number of pi:
>
> 31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
> 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
> 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
> 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
> 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
> 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
> 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
> 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
> 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
> 1907021798 6094370277 0539217176
> 293176752384674818467669405132000

Now the cube root of that number would be 146459189..

So the question is, at digit 10^201 for 10^603, does that cube root
have a 0 ending digit?

I found plenty of specialness about pi at 10^603 for geometry, but if
pi is evenly cube root
at 10^603 would be a beautiful algebra specialness for pi.

[Archimedes Plutonium]

On Jun 15, 5:39 pm, Archimedes Plutonium

Now I did promise to immediately turn back again and focus on physics,
but let me
just do a little more while on this island of mathematics. As the
example above shows
that "e" is a whole number cube root at 27. So we ask whether e and pi
have more such
whole number square roots or cube roots. Now some may think that only
"27" is a whole
number square or cube root of both "e and pi". But I think we can
immediately draw up
a pattern that both pi and e have a lot more of square root and cube
roots.

This reminds me of the proof that pseudosphere area catches up with
attendant sphere area,
due to the three zeroes in a row in pi allow incremental add ons to
the pseudosphere but no
new add ons to the sphere area and so the pseudosphere catches up,
perhaps even surpasses the associated sphere when pi has those three
zeroes in a row.

So the same pattern applies to square root and cube root in that pi
and e will at some moment have
4 zeroes in a row and at some point have 6 zeroes in a row. And the
message here is that when pi has
those many zeroes in a row, that the associated square root or cube
root number will need to be able to
multiply two or three digits (square or cube root) in order to muster
a "0" And the only two way to muster that
zero or string of zeroes is by carryover, or by the digit being "0".
Now that does not constitute a proof but only
a strong indication that 27 = 3^3 is not the only whole cube root
number in both pi and e, and that there are
far more such roots.

What I really want to know is whether that number above of pi at
3x10^603 has a whole number cube root
due to those three zero digits in a row. Now in 2011, the Computer
said no, it is not a even whole number
cube root there. But perhaps the computer made a mistake or was
insufficiently programmed for the task?

[Transfer Principle]

On Jun 15, 11:18 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> And for short, let me call the number simply as 3x10^603.

A better name for the number is pi*10^603.
An even better name for the number is floor(pi*10^603).

> The same sort of question I am asking about pi, if perchance the first

> time that pi is evenly cube rooted is [floor(pi*10^603)].

It isn't a perfect cube. For trial division reveals this number is a
multiple of three, yet not nine. Therefore floor(pi*10^603) can't be a
perfect power at all.

As for perfect powers among the numbers of the form floor(pi*10^n),
let
us begin with squares. Here's a simple heuristic: as d(sqrt(x))/dx is
1/(2sqrt(x)), the probability that a natural number of approximate
size
x is a square would be 1/(2sqrt(x)). Thus the probability that a
number
of the form floor(pi*10^n) is a square is 1/(2sqrt(pi*10^n)), which we
then sum as n ranges from zero to infinity.

This is a geometric series with initial term 1/(2sqrt(pi)) and a
common
ratio of 1/sqrt(10). So the expected number of squares is:

.412556664...

As this is less than one, we shouldn't be surprised if there are _no_
squares among 3, 31, 314, 3141, 31415, 314159, ..., and very surprised
if there were two or more squares.

Replacing pi with e in the formula above doesn't give much
improvement:

.443518052...

As cubes and higher powers are much rarer than squares, we don't even
need to calculate to determine that the expected number of cubes among
3, 31, 314, 3141, 31415, 314159, ..., is close to zero. That 27 is a
cube is a fluke -- we'd be surprised to find any more cubes in the
sequence of 271, 2718, 27182, 271828, ..., and so on.

[Archimedes Plutonium]

On Jun 16, 5:00 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jun 15, 11:18 am, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > And for short, let me call the number simply as 3x10^603.
>
> A better name for the number is pi*10^603.
> An even better name for the number is floor(pi*10^603).
>
> > The same sort of question I am asking about pi, if perchance the first
> > time that pi is evenly cube rooted is [floor(pi*10^603)].
>
> It isn't a perfect cube. For trial division reveals this number is a
> multiple of three, yet not nine. Therefore floor(pi*10^603) can't be a
> perfect power at all.
>

Hi LWalk, extremely clever! I was hung up on the idea that 1000 has a
perfect cube of 10 and is not divisible by 3 or 9, but is divisible by
2 and 5. And so is floor(pi*10^603).

LWalk, since that floor pi has those three zero digits and is evenly
divisible
by both 2 and 5, could it just happen to be that it has a perfect cube
there?

Can you check into it, please?

> As for perfect powers among the numbers of the form floor(pi*10^n),
> let
> us begin with squares. Here's a simple heuristic: as d(sqrt(x))/dx is
> 1/(2sqrt(x)), the probability that a natural number of approximate
> size
> x is a square would be 1/(2sqrt(x)). Thus the probability that a
> number
> of the form floor(pi*10^n) is a square is 1/(2sqrt(pi*10^n)), which we
> then sum as n ranges from zero to infinity.
>
> This is a geometric series with initial term 1/(2sqrt(pi)) and a
> common
> ratio of 1/sqrt(10). So the expected number of squares is:
>
> .412556664...
>
> As this is less than one, we shouldn't be surprised if there are _no_
> squares among 3, 31, 314, 3141, 31415, 314159, ..., and very surprised
> if there were two or more squares.

Clever again! Would you say that such constitutes a proof that pi does
have
perfect cubes on down the line? At least "e" has the counterexample of
27
and phi has 16.

Now some say that phi is not transcendental, but when mathematics has
a borderline
of finite with infinite at floor pi 10^603, the concept of
transcendental sort of
melts away. Transcendental is meaningless in this New Math. So, LWalk,
if we accept
the borderline at 10^603 and find that the number of perfect squares
and perfect cubes
matches the number of them in all three constants-- pi, e, phi. Then
would that be a
argument in favor of the idea that transcendental is a meaningless
concept in math. That
transcendental is a sort of paper mache fakery because infinity was
fakery.

>
> Replacing pi with e in the formula above doesn't give much
> improvement:
>
> .443518052...
>
> As cubes and higher powers are much rarer than squares, we don't even
> need to calculate to determine that the expected number of cubes among
> 3, 31, 314, 3141, 31415, 314159, ..., is close to zero. That 27 is a
> cube is a fluke -- we'd be surprised to find any more cubes in the
> sequence of 271, 2718, 27182, 271828, ..., and so on.

LWalk, can you help on this issue:

Floor pi

314159...

31 = 3^3 + 2^2

314 = 17^2 + 5^2

3141 = 5^5 + 4^2


Now I do not see any clear pattern emerging and I am assuming, which
maybe wrong that
every digit string has a combination perfect square or cube (where 5^5
is (5^3) *(5^2))

Maybe someone in math history has already located this string of cubes
and squares for pi.

LWalk, have you some input on this issue?

I feel the issue of whether floor pi 10^603 is still open, in that it
maybe a perfect cube since 1000 is a perfect cube.

But I am afraid I no longer have computer access as when I had it in
2011. They sort of got tired of the demands I placed on their
machines, and I do not blame them one bit. I
do not think they realized when I started, of how tenacious and
persistent one can be
on a computer. And that is what it takes.

[Archimedes Plutonium]

Alright, the Google calculator shows the cube root of pi as:

cube root(pi) = 1.46459189
More about calculator.


But now, turning that into a floor pi cube root number is

146459189..

And the number we wish to find out is floor pi to 10^603:

31415926535 8979323846 2643383279 5028841971 6939937510 5820974944
5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000

So if I take 14 x 14 x 14 I get 2744

Then I take 146 x 146 x 146 and I get 3112..

So I now have formed the 31 of floor pi.

So what I want to know is when this calculator or computer can go out
to 201 digits and whether that 201 digit is a "0" digit. For if it is
a 0 digit would signify that floor pi 10^603 is a perfect cube.

Now if it is a perfect cube would be very remarkable and cause for
celebration.

But the question would still be open as to whether the borderline of
finite with infinite is required
out of necessity to be a perfect cube? That would be the larger and
more mysterious question.

[Transfer Principle]

On Jun 17, 10:29 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> So what I want to know is when this calculator or computer can go out
> to 201 digits and whether that 201 digit is a "0" digit. For if it is
> a 0 digit would signify that floor pi 10^603 is a perfect cube.

Unfortunately, the 201st digit happens to be five, not zero.

Furthermore, we can prove that floor(pi*10^603) is neither a sum of
two squares, nor a sum of two cubes. As it turns out, the fact that
the number is a multiple of three, but not nine, is fatal for both
types of sums.

For squares, there's a wellknown theorem involving prime factors
equivalent to 3 mod 4. No number expressible as the sum of two
squares can be divisible by a prime p unless it is also divisible
by p^2 (or the next even power). Obviously, three is equivalent to
3 mod 4, and so no number can be the sum of two squares if it's a
multiple of three and not nine, including floor(pi*10^603).

Now the result for sums of two cubes isn't as wellknown. I had to
perform a Google search to find the following page:

Broughan, "Characterizing the Sum of Two Cubes"
http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf

We scroll down to the bottom of page three, where Broughan writes:

"Example 3.1. Let neN be such that n satisfies one of the congruences
listed below.
Then n = x^3 + y^3 has no solution in Z:
2. n == 3; 4; 5 or 6 mod 9, [...]"

And we notice that a number that is divisible by three and not
3^2 must be equivalent to either 3 or 6 mod 9. Both of these are
in Broughan's list of forbidden congruences. And so it turns out
that floor(pi*10^603) can't be the sum of two cubes either. (We
easily calculate that it's equivalent to 6 mod 9.)

Of course, it's still possible that floor(pi*10^603) can be the
sum of two powers, just not two squares or two cubes. Notice that
31, from AP's list above, is a prime equivalent to 3 mod 4 (so it
can't be the sum of two squares) and 4 mod 9 (in Broughan's list,
so it can't be the sum of two cubes either), yet it's the sum of
a cube and and square, as given by AP (31 = 3^3 + 2^2), so there
is no reason that floor(pi*10^603) can't be likewise the sum of a
cube and and square.

Notice that floor(e*10^603) is equivalent to 1 mod 4, as are its
two factors easily found by trial division (seventeen and 14797),
so it might be the sum of two squares. It is equivalent to both
2 mod 7 and 2 mod 9, so it could also be the sum of two cubes. Of
course, those squares and cubes can be found only with great
difficulty -- it's well beyond my computing power.

Now floor(phi*10^603) has only one factor easily found by trial
division, and it's 34327, which is equivalent to 3 mod 4 (and
it's not divisible by 34327^2), so it can't be the sum of two
squares, though it might be the sum of two cubes as it is
congruent to 5 mod 7 and 1 mod 9.

> Would you say that such constitutes a proof that pi does
> have perfect cubes on down the line?

No. It's just a _heuristic_ telling us that perfect cubes in
such a sequence are extremely _rare_. And when one does find a
perfect power early (such as 3^3 for e or 2^4 for phi) there is
unlikely to be any others.

> Maybe someone in math history has already located this string of cubes
> and squares for pi.

I couldn't find this info on Google, but I'll keep looking.

[Transfer Principle]

On Jun 17, 12:32 am, Archimedes Plutonium

OK, I found this page:

http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/index.htm

which contains some factorizations of floor(pi*10^n) and floor(e*10^n)
for n up to 250 -- once again, 603 is too big for most computers.

Not all of the factorizations are complete, but they are sufficient to
conclude that none of the numbers are perfect powers (other than 27).

Here are the links for pi:
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_pi000.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_pi100.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_pi200.htm

And here they are for e:
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_e000.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_e100.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/de_e200.htm

The page also gives factorizations for floor(gamma*10^n), that is,
for Euler's gamma constant.

> 
where dots of the electron-dot-cloud are galaxies- Hide quoted text -
>
> - Show quoted text -

[Archimedes Plutonium]

On Jun 17, 4:30 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jun 17, 10:29 am, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > So what I want to know is when this calculator or computer can go out
> > to 201 digits and whether that 201 digit is a "0" digit. For if it is
> > a 0 digit would signify that floor pi 10^603 is a perfect cube.
>
> Unfortunately, the 201st digit happens to be five, not zero.

Hi LWalk, thanks for the valuable information, and I am sure to keep
it.

I want to know if you can detail that cube root of pi as given by
Google calculator of floor pi

146459189..

So you say that the 201st digit is 5.

LWalk, can you please tell me the five digits before the "5" and the
five digits afterwards? It maybe that we have the cube root in this
bandwidth looking like
this 19820522187 or perhaps looking like this 19822502187 (numbers I
just pulled out of the air). The reason I ask is because
that cube root has the decimal point and maybe the 201st digit is not
the one I need focus on, but rather a digit to the right or left of
the "5" digit. So could you please check to see if there are any
zeroes in that bandwidth around the "5" digit? Thanks

I am very confident that we can walk down the digits of pi and find a
combination
cube and square that will match that digit string. But whether a
pattern emerges is hard to say. Also, I would not be restricted to
just addition but perhaps use subtraction or
multiplication.

[Archimedes Plutonium]

On Jun 17, 4:30 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

> On Jun 17, 10:29 am, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > So what I want to know is when this calculator or computer can go out
> > to 201 digits and whether that 201 digit is a "0" digit. For if it is
> > a 0 digit would signify that floor pi 10^603 is a perfect cube.
>
> Unfortunately, the 201st digit happens to be five, not zero.
>
> Furthermore, we can prove that floor(pi*10^603) is neither a sum of
> two squares, nor a sum of two cubes. As it turns out, the fact that
> the number is a multiple of three, but not nine, is fatal for both
> types of sums.
>
> For squares, there's a wellknown theorem involving prime factors
> equivalent to 3 mod 4. No number expressible as the sum of two
> squares can be divisible by a prime p unless it is also divisible
> by p^2 (or the next even power). Obviously, three is equivalent to
> 3 mod 4, and so no number can be the sum of two squares if it's a
> multiple of three and not nine, including floor(pi*10^603).
>
> Now the result for sums of two cubes isn't as wellknown. I had to
> perform a Google search to find the following page:
>
> Broughan, "Characterizing the Sum of Two Cubes"http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf
>

(snipped)

LWalk, I have not had the time to look at that page, as I was
wrestling with 31415.

So far I have these:

314159...
31 = 3^3 + 2^2
314 = 17^2 + 5^2
3141 = 5^5 + 4^2

Now I found those haphazardly with no rhyme or rhythm.

So I want a technique that will fetch the succeeding digit strings
with little effort.
I found such a technique, only it spoils the demand for a duo combo of
perfect cubes
or perfect squares, which the 5^5 already spoiled.

I have the instinct that "5" is intimately connected with pi, as it is
intimately connected with
phi, the golden ratio.

So we use the 5 in construction and we can only use cubes or squares

31 = 5^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2

314 = 5^3 + 5^3 + 8^2

3141 = 25^2 + 25^2 + 25^ + 25^2 + 25^2 + 4^2

31415 = 125^2 + 25^3 + 5^2 + 5^2 + 9^2 + 3^2

Now I may have made an arithmetic error for working fast.
But the general idea is set there, so I no longer have to haphazard
look in the dark for
fits.

Here, I simply eat away with the 5 as the working engine. In larger
numbers I graduate
from 125 to that of 625 as the stock root base, then to 3125.

Now everyone knows the golden ratio log spiral and pi are circular in
shape and so the 5 in
the golden ratio must be deeply seeded inside of pi itself.

So looking at the next digit string:

314159, where I cannot start using 625 but must be content with still
the 125. And I cannot use
the 125^3 just yet, so have to use 125^2 = 15625 and how many of those
perfect squares are in
314159? There are 20 of them inside 314159 with a remainder of 314159
- 312500 = 1659 to deal with.
Now there are 2 of 625 in 1659 with a remainder of 409 and that would
be 20^2 + 3^2.

So

314159 = 20(125^2) + 2(25^2) + 20^2 + 3^2

Again I am too tired to check the arithmetic. What I wanted was a
method, a pattern and I believe every
digit of pi succumbs to this method.

Now whether this method is already known is unknown to me at this
moment.

What would be extremely exciting is if it can predict what the next
digit is.

[KBH]

>
> So we use the 5 in construction and we can only use cubes or squares
>
> 31 = 5^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2
>
> 314 = 5^3 + 5^3 + 8^2
>
> 3141 = 25^2 + 25^2 + 25^ + 25^2 + 25^2 + 4^2
>
> 31415 = 125^2 + 25^3 + 5^2 + 5^2 + 9^2 + 3^2
>

That looks like a fun game.

5^5 * 10 = 5^6 * 2
5^6 * 20 = 5^7 * 4
5^7 * 40 = 5^8 * 8

[Archimedes Plutonium]

Let me work backwards and design the number I want. It must be evenly
divisible by
2,3,4,5, thus 120 to obey Euler regular polyhedra formula. It must be
a perfect
cube.

Now floor pi 10^603 obeys Euler formula, but is not a perfect cube. So
let us try
1x10^603 for it is a perfect cube but does not obey division by 3.

So, is there any number between 10^603 and floor pi 10^603 that obeys
those
requirements? Let me sleep on it.

[KBH]

>
> That looks like a fun game.
>
> 5^5 * 10 = 5^6 * 2
> 5^6 * 20 = 5^7 * 4
> 5^7 * 40 = 5^8 * 8

5^4 * 05 = 5^5 * 1

5^5 * 10 = 5^6 * 2
5^6 * 20 = 5^7 * 4
5^7 * 40 = 5^8 * 8

.
.
.


Also note

5^3 * .25
5^4 * .50



http://www.kbhscape.com/integer.htm
.

[KBH]

On Jun 18, 6:58 am, KBH <emptyp...@hotmail.com> wrote:
> > That looks like a fun game.
>
> > 5^5 * 10 = 5^6 * 2
> > 5^6 * 20 = 5^7 * 4
> > 5^7 * 40 = 5^8 * 8
>
> 5^4 * 05 = 5^5 * 1
> 5^5 * 10 = 5^6 * 2
> 5^6 * 20 = 5^7 * 4
> 5^7 * 40 = 5^8 * 8
> .
> .
> .
>
> Also note
>
> 5^3 * .25
> 5^4 * .50
>

Further note

5^2 * .125


> http://www.kbhscape.com/integer.htm
> .

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio
news:5b1e43a2-4b3d-4a06...@l32g2000yqc.googlegroups.com...

----------
if i understand well the question...
for me the answer is not
note in the output 41 there is the "." and digits are 800 > 603
the output '42' would be the cube root of 10^603 * pi

(40) -> digits 800
(40) 800
(41) -> a:=10^603*%pi::Float
(41)
3141 5926535897 9323846264 3383279502 8841971693 9937510582 0974944592 307816
4062 8620899862 8034825342 1170679821 4808651328 2306647093 8446095505 822317
2535 9408128481 1174502841 0270193852 1105559644 6229489549 3038196442 881097
5665 9334461284 7564823378 6783165271 2019091456 4856692346 0348610454 326648
2133 9360726024 9141273724 5870066063 1558817488 1520920962 8292540917 153643
6789 2590360011 3305305488 2046652138 4146951941 5116094330 5727036575 959195
3092 1861173819 3261179310 5118548074 4623799627 4956735188 5752724891 227938
1830 1194912983 3673362440 6566430860 2139494639 5224737190 7021798609 437027
7053 9217176293 1767523846 7481846766 9405132000.5681271452 6356082778 577134
2757 7896091736 3717872146 8440901224 9534301465 4958537105 0792279689 258923
5420 1995611212 9021960864 0344181598 1362977477 1309960518 7072113499 999983
7297 8049951059 7317328160 963186
(42) -> a**(1/3)
(42)
14 6459188756 1523263020 1425272637 9039173859 6855627937 1743572559 37138393
64 9798286266 1456820678 2035382089 7503970015 2189932809 2457502599 02118221
94 3750432362 7159598553 3166075444 3485099033 9204508805 0903776065.49491969
93 4265817226 7942315230 2834162918 2121187298 0542964019 3058248338 48538995
00 9235517079 6075783189 3479803665 4785025668 6536710438 7782495366 90301754
85 9121202849 6563990817 1248396095 1880324642 4982640324 3390244800 31842046
36 0937876480 5653694078 1968354840 9250105232 6335031763 5399902875 64969642
50 4994246467 9589940958 8252542525 4116605212 6206206423 0067341943 17929867
59 8992275831 5525603964 8989691314 1491169429 3375481593 3230256788 92911946
42 3582276392 7421769692 4726443812 4058136080 0847286228 6806448822 49259775
53 9704344924 5268883966 4321715038 2072793718 3365550867 7066468961 34278473
11 8735741413 5284808750 75956866
(43) ->
----------

[Bob Masta]

On Mon, 18 Jun 2012 01:15:59 -0700 (PDT), Archimedes
Plutonium <plutonium....@gmail.com> wrote:

>On Jun 17, 4:30=A0pm, Transfer Principle <david.l.wal...@lausd.net>
>wrote:
>> On Jun 17, 10:29=A0am, Archimedes Plutonium

>>
>> <plutonium.archime...@gmail.com> wrote:
>> > So what I want to know is when this calculator or computer can go out
>> > to 201 digits and whether that 201 digit is a "0" digit. For if it is
>> > a 0 digit would signify that floor pi 10^603 is a perfect cube.

<snip>

Methods for computing the nth digit of pi are already known,
where n can be as large as you want. Google for
"Bailey-Borwein-Plouffe formula".

Best regards,


Bob Masta

DAQARTA v6.02
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
Frequency Counter, FREE Signal Generator
Pitch Track, Pitch-to-MIDI
Science with your sound card!

[Archimedes Plutonium]

On Jun 18, 6:25 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:5b1e43a2-4b3d-4a06...@l32g2000yqc.googlegroups.com...

Thanks for that valuable information.

I can see now where LWalk got the "5".

Can your machine run a cube on this number?

14 6459188756 1523263020 1425272637 9039173859 6855627937 1743572559
37138393
64 9798286266 1456820678 2035382089 7503970015 2189932809 2457502599
02118221
94 3750432362 7159598553 3166075444 3485099033 9204508805 09037760

I would like to see how close it comes to floor pi 10^603.

[Archimedes Plutonium]

On Jun 18, 5:20 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> Let me work backwards and design the number I want. It must be evenly
> divisible by
> 2,3,4,5, thus 120 to obey Euler regular polyhedra formula. It must be
> a perfect
> cube.
>
> Now floor pi 10^603 obeys Euler formula, but is not a perfect cube. So
> let us try
> 1x10^603 for it is a perfect cube but does not obey division by 3.
>
> So, is there any number between 10^603 and floor pi 10^603 that obeys
> those
> requirements? Let me sleep on it.
>

Alright I slept on the question and problem and like this outcome.


I can take the number 120 as the baseline number with repeated self
multiplication
to arrive at a number between 10^603 and 10^604. Since 120 is evenly
divisible by
2,3,4,5 or 5!, then I am guaranteed of even divisibility to satisfy
Euler regular
polyhedra formula for geometry. Our borderline must satisfy geometry
and essentially
the Euler formula is one of the most important. There are others, such
as the formula
that pseudosphere and respective sphere have equal area at infinity.
But I think with
all those zeroes the pseudosphere will have equal area.

Now the problem I face here, in engineering this borderline, is
whether it exists
between 10^603 and floor pi 10^603, or whether I have to be more
lenient and require
it to exist between 10^603 and 10^604? Or, whether it even exists is a
question mark?

It is obvious such a engineered number must satisfy Euler formula and
pseudosphere area
and other geometry results, but it is not at all obvious that it must
be a perfect cube.
I have no justification for making this engineered borderline infinity
number a perfect cube, other than to say that Physics is 3rd
dimensional and so volume of Space, finite Space would end at the
borderline and there is no reason for space to prefer the x or y
or z axis over one another. This is the Maxwell Equations in that
space is symmetrical of
3rd dimension. So for that reason, I seek to have the borderline a
perfect cube.

So, does there exist a number that is between 10^603 and 10^604 that
is a root multiple of
120, that is 120^n, and for which it is evenly divisible by 2,3,4,5
and is a perfect cube?
Is there such a number that fits that description and is it a unique
number between 10^603
and 10^604?

[Archimedes Plutonium]

On Jun 18, 6:25 am, "io_x" <a...@b.c.invalid> wrote:

> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:5b1e43a2-4b3d-4a06...@l32g2000yqc.googlegroups.com...

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies

testing why my reply did not get through

[Archimedes Plutonium]

On Jun 18, 1:02 pm, Archimedes Plutonium

Well, the Maxwell Equations are Elliptic geometry where cubes and
squares
do not exist, but rather circles and spheres. This is the Coulomb law
or inverse square law. And that law, or the Maxwell Equations in toto
are
symmetrical. When we translate that into Euclidean geometry, the
sphere
becomes a cube. So I need the borderline of finite with infinite to be
a perfect cube. So if I keep multiplying by only the number 120, do I
end up with a
number that falls between 10^603 and 10^604 and is a perfect cube?

[Archimedes Plutonium]

9.6x10^603 looks promising

Will spend the day thinking on it

[Transfer Principle]

On Jun 18, 11:02 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> So, does there exist a number that is between 10^603 and 10^604 that

> is a [power] of 120, that is 120^n [...]

No, no such number exists. The powers of closest to this range are:

120^290 = 9.174055302...*10^602
120^291 = 1.100886636...*10^605

As 291 is a multiple of three, 120^291 is a perfect cube -- its
cube root would be 120^97.

[Archimedes Plutonium]

On Jun 18, 4:00 pm, Transfer Principle <david.l.wal...@lausd.net>
wrote:

Thanks Transfer, I had some inkling that the gap was going to be
larger than 10^603
by loosening the reins over floor pi but did not expect it to go out
to 10^605. If it
is not too difficult, can you compute the perfect cube 120^n
immediately below
120^291? Maybe it is 10^600???

If I have to accept 120^291, that leaves me a problem of reconciling
pi for the
sphere pseudosphere area. They have to be equal in area at infinity.
So if infinity
starts at 120^291, I would be required to use pi at 10^-605

Pi = 3.
1415926535 8979323846 2643383279 5028841971 6939937510  : 50
5820974944 5923078164 0628620899 8628034825 3421170679  : 100
8214808651 3282306647 0938446095 5058223172 5359408128  : 150
4811174502 8410270193 8521105559 6446229489 5493038196  : 200
4428810975 6659334461 2847564823 3786783165 2712019091  : 250
4564856692 3460348610 4543266482 1339360726 0249141273  : 300
7245870066 0631558817 4881520920 9628292540 9171536436  : 350
7892590360 0113305305 4882046652 1384146951 9415116094  : 400
3305727036 5759591953 0921861173 8193261179 3105118548  : 450
0744623799 6274956735 1885752724 8912279381 8301194912  : 500
9833673362 4406566430 8602139494 6395224737 1907021798  : 550
6094370277 0539217176 2931767523 8467481846 7669405132  : 600
0005681271 4526356082 7785771342 7577896091 7363717872  : 650

I would be required to use the --00056--

So say we had a sphere of volume 1, would we have a pseudosphere
volume
of exactly 1/2 and both areas equal by using pi at those 605 digits
and where infinity is this 1.1 x 10^605.

Previously I was counting on the three zero digits to allow for the
pseudosphere
to catch up but now I need to tag on that extra two digits of 56.

The trouble here in mathematics, is that we have no computer or
formula to
compute the area and volume of pseudosphere in the 10^603 to 10^605
range.

I need to know if the pseudosphere area catches up and matches the
attendant
sphere in that range.

[Archimedes Plutonium]

On Jun 18, 4:51 pm, Archimedes Plutonium

Now I am hoping and thinking that 10^600 would be the infinity
borderline for Euclidean geometry and that floor pi 10^603 is infinity
for elliptic geometry. So that pi ending
digits are those three zero digits in a row.

So that we can find that the pseudosphere area equals the sphere area
at this borderline.

A question arises that the perfect cube 10^600 would then fit inside
the 10^603 sphere by
a specific factor. Is that factor something special?

[io_x]

"Transfer Principle" <david.l...@lausd.net> ha scritto nel messaggio
news:fb68a4b8-16c9-438c...@st3g2000pbc.googlegroups.com...

#for the number floor(pi*10^603) as factor i found 131060899

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio

news:6323a11b-e78a-4bb0...@z2g2000yqf.googlegroups.com...

-----------------
i insert only that number in b output (27) and doing b**(1.0/3.0)
output (28)

but that number is not near the number ((10^603*%pi::Float)^(1/3))^(1/3)
output (31)

(26) -> digits 800
(26) 800
(27) -> b:=146459188756152326302014252726379039173859685562793717435725593713839
36497982862661456820678203538208975039700152189932809245750259902118221943750432
362715959855331660754443485099033920450880509037760
(27)
1464591887561523263020142527263790391738596855627937174357255937138393649798_
286266145682067820353820897503970015218993280924575025990211822194375043236_
2715959855331660754443485099033920450880509037760
(28) -> b**(1.0/3.0)
(28)
2446652 0354280897 9336744946 3918665943 0104323739 6049954317 2575145319.891
9266173 9390374742 1435477755 4171241173 6656378832 2531450288 3518478018 846
8506027 3497026176 1701489045 9852604278 9393759218 4240046269 8347830394 218
9685351 2452935876 1645702412 2539436797 0414819011 0567066667 0925503351 033
6614299 3798901299 9806249227 9725824201 6324086968 5156490433 4998507854 955
8367489 3063296392 8058874944 8528052201 1916399649 1130063166 9308676033 133
6962725 6553096832 5212877367 5632733986 4421667209 2538342334 7340047541 096
9220205 4160868669 0978955811 5976062861 3557839431 0388754966 6450992232 930
7262574 6635744440 4986153505 0702652270 4428655600 3361544509 6429724237 715
3463170 8933193408 3640767034 4467534229 0517305776 7831578952 8632548208 936
8211026 5260292573 1135457431 9130367941 2428521799 1230257648 5923921981 444
5824647 9377097741 1850711547 758
(29) -> a:=10^603*%pi::Float
(29)

3141 5926535897 9323846264 3383279502 8841971693 9937510582 0974944592 307816
4062 8620899862 8034825342 1170679821 4808651328 2306647093 8446095505 822317
2535 9408128481 1174502841 0270193852 1105559644 6229489549 3038196442 881097
5665 9334461284 7564823378 6783165271 2019091456 4856692346 0348610454 326648
2133 9360726024 9141273724 5870066063 1558817488 1520920962 8292540917 153643
6789 2590360011 3305305488 2046652138 4146951941 5116094330 5727036575 959195
3092 1861173819 3261179310 5118548074 4623799627 4956735188 5752724891 227938
1830 1194912983 3673362440 6566430860 2139494639 5224737190 7021798609 437027
7053 9217176293 1767523846 7481846766 9405132000.5681271452 6356082778 577134
2757 7896091736 3717872146 8440901224 9534301465 4958537105 0792279689 258923
5420 1995611212 9021960864 0344181598 1362977477 1309960518 7072113499 999983
7297 8049951059 7317328160 963186

(30) -> a**(1.0/3.0)

(30)

14 6459188756 1523263020 1425272637 9039173859 6855627937 1743572559 37138393
64 9798286266 1456820678 2035382089 7503970015 2189932809 2457502599 02118221
94 3750432362 7159598553 3166075444 3485099033 9204508805 0903776065.49491969
93 4265817226 7942315230 2834162918 2121187298 0542964019 3058248338 48538995
00 9235517079 6075783189 3479803665 4785025668 6536710438 7782495366 90301754
85 9121202849 6563990817 1248396095 1880324642 4982640324 3390244800 31842046
36 0937876480 5653694078 1968354840 9250105232 6335031763 5399902875 64969642
50 4994246467 9589940958 8252542525 4116605212 6206206423 0067341943 17929867
59 8992275831 5525603964 8989691314 1491169429 3375481593 3230256788 92911946
42 3582276392 7421769692 4726443812 4058136080 0847286228 6806448822 49259775
53 9704344924 5268883966 4321715038 2072793718 3365550867 7066468961 34278473
11 8735741413 5284808750 75956866

(31) -> (a**(1.0/3.0))**(1.0/3.0)

(31)
11356352 7673789986 8379811464 5330918480 6238366027 9853028041 8207465619.20
75351096 7627162811 7042099802 1832220839 7560208114 9607507587 1721239114 66
85788400 7765724374 0467944903 7945453750 5375472175 5751689119 4939728323 78
93464238 5009480141 0449597179 9475254743 7899069920 4283536483 6462727302 88
77144197 6703033310 9387423243 6772924420 2520125893 8548521069 9519410499 85
75409612 7267049308 3731277286 8360379668 7105783180 4843635515 4950009573 78
82797335 5025424910 3469913436 8688838804 5899749289 0864360070 4326516517 69
91752867 8930333257 2672405790 8503826000 4837712333 5542027320 0037201078 83
08077547 0742846358 5304544243 8254513816 2132086270 6922454397 4740626414 13
10931172 8475691560 3693421238 6740604858 3606471136 4753391023 7951574678 74
33675974 4306342700 5253350749 2565067173 0777011768 3276960840 2350927107 28
60135938 0966170402 4382788701 32
(32) ->

-----------------

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio

news:884c6b19-bb2b-43a1...@v33g2000yqv.googlegroups.com...

On Jun 18, 6:25 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel
> messaggionews:5b1e43a2-4b3d-4a06...@l32g2000yqc.googlegroups.com...

>testing why my reply did not get through

i find 2 your replies this too to my post

[Archimedes Plutonium]

On Jun 19, 1:49 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:884c6b19-bb2b-43a1...@v33g2000yqv.googlegroups.com...

> On Jun 18, 6:25 am, "io_x" <a...@b.c.invalid> wrote:
>
> > "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel
> > messaggionews:5b1e43a2-4b3d-4a06...@l32g2000yqc.googlegroups.com...
> >testing why my reply did not get through
>
> i find 2 your replies this too to my post

Hi, Io, I find some posts not making their way to the poster board in
a
efficient timely manner. Whenever that happens I post a test and keep
track
of the time. So if there is prank or censoring, I keep a record. The
post
finally reached the poster board but it was delayed.

But let us get back to math.

LWalk gives this:

> 120^290 = 9.174055302...*10^602

> 120^291 = 1.100886636...*10^605

Which implies that 120^288 is the perfect cube immediately below
120^291

120^288 = approx 10^599

Now that is bad news for pi since the zeroes are from 10^600 to
10^603.

What I need now, is to see if there is a perfect cube near floor
pi*10^603
that is factorable by 120.

So I say to myself, suppose I had 120^291 which is factorable by 120
and
a perfect cube. And I wanted to make it smaller yet still be
factorable
by 120 and a perfect cube.

So I propose dividing by 9 x 9 = 81. The reasoning here is that if I
had 1000
a perfect cube and multiplied by 9 it would still be a perfect cube.
And dividing
by 81, I think, would not affect the factorability by 120.

So, I propose that (120^291)/ 81 will deliver to me a number that is
divisible by
2,3,4,5 and be a perfect cube and be a number very near floor pi
10^603.

What do you say Io?

Archimedes Plutonium

[Archimedes Plutonium]

On Jun 19, 1:48 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:6323a11b-e78a-4bb0...@z2g2000yqf.googlegroups.com...

Why is this starting out at 2446 when it should start with 3141

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

[Archimedes Plutonium]

On Jun 19, 2:08 am, Archimedes Plutonium

Alright, I know what is wrong with that proposal, that 120^291 is not
divisible by 81 to drop it lower.

So I have to take the opposite direction and multiply 120^288 to get
it
up to the 10^603 region as close as possible to floor pi 10^603. Now
that
number is somewhere about 10^598. So that means I need to multiply in
succession
by 9 for about 5 times. 9x9x9x9x9

Now I think, but I maybe wrong on this plan, that I save the 120
factorability
and I save the perfect cube. But I am a bit uneasy as to whether I
really save those
features.

If I do save those features, then I want to know what is the perfect
cube that is the
very closest to floor pi and factorable by 120.

[Archimedes Plutonium]

On Jun 19, 2:32 am, Archimedes Plutonium

Bad night for me with mistakes, I was not thinking cubes but squares,
and
I need to think of cubes. So scratch the 9 and replace with 27.

So I need five more zeros to get me from 10^598 up to 10^603. Now
instead
of getting as close as possible to floor pi 10^603, what I really want
is getting
as close as possible to 10^604 but still remaining in 10^603.
Something like
9*10^603.

I should go to bed rather than make more mistakes.

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio

news:e186fca2-e163-4d7f...@y41g2000yqm.googlegroups.com...

On Jun 19, 1:49 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel
> messaggionews:884c6b19-bb2b-43a1...@v33g2000yqv.googlegroups.com...
> On Jun 18, 6:25 am, "io_x" <a...@b.c.invalid> wrote:
>
>> > "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel

>> > messaggionews:5b1e43a2-4b3d-4a06-90d6->9ca641...@l32g2000yqc.googlegroups.com...

>> >testing why my reply did not get through
>>
>> i find 2 your replies this too to my post

>Hi, Io, I find some posts not making their way to the poster board in
>a
>efficient timely manner. Whenever that happens I post a test and keep
>track
>of the time. So if there is prank or censoring, I keep a record. The
>post
>finally reached the poster board but it was delayed.

>But let us get back to math.

>LWalk gives this:

>> 120^290 = 9.174055302...*10^602

?>> 120^291 = 1.100886636...*10^605

>Which implies that 120^288 is the perfect cube immediately below
>120^291

120^288 is a perfet cube because 120^288=120^(96*3)=(120^96)^3

>120^288 = approx 10^599

>Now that is bad news for pi since the zeroes are from 10^600 to
>10^603.

>What I need now, is to see if there is a perfect cube near floor
>pi*10^603

x^603 is a perfect cube because x^603=(x^201)^3

>that is factorable by 120.

>So I say to myself, suppose I had 120^291 which is factorable by 120
>and
>a perfect cube.

yes above

> And I wanted to make it smaller yet still be
>factorable
>by 120 and a perfect cube.

i find one number about what you write above is:
120^(291-3)=120^288

>So I propose dividing by 9 x 9 = 81. The reasoning here is that if I
>had 1000
>a perfect cube and multiplied by 9 it would still be a perfect cube.
>And dividing
>by 81, I think, would not affect the factorability by 120.

>So, I propose that (120^291)/ 81 will deliver to me a number that is
>divisible by
>2,3,4,5

here this is ok
(16) -> c:=(120 ^ 291)/ 81

(16)
1359119304133287719278562480882272733305848379173479405232879163435186918999_
690585504819382216526236402082584790197171041378399859102382461453415865321_
289342578352265056153572709886088953858778381318346248947192203901956277713_
062733148570612814040537186033993596502486544301179402174908486464784319131_
505496424448000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000
(17) -> c/3.0

(17)
453 0397680444 2923975952 0826960757 5777686161 2639115980 1744293054 4783956
396 6656352850 1606460738 8420788006 9419493006 5723680459 4666197007 9415381
780 5288440429 7808594507 5501871785 7569962029 6512862594 6043944874 9649064
067 9673187592 3768757771 6190204271 3468457286 7799786550 0828848100 3931340
583 0282882159 4773043835 1654748160 0000000000 0000000000 0000000000 0000000
000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000
000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000
000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000
000 0000000000 0000000000 0000000000 0000000000.0
(18) -> c/13.0

(18)
104 5476387794 8367071373 5575452482 5179466037 2147488303 1171759935 6488605
322 3074542965 4216875555 1174028001 6019883001 5167003182 9538353155 6788165
026 2758870868 4109675655 5885047335 1746914314 5349122137 2163987278 8380553
246 4539966367 4715867178 0659277908 7723490143 1030719973 0960503407 7830309
365 3142203575 2639933192 7304941883 0769230769 2307692307 6923076923 0769230
769 2307692307 6923076923 0769230769 2307692307 6923076923 0769230769 2307692
307 6923076923 0769230769 2307692307 6923076923 0769230769 2307692307 6923076
923 0769230769 2307692307 6923076923 0769230769 2307692307 6923076923 0769230
769 2307692307 6923076923 0769230769 2307692307.6923076923 0769230769 2307692
307 6923076923 0769230769 2307692307 6923076923 0769230769 2307692307 6923076
923 0769230769 2307692307 6923076923 0769230769 2307692307 6923076923 0769230
769 2307692307 6923076923 0769230769 2307692307 6923076923 0769230769 2307692
307 6923076923 0769230769 2307692307 6923076923 0769230769 2307692
(19) -> c

(19)
1359119304133287719278562480882272733305848379173479405232879163435186918999_
690585504819382216526236402082584790197171041378399859102382461453415865321_
289342578352265056153572709886088953858778381318346248947192203901956277713_
062733148570612814040537186033993596502486544301179402174908486464784319131_
505496424448000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000

>and be a perfect cube

this not, but near [there is a point search for "334.97"]

(20) -> c^(1.0/3.0)

(20)
11 0769244509 4953813000 7317576490 8800613087 6581618073 0079610568 29165045
76 5608141172 1200768895 5768835551 7732322756 3986047983 8802418388 79652761
37 6218804743 1403248621 4209983034 8762319742 9854552279 5682315334.97402795
57 9308371935 0288957133 4208929407 6573567592 0547800032 8483941753 19278608
04 6426203988 5333588135 9457405999 8818333031 8591313300 3663401875 83320580
34 6643593441 3488300275 8891966806 7876509613 5323671553 5104973033 83498389
94 8589307232 5600786890 6673321392 1439770924 0172726230 1465261035 57273716
20 0839661821 5911003520 0891828909 2533436478 6841068632 9963502690 16560468
73 0738964795 0241247446 4983372083 6802368106 1560938978 4066895735 33649348
25 3075256398 7355875071 9357437474 3919021479 9362682925 4680791095 28283165
81 3089058792 9695731277 1884259483 8319931720 2998616491 6569733162 55049585
68 0763626781 6842342186 8974391430 4848082947 2775548165 6917138749 18145855
65 8025588041 1035787377 8528033294 1593969805 3594064654 88660272
(21) ->

>and be a number very near floor pi
>10^603.

(21) -> a:=floor(10^603*%pi::Float)::Integer
(21)
3141592653589793238462643383279502884197169399375105820974944592307816406286_
208998628034825342117067982148086513282306647093844609550582231725359408128_
481117450284102701938521105559644622948954930381964428810975665933446128475_
648233786783165271201909145648566923460348610454326648213393607260249141273_
724587006606315588174881520920962829254091715364367892590360011330530548820_
466521384146951941511609433057270365759591953092186117381932611793105118548_
074462379962749567351885752724891227938183011949129833673362440656643086021_
394946395224737190702179860943702770539217176293176752384674818467669405132_
000

is not near c
(19) -> c
(19)
1359119304133287719278562480882272733305848379173479405232879163435186918999_
690585504819382216526236402082584790197171041378399859102382461453415865321_
289342578352265056153572709886088953858778381318346248947192203901956277713_
062733148570612814040537186033993596502486544301179402174908486464784319131_
505496424448000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000000000000000000000000000000000000000000000000000000000000000000000000000_
000

[Archimedes Plutonium]

On Jun 19, 3:47 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:e186fca2-e163-4d7f...@y41g2000yqm.googlegroups.com...

> On Jun 19, 1:49 am, "io_x" <a...@b.c.invalid> wrote:
>
>

>
> here this is ok
> (16) -> c:=(120 ^ 291)/  81
>
>    (16)
>   135911930413328771927856248088227273330584837917347940523287916343518691899 9_
>    690585504819382216526236402082584790197171041378399859102382461453415865321 _
>    289342578352265056153572709886088953858778381318346248947192203901956277713 _
>    062733148570612814040537186033993596502486544301179402174908486464784319131 _
>    505496424448000000000000000000000000000000000000000000000000000000000000000 _
>    000000000000000000000000000000000000000000000000000000000000000000000000000 _
>    000000000000000000000000000000000000000000000000000000000000000000000000000 _
>    000000000000000000000000000000000000000000000000000000000000000000000000000 _
>    000

Io, division will not work because it leads to fractions, so you have
to go to a
smaller perfect cube from 10^604 and build up a perfect cube in
10^603.

120^288 = 6.37087 x 10^598

Now we want to build that up to where it is a maximum in 10^603 but
just shy of
10^604.

So we multiply by 27 x 27 x 27 x 8 = 157464

and so we have 1.5x10^5 x 6.3x10^598 = 9.4x10^603

Now I do not know if that is the maximum for there are many other
combinations
of perfect cube multipliers than the one chosen.

There maybe a 9.999.. x 10^603 type of perfect cube factorable by 120.

[Archimedes Plutonium]

120^288 = 6.37087 x 10^598

Looks like 9.4 x 10^603 is the largest number that is a perfect cube
and divisible by 120
in 10^603.

54^3 = 157464
53^3 = 148877

6.37 x 1.48877 = 9.48
6.37 x 1.57464 = 10.0009 slightly over 10^603

So that ends this episode in an optimistic condition for we have
infinity as 9.4 x 10^603
And we use pi as 3.14159..32000 as infinity pi

So we have these two conditions baked into the numbers:

(i) Euler regular polyhedra formula is good to infinity, divisible by
2,3,4,5
(ii) infinity is a perfect cube, ie, space is symmetrical of x, y, z
axis, ie, Coulomb
inverse square law

Now we need as a final proof that the above is all correct. We need to
show that
as you take a unit pseudosphere out to 9.4x10^603 that its area is
exactly equal
to the area of unit sphere, where pi is ending in --32000. Reasoning:
the unit sphere
has larger area as we move down the digits of pi until the 601, 602,
and 603 place
value digits of pi where the pseudosphere is still adding on more
area, whilst the
sphere has stopped adding on area at 600 place value.

So if we can show that the pseudosphere and sphere have equal area for
9.4x10^603
borderline of finite with infinite, then we are finished.

Now the last time I pursued this issue of pseudosphere finds there is
no good mathematics
for area of such a large number as 10^604, and the computers are not
good this comparing
of area, either. Perhaps only a supercomputer, specially programmed
would be able to
tackle this question of whether the sphere and pseudosphere become
equal in area at the
horizon between 10^603 and 10^604.

[Transfer Principle]

On Jun 19, 3:19 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Jun 19, 3:47 am, "io_x" <a...@b.c.invalid> wrote:
> > here this is ok
> > (16) -> c:=(120 ^ 291)/  81
> >    (16)
> >   135911930413328771927856248088227273330584837917347940523287916343518691899 9_
> >    690585504819382216526236402082584790197171041378399859102382461453415865321 _
> >    289342578352265056153572709886088953858778381318346248947192203901956277713 _
> >    062733148570612814040537186033993596502486544301179402174908486464784319131 _
> >    505496424448000000000000000000000000000000000000000000000000000000000000000 _
> >    000000000000000000000000000000000000000000000000000000000000000000000000000 _
> >    000000000000000000000000000000000000000000000000000000000000000000000000000 _
> >    000000000000000000000000000000000000000000000000000000000000000000000000000 _
> >    000
> Io, division will not work because it leads to fractions

Normally, division will indeed lead to fractions, but not in
this case, because we can take advantage of powers.

Notice that 120 factors as 2^3*3*5. This means that we have:

120^291 = 2^(3*291) * 3^291 * 5^291
= 2^873 * 3^291 * 5^291

So there are lots of factors of three around, making division
very simple indeed. Of course, AP subsequently corrected his
mistake, and so we want to divide by 27, not 81 here:

120^291/27 = 2^873 * 3^288 * 5^291

This number is a perfect cube and is exactly three times the
number that Io produced above.

> There maybe a 9.999.. x 10^603 type of perfect cube factorable by 120.

Indeed there is. Notice that if a number is divisible by 2*3*5
(which equals thirty), its cube will have 2^3 * 3^3 * 5^3 as a
factor, and therefore 2^3 * 3 * 5 (which is 120) as well. This
gives us a recipe for finding a cube of the form 9.999...*10^603
that has 120 as a factor:

-- Begin with the number 10^604.
-- Take its cube root and discard everything after the decimal.
-- Divide by thirty and discard everything after the decimal.
-- Multiply it by thirty.
-- Cube it.

The resulting number will be the largest cube less than 10^603
with 120 as a factor. As it turns out, this number is approx.
10^604 - 2.111213954...*10^404, so there are 199 nines before
the first digit other than nine. (This is expected since the
d/dx(x^3) is 3x^2, so we expect the largest cube less than
10^604 to be on the order of 10^402 away. The extra two orders
taking us to 10^404 is due to the additional constraint that
the cube have 120 as a factor.)

Of course, most of AP's cubes weren't divisible merely by 120,
but by large powers of 120. It turns out that the number that
we just found happens to have 120^2 as a factor, but no more
factors of 120. (Due to excess powers of two, finding cubes
divisible by 120^2 or 120^3 are not uncommon by this method,
but we must be especially lucky to find 120^4.)

Most likely, the cube that we just found won't help AP much,
but it does provide an answer to his question.

Meanwhile, I noticed that Io started a thread looking for
factors of floor(pi*10^603) beyond the ones that he found (I
presume by trial division). Such factors will be difficult,
since the number to be factored is only a few orders shy of
the RSA numbers. (This is why the link that I provided earlier
stops at 250, with about half of those incompletely factored.)

[Archimedes Plutonium]

On Jun 19, 4:45 pm, Transfer Principle <david.l.wal...@lausd.net>

It helps me a lot.

Thanks for the data and information.

So I can say we have infinity borderline of 9.9999.. x 10^603

So two of the three tasks are done. We have 120 factorability
for Euler formula of regular polyhedra, and we have a perfect
cube. So the third task remains.

At infinity, it is said that the pseudosphere is equal in area to the
respective sphere and is 1/2 the volume of the respective sphere.

So we have Infinity as ((9.9999 x 10^603)+1) and we use that precise
exact
number for the arms of the pseudosphere. And we use pi as
3.1415..32000. That
is the only pi we are allowed to use.

Now we integrate for unit pseudosphere area with arms of that length
at infinity.
We compare this area with the unit sphere area. In every digit, except
the last three
digits of pi, is the area of sphere larger than pseudosphere. As we
add on the area of
the last three digits of pi --000 they contribute a good sizeable
amount of area, allowing the pseudosphere area to catch up with the
sphere area for the first time.

Now pi has two zero digits in a row earlier on than the 601 place
value, but the amount of area of the sphere was far ahead that two
zeros in a row could not catch up. When pi has those three zeroes in a
row, the pseudosphere caught up and became equal to the sphere area.

Can you help on this last task, LWalk? To show that the area of
pseudosphere equals the sphere when infinity is 9.9..*10^603 and pi is
ending in --32000.

Or, can you think of another place in mathematics that has such a
specific demand at
infinity. The demand of sphere and pseudosphere area being equal at
infinity is about the
only such demand that I know of in mathematics. There are some demands
of "point at infinity" but those have no numerics to them.

[Archimedes Plutonium]

On Jun 19, 5:21 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Jun 19, 4:45 pm, Transfer Principle <david.l.wal...@lausd.net>
> wrote:

>
> It helps me a lot.
>
> Thanks for the data and information.
>
> So I can say we have infinity borderline of 9.9999.. x 10^603
>
> So two of the three tasks are done. We have 120 factorability
> for Euler formula of regular polyhedra, and we have a perfect
> cube. So the third task remains.
>
> At infinity, it is said that the pseudosphere is equal in area to the
> respective sphere and is 1/2 the volume of the respective sphere.
>
> So we have Infinity as ((9.9999 x 10^603)+1) and we use that precise

Mistake by me: that infinity is the perfect cube 9.99..x10^603, not a
one added to it for it no longer is a perfect cube.

I need to use that number to show that the pseudosphere equals the
area of sphere at that number and truncated pi of just 10^-603 ending
in --32000.

The question is, is there a computer on Earth at this time good enough
to go up to 10^604 on tractrix or pseudosphere to prove the area is
equal for the first time in 10^603?

I needed to brush up on where I had ventured on showing the
pseudosphere
equals the sphere:

Newsgroups: sci.math, sci.physics, sci.logic
From: Archimedes Plutonium <plutonium.archime...@gmail.com>
Date: Wed, 18 May 2011 01:03:48 -0700 (PDT)
Local: Wed, May 18 2011 3:03 am
Subject: Chapt2 area of pseudosphere equalling and exceeding area of
sphere at 10^603 thus infinity #484 Correcting Math 3rd ed
Reply | Reply to author | Forward | Print | Individual message | Show
original | Remove | Report this message | Find messages by this author
The error of Old Math, is they do not give precision to "infinity".
To
make it precise
you must have a border between finite and infinite, and without that
border specification
one cannot utter the concept of "infinity".
About that Monotone concept:
The Computer gave this data: 
Each Out line gives (1/x,Pi,y} to have
the 1/x and y needed to have 
that value of "Pi". Thus x=1/163.581,
y=4.79046 to have an area of 
3.0
Out[53]= {163.581,3.,4.79046}
Out[54]= {696.154,3.1,6.23872}

Out[55]= {27405.1,3.14,9.91163}

Out[56]= {80956.6,3.141,10.9948}
Out[57]= {604713.,3.1415,13.0057}

Out[58]= {27017274.333, 3.14159, 16.805}
Out[59]= {118814294.732, 3.141592, 18.286}

Out[60]= {1659502386.428, 3.1415926, 20.922}
I reconfigured it to be: 
So now the above tractrixes on x-axis with
asymptote on x-axis of unit 
radius:

tractrix arms go -163.581 to +163.581 distance for a area of 3.
tractrix arms go -696.154 to +696.154 distance for a area of 3.1

tractrix arms go -27405.1 to +27405.1 for a area of 3.14

tractrix arms go -80956.6 to +80956.6 for a area of 3.141
tractrix arms go -604713. to +604713. for a area of 3.1415
tractrix arms go -27017274.333  to +27017274.333 for a area of
3.14159

tractrix arms go -118814294.732 to +118814294.732 for a area of
3.141592
tractrix arms go -1659502386.428  to +1659502386.428 for a area of
3.1415926
The above is my sense of monotone. In that the tractrix area is
steadily going along at about a pace of 10^3 in matching the circle
area of truncated pi. I mean, the tractrix goes to 163 to match the
circle at 3; and 
then the tractrix goes to 696 to match the circle at
3.1 
and this steady monotone following is approximately a factor of
about 
10^3 in arm 
length. So 10^3 is a nice steady monotone
following.
I believe that the Computer gave its best data in the above of
showing 
this monotone 
correspondence of the arms of the tractrix
following the slight 
increase in circle area and the tractrix area
just slightly less as pi 
goes out in digits.
And so, to find that borderline of Infinity, when pi has zero digits
in a row, the circle has no more contributions to area and stagnates,
whereas the tractrix arms are still piling on 
more area in this zero
digits zone. 10^603, gives the tractrix the 
opportunity for the first
time to surpass, briefly the area of the 
circle and that would be the
Border of infinity, and end of my work 
on finding that border.
Perhaps the Computer can go a distance more out, although the above
took alot of time.
Notice also, that the arms have to go out further when the 
digits in
pi are 5 to 9, rather than 
when the digits of pi are 4 or less.

[Archimedes Plutonium]

I suppose a second proof is the Riemann zeta with Euler zeta, which
pretty much parallels
what happens with tractrix/pseudosphere and circle/sphere. In both
cases, one is always
behind the other except at 10^601 through 10^603 where the laggard
catches up and equals.

For the tractrix and pseudosphere the three zeroes in a row in pi
allow the tractrix to make
a catching up. In the case of the Zetas, the spacing of primes
thinning out from 10^601 through
10^603 allows the addition zeta to catch up and equal. However, there
is a important distinction
in that in New Math with this borderline for infinity, the Zetas are
never really equal so the Riemann
Hypothesis was not well-defined since they never were equal zetas.
However, in the tractrix and
pseudosphere there really is a moment of equality. The zetas
crossover, but that is not really
equality, whereas in the tractrix there is equality.

I am kind of puzzled as to the paucity of examples of items in math
that are numeric and a
meeting at infinity. Here we have tractrix and zetas, and it should be
where math has hundreds of
these items of a numeric at infinity. Maybe there are, and I am just
not aware of it. Perhaps every
function with an asymptote is a numeric meeting at infinity. So when
we have the Function F(x) = 1/x
we know what happens at infinity for we have 9.99..x10^603, and at 0
of 1/0 we have that number again.

[io_x]

"Transfer Principle" <david.l...@lausd.net> ha scritto nel messaggio
news:cc38cd78-bf9d-46c5...@x6g2000pbh.googlegroups.com...

#io_x
#i not understand all...i have problem in understand...
#i not say all you are not clear, i say i'm not too much smart...
#
#if the game if find the biggest number 'in' 120^291
#that it is divisible by 120 and a perfet cube...
#where is the problem with the number 120^291 / 8=
#=(2^(3*291) * 3^291 * 5^291)/ 2^3=
#=2^(3*291-3) * 3^291 * 5^291 = 2^(3*290) * 3^291 * 5^291
#that is a cube, it is divisible for 120
#and it is the max that has this
# [if i not make wrong the math...]

This number is a perfect cube and is exactly three times the
number that Io produced above.

#and it is > 3 times bigger the one *you* produce :)

#possible it is easy factorizable as it says 'biofilm'; it is
#a test...

[Archimedes Plutonium]

Alright, I need to get back to pure physics very soon, but cannot
leave this
without making this vast progress. The further demand on infinity
border that
it be a perfect cube (symmetry of space coordinates x,y,z) allows for
this
added progress.

A year ago, when I was doing this, I ran into the problem that no
computer could
integrate large numbers of 10^603 and the integration of the
pseudosphere was
messy.

But what I have now and not then is more information. The information
that the
infinity border is a perfect cube and divisible by 120 and pi is
divisible by 120 of
the exponent of infinity.

I can state the proof problem differently this time, than a year ago.
A year ago, the
reasoning for the proof was that area of sphere was sleeping whenever
it encountered
zeroes in pi digits from increasing in area while the pseudosphere
kept tacking on more
area even though pi digits were zero. And when pi reached three zeroes
in a row for the
first time, the pseudosphere area caught up and equalled the sphere
area. So where the
sphere and pseudosphere area equalled one another for the first time
is the border of
finite and infinite.

But now I have a new reasoning that may allow me to bypass the messy
integration calculus.
The new reasoning allows me to say that the sphere and pseudosphere
and tractrix are discrete
or granular at some small number, such as 10^603 or 10^604, like a
computer graphic such as
the picture of a tractrix in Wikipedia showing "Catenary as involute
of a tractrix"

Now looking at the tractrix in Wikipedia, they look like curves, but
actually they are straight lines
drawn so small that we see them as curves. Therein lies the proof I
seek.

So we ask the question, at what number for infinity do we have floor
pi*10^603 so that the tractrix
area in first quadrant only is exactly equal to the area of the 1/4
circle area, both figures, mind you
are granular.

If you look at that picture in Wikipedia, the complement of the
tractrix inside the 1/4 circle of 1st
quadrant is a "fish shaped object".

The tractrix is looking like this:


\
\
_

And with the 1/4 circle this:

\ -
\ -
_ -


It looks like a "fish shaped object" that is the complement set of
circle versus tractrix.

It is that complement that we take and devolve into thin rectangles
and lay them out and see
how far they go to construct the arm of that tractrix out to infinity.
The distance they go out
is the borderline of infinity because they would make the area of the
tractrix equal to the area
of the circle.

So these rectangles are 10^604 wide, more precisely 9.999..*10^603.

So the proof would be, that for the very first time in the digits of
pi, at 10^603 we have 120
even divisibility. Then we find the perfect cube of 10^603 which also
has 120 even divisibility.

Finally, we check to see the tractrix area, 1/4 circle and take the
complement set and
devolve it into thin rectangles of width 9.999...*10^603. The proof
would be that the
devolved set of thin rectangles reaches out to exactly 9.999..*10^603
and no further.

Now a similar proof would be for the pseudosphere volume, only the
devolved thin rectangles
would be close to 10^-604 times 10^-604 by however long. And here the
pseudosphere would
exactly match 1/2 the volume of sphere at 9.999..*10^603.

Maybe I am wrong and the Calculus and modern day computers are up to
par in doing this task.
If not, the above is a alternative.

And then again, there probably is a fine analysis trick which I have
not thought of, that immediately
tells us that pi at floor 10^603 with the perfect cube infinity at
9.999..*10^603 must obey the
area equal and volume 1/2. Something about the even divisibility of
both by 120. I bet there is an
analysis trick that forgoes all of this calculus and tells us that pi
at floor 10^603 with its attendant
infinity border 9.99..*10^603 is equal in area and 1/2 volume.

[Archimedes Plutonium]

On Jun 20, 2:27 pm, Archimedes Plutonium

Sorry, typing to fast, that should have read 10^-603 and 10^-604 and
changed
on original with a (sic) sign

> computer graphic such as
> the picture of a tractrix in Wikipedia showing "Catenary as involute
> of a tractrix"
>
> Now looking at the tractrix in Wikipedia, they look like curves, but
> actually they are straight lines
> drawn so small that we see them as curves. Therein lies the proof I
> seek.
>
> So we ask the question, at what number for infinity do we have floor
> pi*10^603 so that the tractrix

that should be 10^-603

> area in first quadrant only is exactly equal to the area of the 1/4
> circle area, both figures, mind you
> are granular.
>
> If you look at that picture in Wikipedia, the complement of the
> tractrix inside the 1/4 circle of 1st
> quadrant is a "fish shaped object".
>
> The tractrix is looking like this:
>
> \
>    \
>        _
>
> And with the 1/4 circle this:
>
> \   -
>    \    -
>        _ -
>
> It looks like a "fish shaped object" that is the complement set of
> circle versus tractrix.
>
> It is that complement that we take and devolve into thin rectangles
> and lay them out and see
> how far they go to construct the arm of that tractrix out to infinity.
> The distance they go out
> is the borderline of infinity because they would make the area of the
> tractrix equal to the area
> of the circle.
>
> So these rectangles are 10^604 wide, more precisely 9.999..*10^603.

10^-604 and 10^-603

>
> So the proof would be, that for the very first time in the digits of
> pi, at 10^603 we have 120
> even divisibility. Then we find the perfect cube of 10^603 which also
> has 120 even divisibility.
>
> Finally, we check to see the tractrix area, 1/4 circle and take the
> complement set and
> devolve it into thin rectangles of width 9.999...*10^603. The proof

here again 10^-603

> would be that the
> devolved set of thin rectangles reaches out to exactly 9.999..*10^603
> and no further.
>
> Now a similar proof would be for the pseudosphere volume, only the
> devolved thin rectangles
> would be close to 10^-604 times 10^-604 by however long. And here the
> pseudosphere would
> exactly match 1/2 the volume of sphere at 9.999..*10^603.
>
> Maybe I am wrong and the Calculus and modern day computers are up to
> par in doing this task.
> If not, the above is a alternative.
>
> And then again, there probably is a fine analysis trick which I have
> not thought of, that immediately
> tells us that pi at floor 10^603 with the perfect cube infinity at
> 9.999..*10^603 must obey the
> area equal and volume 1/2. Something about the even divisibility of
> both by 120. I bet there is an
> analysis trick that forgoes all of this calculus and tells us that pi
> at floor 10^603 with its attendant
> infinity border 9.99..*10^603 is equal in area and 1/2 volume.
>

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

[Archimedes Plutonium]

I spent the day thinking about the problem and perhaps I was too hasty
in emitting overboard
the Calculus on tractrix or pseudosphere.

Here is pi to 603 digits rightward:
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944

5923078164 0628620899 8628034825 3421170679 8214808651 3282306647
0938446095 5058223172 5359408128 4811174502 8410270193 8521105559
6446229489 5493038196 4428810975 6659334461 2847564823 3786783165
2712019091 4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436 7892590360
0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724
8912279381 8301194912 9833673362 4406566430 8602139494 6395224737
1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000

Now let me state the question as clearly as I can. We think infinity
border is
9.99..x 10^603 but have to prove it. The way to prove it is that
pseudosphere
has the same exact area as the respective sphere at infinity. We can
use
the tractrix in 2 dimension which is easier, and in fact the first
quadrant is all
we need. Now to prove that 9.99..x10^603 is the borderline we have to
show that
the tractrix is always smaller in area than the circle. We use unit
circle and unit
tractrix. Of course we only pay attention to the first quadrant.

So the unit circle at infinity if our number is the border of
infinity, the unit circle area
is that number for pi given above. (Of course 1/4 for 1st quadrant.)

Now the tractrix is always a laggard and so we know for sure that the
circle at
10^600 is the same area as the circle at 10^601 through 10^603 since
there
are zeroes and so the circle cannot build anymore area during that
ride from
10^601 through 10^603. However the tractrix can add more area during
the ride
from 10^601 through 10^603.

So let us say that the ending last six digits of the tractrix at
10^600 is this ending:
--404011 whereas the circle at 10^600 is exactly --405132. So at
10^600 the tractrix
laggs behind the circle by the amount of area of 405132 subtract
404011 which is 1121 amount
of area.

So the question is, since the tractrix has to wander down three
exponents of 10^601, 10^602,
and 10^603, can it pick up enough area of that asymptotic curve to add
up to 1121 amount
of area? I believe so, and that it does in fact pick up the area so as
to equal the circle area
for the first time. And I believe the exact number for the x distance
of the tractrix that picks up
that area is the number 9.999..x10^603.

Now I believe calculus can do this task because we can segment the
calculus. We can segment
that portion of the tractrix from 10^601 through 10^603. And we can
segment other portions of the
tractrix and in the end piece together all the segments for the final
area. So I believe someone with
a supercomputer and good program can actually do the above task. Find
what specific number that the
tractrix area exactly equals the circle area. Is it the number
9.999..x10^603 the maximum perfect cube
divisible by 120 in the interval of 10^603 to 10^604.

Now there maybe an analysis trick also that is supporting evidence
that the number where the areas are
equal is 9.999..x10^603 do to the fact that both pi and this border
number are both divisible by 120 and one is a perfect cube. The
perfect cube is important because when we need exactly 1/2 the volume
for the pseudosphere versus the sphere of 4/3 pi r^3 so the unit
sphere volume involves a 4/3 factor and is handled
by the 120 divisibility.

Archimedes Plutonium

[Jos Bergervoet]

On 6/21/2012 8:23 AM, Archimedes Plutonium wrote:
> I spent the day thinking about the problem and perhaps I was too hasty
> in emitting overboard
> the Calculus on tractrix or pseudosphere.
>
> Here is pi to 603 digits rightward:

You have to list upto the 767th digit, otherwise you
will never see the pattern!

http://en.wikipedia.org/wiki/Feynman_point

--
Jos

[Archimedes Plutonium]

On Jun 21, 1:23 am, Archimedes Plutonium

Let me describe the second method proof that I started a few days ago.
Just
in case Calculus cannot prove that the pseudosphere exactly matches
the
area of sphere with pi truncated at 10^-603 and where infinity starts
exactly
at 9.999..x10^603.

What we do in the second method is create what is going to be the
smallest
bit for 2nd dimension and 3rd dimension. The number 1/9.99..x10^603 is
the smallest
nonzero positive number. It is close to 10^-604 but larger by a small
amount.

So that the smallest bit in this second method is for area that of
1.1..x10^-604 x(10^-604)
or area approx 1.1x10^-1208.

Now we look at the unit circle with pi ending in digits --32000 and
for radius 1 the unit circle
area would be that pi. It is evenly divisible by 4 since we want only
the first quadrant. And
we divide it by 4 and is approx 0.785..

Now we ask, how many bits of area of 1.1x10^-1208 are contained in the
first quadrant unit circle.
and so divide that into 0.785..

Now we find the complement set, the set that is of area of the unit
circle but not of the unit tractrix.

So that of that 0.785.. unit circle area is the complement set of
circle with tractrix. So how many of
these tiny rectangles of 1.1x10^-1208 area are contained in the
complement set?

Now we do construction. For we take the number of those complement
rectangles and build out the
rest of the arm of the tractrix.

If my hunch is correct then those complement rectangles will reach a
maximum distance of 9.99..x10^603
distance and run out of these tiny rectangles.

That is the essence of the second method, provided if the Calculus,
modern day Calculus is unable to
do the proof.

The second method will depend on the "thickness of the arm" as it
reaches each exponent and leaves that
exponent to construct the tractrix arm.

I am optimistic of this method because a trick of analysis would apply
that can prove the entire conjecture
that 9.99..x10^603 is where the distance makes the areas exactly
equal.

A similar type of argument would also prove the volume of unit
pseudosphere equals exactly 1/2 of the unit sphere volume at that
distance.

Things are looking up as roses, lillies here.

[Archimedes Plutonium]

On Jun 21, 2:15 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Jun 21, 1:23 am, Archimedes Plutonium
>

(snipped)

This second method is similar to the picket fence construction of
Calculus with slender
rectangles and a triangle atop the rectangle for angle negotiation.

Here we have just slender but tiny rectangles. Here we have rectangles
that are almost
squares however, one side is smaller than the other side.

So what we do is draw the unit circle and unit tractrix in 1st
quadrant we find how many
of these microscopic rectangles fit into the 1/4 unit circle and then
find the complement
set of those rectangles. We then use those complement rectangles to
construct the long
arm of the tractrix. If my conjecture is correct then at 9.99..x10^603
distance with pi as
truncated 3.1415..32000 that the tractrix area matches **exactly** the
circle area.

Now I called it the "cubing effect" in 2011 when I had the Computer
deliver this data

As we can see the cubing effect is pretty much enforced in that the
tractrix is a
laggard of approximately 10^3.

So that as we reach 10^-603 in the digits of pi, we have those three
zero digits in
a row and that is enough to allow the tractrix arm to have an area
that is **exactly**
equal in value to the circle area for there is no more area to add to
the circle from
exponent 10^-601 through 10^-603 for they are all zeroes.

So the only thing left to do is to see if the Computer can show that
the areas are equal
at the 10^603 (10^-603) level, and whether that number is precisely
the 9.999..x10^603
perfect cube divisible by 120.

Seems like a simple task to ask for the Computer to perform, however,
the tractrix integration
is bizarrely complicated and the best computers in the world have a
tough time with the order
of 10^603. So my guess is that it is 50 to 50 chance that computers
are sophisticated enough
to see if what I said is true or not.

However, again, there maybe a analysis trick that proves part of the
conjecture is true, that somewhere
in that territory of 10^603 is a number that matches the circle
exactly.

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio

news:6c42fc65-6d76-4e0c...@30g2000yqi.googlegroups.com...

On Jun 21, 2:15 pm, Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> On Jun 21, 1:23 am, Archimedes Plutonium

Now I called it the "cubing effect" in 2011 when I had the Computer
deliver this data
About that Monotone concept:

The Computer gave this data:

Each Out line gives (1/x,Pi,y} to have ?the 1/x and y needed to have

that value of "Pi".

#io_x
#ok so i consider x, y free

Thus x=1/163.581, ?y=4.79046 to have an area of ?3.0

#how calculate the area that you say it is... it is: pi*r^2
#it is x*y it is etc

Out[53]= {163.581,3.,4.79046}

#where do you find 163.581?
#where do you find 4.79046?

[Archimedes Plutonium]

On Jun 22, 2:38 am, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:6c42fc65-6d76-4e0c...@30g2000yqi.googlegroups.com...

Hyperbolic functions are too tough to integrate so use 1/x, however we
have two
asymptotes not one, and no cusp of tractrix.

Here is the Computer data of May 2011, posted to sci.math


Use 4* the integral in one quadrant to make up all four quadrants.
In[15]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/2,1}] 
Out[15]= 0.326505
So from x=1/2 to x=1 the area is only .326505. 
Try smaller lower
bound.
In[14]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/4,1}] 
Out[14]= 1.05692
In[16]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/8,1}] 
Out[16]= 1.75465
In[17]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/16,1}] 
Out[17]= 2.27508
As the lower bound on x gets smaller we are getting closer to Pi,
keep 
going.
In[18]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/32,1}] 
Out[18]= 2.62172
In[20]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],
{x, 
1/128,1}] 
Out[20]= 2.96831
Let the computer hunt for a lower bound so the area = 3.
In[27]:= a:=1;For[i=128,i<256,i++, 
    If[4*NIntegrate[a Log[(a
+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i, 
1}]>3.,Break[]]]; 
{i,
4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i,1}]}
Out[27]= {164,3.0003}
so for x=1/164 to x=1 the area=3.0003 but what is y?
In[53]:= a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2]/.x->1/164//N
Out[53]= 4.79302
For  -4.79<=y<=4.79 that the area=3.

Now look for 3.1 area
In[28]:= a:=1;For[i=164,i<1024,i++, 
    If[4*NIntegrate[a Log[(a
+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i, 
1}]>3.1,Break[]] 
    ]; 
{i,
4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i,1}]}
Out[30]= {697,3.10004}
In[54]:= a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2]/.x->1/697//N
Out[54]= 6.23993
Then -6.2399<=y<=6.2399 gives an area of 3.1

In[50]:= a:=1;For[i=16384,i<32768,i++, 
    If[4*NIntegrate[a Log[(a
+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i, 
1}]>3.14,Break[]]]; 
{i,
4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i,1}]}
Out[52]= {27406,3.14}
In[55]:= a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2]/.x->1/27406//N
Out[55]= 9.91166
So then -9.91166<=y<=9.91166 gives an area of 3.14

Newsgroups: sci.math, sci.physics, sci.logic
From: Archimedes Plutonium <plutonium.archime...@gmail.com>

Date: Sat, 14 May 2011 11:21:07 -0700 (PDT)
Local: Sat, May 14 2011 1:21 pm
Subject: Computer weighs in on Tractrixes (pseudosphere in 2D) #480

Correcting Math 3rd ed
Reply | Reply to author | Forward | Print | Individual message | Show
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On May 13, 6:11 pm, Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> On May 13, 12:46 pm, Archimedes Plutonium

(snipped)

- Show quoted text -
Some more Computer results:

Each Out line gives (1/x,Pi,y} to have the 1/x and y needed to have
that value of "Pi". 
Thus x=1/163.581, y=4.79046 to have an area of
3.0
Out[53]= {163.581,3.,4.79046}

Out[54]= {696.154,3.1,6.23872}
Out[55]= {27405.1,3.14,9.91163}
Out[56]= {80956.6,3.141,10.9948}
Out[57]= {604713.,3.1415,13.0057}
Out[58]= {27017274.333, 3.14159, 16.805}
Out[59]= {118814294.732, 3.141592, 18.286}
Out[60]= {1659502386.428, 3.1415926, 20.922}

[Archimedes Plutonium]

On Jun 22, 3:23 am, Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)

> Out[53]= {163.581,3.,4.79046}
> Out[54]= {696.154,3.1,6.23872}
> Out[55]= {27405.1,3.14,9.91163}
> Out[56]= {80956.6,3.141,10.9948}
> Out[57]= {604713.,3.1415,13.0057}
> Out[58]= {27017274.333, 3.14159, 16.805}
> Out[59]= {118814294.732, 3.141592, 18.286}
> Out[60]= {1659502386.428, 3.1415926, 20.922}
>

I maybe wrong, but on second reflection of the above data,
I believe it proves the problem, and even though it is
not the tractrix but function 1/x.

If you notice in the above the "cubing effect" is obeyed, since
none of the pi to x results are greater than three decimal place
values.

For example 3.14 is two place value and 27405 is five place value
3.141 is three place value and 80956 is again five place value. So
none of them exceed three place values, or the cubing effect holds.

That would mean at 10^-603 with the three zero digits in a row in pi
makes the circle area equal to the tractrix area.

So I think that such is a proof, even though it is not the tractrix,
for
function 1/x bounds the tractrix.

It does not prove that 9.999..x10^603 is where the x distance reaches
the
equality but it does imply that some number in 10^603 is the equality
number.

[io_x]

"Archimedes Plutonium" <plutonium....@gmail.com> ha scritto nel messaggio

news:1454c6ca-bb85-4d3c...@v33g2000yqv.googlegroups.com...

{x, ?1/2,1}] ?Out[15]= 0.326505
So from x=1/2 to x=1 the area is only .326505. ?Try smaller lower

bound.
In[14]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],

{x, ?1/4,1}] ?Out[14]= 1.05692

In[16]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],

{x, ?1/8,1}] ?Out[16]= 1.75465

In[17]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],

{x, ?1/16,1}] ?Out[17]= 2.27508

As the lower bound on x gets smaller we are getting closer to Pi,

keep ?going.

In[18]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],

{x, ?1/32,1}] ?Out[18]= 2.62172

In[20]:= a:=1;4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],

{x, ?1/128,1}] ?Out[20]= 2.96831

Let the computer hunt for a lower bound so the area = 3.

In[27]:= a:=1;For[i=128,i<256,i++, ? If[4*NIntegrate[a Log[(a
+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i, ?1}]>3.,Break[]]]; ?{i,

4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i,1}]}
Out[27]= {164,3.0003}
so for x=1/164 to x=1 the area=3.0003 but what is y?
In[53]:= a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2]/.x->1/164//N
Out[53]= 4.79302
For -4.79<=y<=4.79 that the area=3.

Now look for 3.1 area

In[28]:= a:=1;For[i=164,i<1024,i++, ? If[4*NIntegrate[a Log[(a
+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i, ?1}]>3.1,Break[]] ? ]; ?{i,

4*NIntegrate[a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2],{x,1/i,1}]}
Out[30]= {697,3.10004}
In[54]:= a Log[(a+Sqrt[a^2-x^2])/x]-Sqrt[a^2-x^2]/.x->1/697//N
Out[54]= 6.23993
Then -6.2399<=y<=6.2399 gives an area of 3.1

-----------------
this is io_x:
i not follow very well...
all i can show is this for case "area 3.1" when i not know the area of what...:
"S mean integrate"

(23) -> r(x) == log( (1+sqrt(1-x^2))/x) -sqrt(1-x^2)
Compiled code for r has been cleared.
1 old definition(s) deleted for function or rule r
(24) -> r(x) == numeric( log( (1+sqrt(1-x^2))/x) -sqrt(1-x^2) )
1 old definition(s) deleted for function or rule r
(25) -> g(a:Integer):Float==4*numeric(S(log( (1+sqrt(1-x^2))/x) -sqrt(1-x^2) ,x
=1/a..1, "noPole") )
Function declaration g : Integer -> Float has been added to
workspace.
Compiled code for g has been cleared.
1 old definition(s) deleted for function or rule g
(26) -> for h in 690..700 repeat output[h, g(h), r(1/h)]
There are 4 exposed and 0 unexposed library operations named numeric
having 1 argument(s) but none was determined to be applicable.
Use HyperDoc Browse, or issue
)display op numeric
to learn more about the available operations. Perhaps
package-calling the operation or using coercions on the arguments
will allow you to apply the operation.
Cannot find a definition or applicable library operation named
numeric with argument type(s)
Union(f1: OrderedCompletion Expression Integer,f2: List OrderedCompletion Expres
sion Integer,fail: failed,pole: potentialPole)

Perhaps you should use "@" to indicate the required return type,
or "$" to specify which version of the function you need.
AXIOM will attempt to step through and interpret the code.
Compiling function g with type Integer -> Float
Compiling function r with type Fraction Integer -> Float
[690.0,3.0996805664 029717561,6.2298393032 511571967]
[691.0,3.0997328594 606100318,6.2312875279 0879529]
[692.0,3.0997849722 829155072,6.2327336582 466475413]
[693.0,3.0998369639 135987746,6.2341777003 132751606]
[694.0,3.0998887994 104812058,6.2356196601 310740791]
[695.0,3.0999405453 466507909,6.2370595436 964256488]
[696.0, 3.0999920894 807095502, 6.2384973569 798462589]
[697.0, 3.1000435450 426582761, 6.2399331059 261358768]

[698.0,3.1000948531 346322315,6.2413667964 545255257]
[699.0,3.1001460082 028346694,6.2427984344 588237059]
[700.0,3.1001970830 840629751,6.2442280258 075617694]
(27) ->

[Narasimham G.L.]

On Jun 25, 1:28 pm, "io_x" <a...@b.c.invalid> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> ha scritto nel messaggionews:1454c6ca-bb85-4d3c...@v33g2000yqv.googlegroups.com...

At any saddle point,every surface of negative Gauss curvature can
be partitioned into two pairs of normal curvature, one negative
and other positive.They are separated by two conjugate asymptotes.

> and no cusp of tractrix.

No.

The most important feature of the pseudosphere is its circumferential
cuspidal edge

[Transfer Principle]

On Jun 20, 1:06 am, "io_x" <a...@b.c.invalid> wrote:
> "Transfer Principle" <david.l.wal...@lausd.net> ha scritto nel messaggionews:cc38cd78-bf9d-46c5...@x6g2000pbh.googlegroups.com...

> On Jun 19, 3:19 am, Archimedes Plutonium

> Meanwhile, I noticed that Io started a thread looking for
> factors of floor(pi*10^603) beyond the ones that he found (I
> presume by trial division). Such factors will be difficult,
> since the number to be factored is only a few orders shy of
> the RSA numbers. (This is why the link that I provided earlier
> stops at 250, with about half of those incompletely factored.)
> #possible it is easy factorizable as it says 'biofilm'; it is
> #a test...

With today being Tau Day (tau = 2pi = 6.2831853...), it got me
wondering about floor(tau*10^603) and its factors.

Notice that the next digit after the -000- in pi is five. This
means that the 601st through 603rd digits in tau are -001-, not
-000- (otherwise floor(tau*10^603) would be 2*floor(pi*10^603)
and so the factorization of floor(tau*10^603) would be no more
interesting than that of floor(pi*10^603)).

Trial division of floor(tau*10^603) reveals no small factors
(and I think my calculations go up to at least 10^6, possibly
even 10^7 or so). It would be interesting if this number turned
out to be prime...

[James Waldby]

On Thu, 28 Jun 2012 17:57:12 -0700, Transfer Principle wrote:
[I adjusted attributions, which looked jumbled, and snipped
two of the three newsgroups irrelevant for the question]

> On Jun 20, 1:06 am, "io_x" <a...@b.c.invalid> wrote:

>> "Transfer Principle" <david.l.wal...@lausd.net> ha scritto...

>>> Meanwhile, I noticed that Io started a thread looking for
>>> factors of floor(pi*10^603) beyond the ones that he found (I
>>> presume by trial division). Such factors will be difficult,
>>> since the number to be factored is only a few orders shy of
>>> the RSA numbers. (This is why the link that I provided earlier
>>> stops at 250, with about half of those incompletely factored.)

...
> With today being Tau Day (tau = 2pi = 6.2831853...), it got me
> wondering about floor(tau*10^603) and its factors.
>
> Notice that the next digit after the -000- in pi is five. This
> means that the 601st through 603rd digits in tau are -001-, not
> -000- (otherwise floor(tau*10^603) would be 2*floor(pi*10^603)
> and so the factorization of floor(tau*10^603) would be no more
> interesting than that of floor(pi*10^603)).
>
> Trial division of floor(tau*10^603) reveals no small factors
> (and I think my calculations go up to at least 10^6, possibly
> even 10^7 or so). It would be interesting if this number turned
> out to be prime...

Whether interesting or not, it would be surprising. But that
isn't my reason for replying, which is to ask: How is it that
you don't know more specifically a bound on the small factors
tested by your calculations? How did you run the calculation?

--
jiw

[Transfer Principle]

On a cheap online calculator. I just enter the number, click
"factor," and wait an hour. All I have to go by are the factors
found for previous numbers entered and they were around the order
of 10^6 or 10^7.

I'm aware that professional mathematicians would use Mathematica
or Python to trial divide, so they would input the exact bound for
the trial division, but I don't have those. And of course, more
advance methods like Fermat or elliptic curves would be out of the
question for me. That's why that post was directed to _Io_, who
presumably does have access to that software (or at least better
software than I have).

Anyway, Happy Tau Day, James!

[quasi]

Transfer Principle wrote:
>
>With today being Tau Day (tau = 2pi = 6.2831853...), it got
>me wondering about floor(tau*10^603) and its factors.

>...

>It would be interesting if this number turned out to be prime.

It's composite.

There are only 3 positive integers n less than 1000 such that
floor(tau*10^n) is prime, namely 344, 382, 521.

There's nothing special about 10^603 for this kind of question
and I suspect you know that. Even AP knows that 10^603 has no
magical properties, but he also knows we can't actually prove
that 10^603 is _not_ special, hence, as long as his 10^603
myth gets him sufficient attention, it serves its purpose.

So why are you playing along with AP's nonsense?

Do you enjoy feeding trolls?

quasi

[io_x]

"Transfer Principle" <david.l...@lausd.net> ha scritto nel messaggio

news:72334415-0fa0-47db...@q5g2000pba.googlegroups.com...

With today being Tau Day (tau = 2pi = 6.2831853...), it got me
wondering about floor(tau*10^603) and its factors.

#for floor(2*%pi*10^603)
#ecm find one factor i report here as
#1378212755185547

Notice that the next digit after the -000- in pi is five. This
means that the 601st through 603rd digits in tau are -001-, not
-000- (otherwise floor(tau*10^603) would be 2*floor(pi*10^603)
and so the factorization of floor(tau*10^603) would be no more
interesting than that of floor(pi*10^603)).

Trial division of floor(tau*10^603) reveals no small factors
(and I think my calculations go up to at least 10^6, possibly
even 10^7 or so). It would be interesting if this number turned
out to be prime...

#it would be not prime...