Collatz Series 3*C + 1; C/2
bill
even numbers.
Let O_1 be the first odd number in a sequence or subsequence and O_o
is the last
odd number. Then (O_o)/O_1) is a function of the distribution of the
even numbers.
That is, the number of evens between successive odds.
Consider the following CS;
255 766 383 1150 575 1726 863 2590 1295 3886 1943 5830
2915 8746 4373 13120 6560 3280 1640 820 410 205 616 308
154 77 232 116 58 29 88 44 22 11 34 17 52
26 13 40 20 10 5
There are 15 odds and 14 sets of evens; distributed as follows;
e N_e
Number of Evens Number
in Set of Sets
1 8
2 1
3 4
4 0
5 0
6 1
The ratio of two successive odds; ie O_i and O_k is
(O_k) / (O_i) = 3.0 / (2^e) [rounded to the nearest odd
number]
Then (O_o) / (O_1) is the product of the individual ratios, or
(O_o) / (O_1) = PROD { [ 3.0 / (2^e) ] ^ (N_e) } .
For the sample sequence;
(O_o) / (O_1) = (1.5 ^ 8) * (0.75 ^ 1) * (.0375 ^ 4) *
( 0.046875 ^ 1)
= 25.63 * 0.75 * .0198 *
0.046875
= 0.017817948013544
5 / 255 = 0.019607843137254902
This information lets us speculate as to what distributions of the
evens are necessary for a
Collatz sequence to loop, oscillate in a finite range, or to expand
indefinitely. These can
be compared to the expected distribution. The expected distribution
is,
e N_e
1 M/2
2 M/4
3 M/8
| |
x M /
(2^x)
M = total number of intervals between successive odd numbers.
For M = 128, the expected distribution gives,
(O_o) / (O_1) = 4.1470896E-32
For M = 16,384, the expected distribution gives,
(O_o) / (O_1) = 0.369378054E -2050
This will more than cancel a CS that starts with a seed of a
(GOOGOLPLEX) ^ 2.
Another that mitigates against an infinite CS is 2^n.
Hypothesis: There is a CS that oscillates in a finite range, but does
not loop..
Every time the sequence goes through the range, it must generate a new
set of
even numbers (unless it loops) Eventually all of the evens are
exhausted and the
CS must leave the range.
Hypothesis: There is a CS that oscillates in a constantly expanding
finite range,
but does not loop..
This is the same as saying, "The CS will expand to infinitely or
collapse to zero."
mensa...@aol.com
On Oct 29, 4:21 pm, bill <b92...@yahoo.com> wrote:
Made some format edits for spacing purposes.
This is only an approximation, not exact.
The exact relationship is:
(0_k)/(0_i) = (9*(0_i) + 3.0) / ((2**e)**2*(0_k) - 2**e)
as demonstated here:
bill me actual
1.500000 1.501961 1.501961
1.500000 1.501305 1.501305
1.500000 1.500870 1.500870
1.500000 1.500579 1.500579
1.500000 1.500386 1.500386
1.500000 1.500257 1.500257
1.500000 1.500172 1.500172
0.046875 0.046879 0.046879
0.375000 0.375610 0.375610
0.375000 0.376623 0.376623
0.375000 0.379310 0.379310
1.500000 1.545455 1.545455
0.750000 0.764706 0.764706
0.375000 0.384615 0.384615
> Then (O_o) / (O_1) is the product of the individual ratios, or
>
> (O_o) / (O_1) = PROD { [ 3.0 / (2^e) ] ^ (N_e) } .
>
> For the sample sequence;
>
> (O_o) / (O_1) = 1.5^8 * 0.75^1 * 0.0375^4 * 0.046875^1
> = 25.63 * 0.75 * 0.0198 * 0.046875
> = 0.017817948013544
> 5 / 255 = 0.019607843137254902
Which clearly illustrates that you have only an approximation.
Watch what happens when we take the product of the true ratios.
bill 0.0178179480135
me 0.0196078431373
actual 0.0196078431373
No loss of precision whatsoever.
>
> This information lets us speculate as to what distributions
> of the evens are necessary for a Collatz sequence to loop,
Only partly to do with the distribution of evens. Loop requirements
are well established and depend solely on the factorization of the
Crossover Point:
CP = Z/(X-Y)
X is 2**sum(even_sets)
Y is 3**count(even_sets)
Z is based on evens distribution, but includes factors of 3:
even_sets = [1,2,3,4]
Z = 3**0*2**(1+2+3) + 3**1*2**(1+2) + 3**2*2**(1) + 3**3*2**(0)
> oscillate in a finite range,
What does that mean? Oscillating in a finite range is impossible.
There is a finite quantity of numbers and no number in that range
can appear twice (otherwise you have a loop cycle). So every
successor must be different and so you would eventually run out
of distinct numbers (as your range is finite). If you expanded the
range, you would still eventually run out of numbers which can only
be accomodated by an infinite number of range expansions.
Thus, your "oscillating in a finite range" is EITHER a loop cycle
OR a run to infinity. There can't be anything in between.
> or to expand indefinitely.
> These can be compared to the expected distribution.
>
> The expected distribution is,
>
> e N_e
> 1 M/2
> 2 M/4
> 3 M/8
> | |
> x M/(2^x)
>
> M = total number of intervals between successive odd numbers.
>
> For M = 128, the expected distribution gives,
>
> (O_o) / (O_1) = 4.1470896E-32
>
> For M = 16,384, the expected distribution gives,
>
> (O_o) / (O_1) = 0.369378054E -2050
>
> This will more than cancel a CS that starts with a seed of a (GOOGOLPLEX) ^ 2.
I'll have to get back to you on this part.
>
> Another that mitigates against an infinite CS is 2^n.
>
> Hypothesis: There is a CS that oscillates in a finite range, but does
> not loop..
Not possible for reasons given above.
>
> Every time the sequence goes through the range, it must generate a new
> set of even numbers (unless it loops) Eventually all of the evens are
> exhausted and the CS must leave the range.
So why even bring it up?
>
> Hypothesis: There is a CS that oscillates in a constantly expanding
> finite range, but does not loop..
No such thing as "a constantly expanding finite range". The only
choices are finite range (loop cycle) or infinite range.
>
> This is the same as saying, "The CS will expand to infinitely or
> collapse to zero."
Don't you mean 1?
bill
The purpose of this effort is to suggest a set of criteria
necessary to reach a specific goals; ie "looping" or
"constant
expansion". These criteria are designed so that their
probabalities can be estimated.
If there is a loop cycle, then (O_o) / (O_1) = 1.0000000.
Even
with accurate ratios, this is very hard to achieve. But my
approximations give a hint as to what evens distributions are
necessary. One approximate distribution is A 1's and B 2's,
where
0.75^(A + B) = 1 / (2 ^ A). This reduces to
(A + B)/A = 2.40942083965320901 or
B/A = 1.40942083965320901
A = 1000 and B = 1409 are very accurate (but not exact)
solutions.
The expected distribution for M = 2409 is appx 1205 1's, appx 602 2's
and, appx 603 3's and larger.
Intuitively, the problems with other appropriate distributions
are equally troublesome.
But there is still hope. There is an almost infinite number
of candidate seeds.
If the CS expands infinitely, then (O_o) / (O_1) > 1 )for
all seeds. Here, the approximations give a satisfactory
picture of the evens distributions necessary to accomplish
this goal.
A sustained ratio of 5 1's for every 4 2's with no e's larger
than 2 will achieve astounding growth in the CS. For 150 1's
and 150 2's; (O_o) / (O_1) = 199,853,840,664. But this can
be
wiped out by an expected sequence of M = 96.
Bill J
mensa...@aol.com
Since B is 2 evens, you should be mre concerned
about the ratio
(2*B+A)/M
which has to be > log(3)/log(2) if you want your
loop to be in the positive domain.
>
> A = 1000 and B = 1409 are very accurate (but not exact)
> solutions.
Which, unfortunately, is slightly less than log(3)/log(2).
> The expected distribution for M = 2409 is appx 1205 1's, appx 602 2's
> and, appx 603 3's and larger.
A geometric distribution in other words.
> Intuitively, the problems with other appropriate distributions
> are equally troublesome.
>
> But there is still hope. There is an almost infinite number
> of candidate seeds.
> If the CS expands infinitely, then (O_o) / (O_1) > 1 )for
> all seeds. Here, the approximations give a satisfactory
> picture of the evens distributions necessary to accomplish
> this goal.
Well, you won't get expansion to infinity with a
geometric distribution as they are convergent.
>
> A sustained ratio of 5 1's for every 4 2's with no e's larger
> than 2 will achieve astounding growth in the CS. For 150 1's
> and 150 2's;
But that's not 5 1's to 4 2's. Shouldn't it be 150 & 120?
> (O_o) / (O_1) = 199,853,840,664.
How did you calculate that? 150 & 150 is WAY off:
Sequence Vector:
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
Hailstone Function Parameters:
X: 290735489718242756219729523155201813741456544
274927224112596079672255715245359169330476420285
5054262243050086425064711734138406514458624
Y: 13689147905858837599132602738208831596646369
56253374364714801900783689971774990765938002061
55688941388250484440597994042813512732765695774
566001
Z: 13689148196594327317375358852325312433016475
71957842038087273863336122788846968139992209198
79686341973162931301528462995460299889486687083
701873
Seed:
g0: 483895683327166180094467091729140604898258
1705105610097935997669028843100336836481089489
3428277476133592314424000833872983421417737420
7
Hailstone:
a0: 227840075062448251775347280217170562066637
7423974973344817573813936099580337132063123965
6008983172093795334052907540519209690740677829
382900045
Hailstone/Seed ratio:
47,084,543
The 150 & 120 at least the same order of
magitude:
Sequence Vector:
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
Hailstone Function Parameters:
X: 252172839656924666958585856640919128352510
331330978858674869077787172619337582147913051
3040312634601011624191379636224
Y: 664873118457859565327991287081559533294912
222632819802610568135499305204269921340346297
290809262484944930020118813628984214441849
Z: 664873118460381293724560528621587783217887
573630177039972373307916992497106798100353107
354681239039171794097186101435724298997625
Seed:
g0: 17735229142644974475900473204198246407021
643047170136326167027781894572118020137728937
11609918910497855150243746676735
Hailstone:
a0: 46760297907884846087823929220171194228429
639207505395637692021586436067602369289721809
3998574539798644991323763721240162094938485
Hailstone/Seed ratio:
263,657,703,725
> But this can be
> wiped out by an expected sequence of M = 96.
But you seem to be missing a cruial point. The
Hailstone/Seed ratio being 1 is a CONSEQUENCE
of a loop cycle, not the CAUSE.
The cause is determined by whether the Hailstone
Function Parameters shown above make the Crossover
Point Z/(X-Y) an integer.
X is 2**sum(SequenceVector) or 2**(2*B+A)
Y is 3**countSequenceVector) or 3**M
But Z is determied by the PERMUTATION of the
even sets, so although X-Y is invariant, Z
changes with each permutation and loops form
if the factorization of Z cancels all the
factors of X-Y.
Have you read my paper where this is all
explained?
>
> Bill J
mensa...@aol.com
>
> > The purpose of this effort is to suggest a set of criteria
> > necessary to reach a specific goals; ie "looping" or
> > "constant expansion". These criteria are designed so that
> > their probabalities can be estimated.
> >
> > Even
> > with accurate ratios, this is very hard to achieve. =A0But my
> > approximations give a hint as to what evens distributions are
> > where
> > =A0 =A0 =A00.75^(A + B) =3D 1 / (2 ^ A). =A0This reduces to
> > =A0 =A0 =A0(A + B)/A =3D 2.40942083965320901 or
> > =A0 =A0 =A0 =A0 B/A =A0 =A0=3D 1.40942083965320901
>
> Since B is 2 evens, you should be mre concerned
> about the ratio
>
> (2*B+A)/M
>
> which has to be > log(3)/log(2) if you want your
> loop to be in the positive domain.
>
> >
> > solutions.
>
> Which, unfortunately, is slightly less than log(3)/log(2).
>
> > and, appx 603 3's and larger.
>
> A geometric distribution in other words.
>
> > Intuitively, the problems with other appropriate distributions
> > are equally troublesome.
> >
> > But there is still hope. There is an almost infinite number
> > of candidate seeds.
> > If the CS expands infinitely, then (O_o) / (O_1) > 1 )for
> > picture of the evens distributions necessary to accomplish
> > this goal.
>
> Well, you won't get expansion to infinity with a
> geometric distribution as they are convergent.
>
> >
> > A sustained ratio of 5 1's for every 4 2's with no e's larger
> > and 150 2's; =A0
>
> But that's not 5 1's to 4 2's. Shouldn't it be 150 & 120?
>
>
> How did you calculate that? 150 & 150 is WAY off:
>
>
> Hailstone/Seed ratio:
> 47,084,543
>
>
> The 150 & 120 at least the same order of
> magitude:
>
>
> Hailstone/Seed ratio:
> 263,657,703,725
>
> > But this can be
>
> But you seem to be missing a cruial point. The
> Hailstone/Seed ratio being 1 is a CONSEQUENCE
> of a loop cycle, not the CAUSE.
>
> The cause is determined by whether the Hailstone
> Function Parameters shown above make the Crossover
> Point Z/(X-Y) an integer.
>
> X is 2**sum(SequenceVector) or 2**(2*B+A)
> Y is 3**countSequenceVector) or 3**M
>
> But Z is determied by the PERMUTATION of the
> even sets, so although X-Y is invariant, Z
> changes with each permutation and loops form
> if the factorization of Z cancels all the
> factors of X-Y.
To elaborate on this a bit, take the even sets (in
this order)
[1, 1, 1, 2, 1, 1, 4]
and we get a Hailstone/Seed ratio of
Hailstone/Seed ratio:
1.06843919252
If I change the permutaion of the even sets to
[1, 1, 1, 4, 1, 1, 2]
which has the same distribution, the ratio is then
Hailstone/Seed ratio:
1.0690887433
which is close, but not the same. And in neither
case is the ratio 1. Yet, the first even set IS
a loop cycle, albeit in the negative domain, at -17.
This Hailstone/Seed ratio is a bit wanting. Without
the critical permutation data, you may have a loop
cycle and not even know it.
bill
On Nov 3, 7:48 pm, "mensana...@aol.com" <mensana...@aol.com> wrote:
> On Oct 31, 5:18?pm, bill <b92...@yahoo.com> wrote:
>
> > On Oct 29, 5:29 pm, "mensana...@aol.com" <mensana...@aol.com> wrote:
>
> > > On Oct 29, 4:21 pm, bill <b92...@yahoo.com> wrote:
>
> > > Made some format edits for spacing purposes.
>
> > > > A Collatz sequence consists of odd numbers separated by one
> > > > or more even numbers.
>
> > > > Let O_1 be the first odd number in a sequence or subsequence
> > > > and O_o is the last odd number. Then (O_o)/O_1) is a function
> > > > of the distribution of the even numbers.
>
> > > > That is, ?the number of evens between successive odds.
>
> > > > Consider the following CS;
>
> > > > ? ?255 ? 766 ?383 ?1150 ? 575 ?1726 ? 863 ?2590 ?1295 ?3886
> > > > ? 1943 ?5830 2915 ?8746 ?4373 13120 ?6560 ?3280 ?1640 ? 820
> > > > ? ?410 ? 205 ?616 ? 308 ? 154 ? ?77 ? 232 ? 116 ? ?58 ? ?29
> > > > ? ? 88 ? ?44 ? 22 ? ?11 ? ?34 ? ?17 ? ?52 ? ?26 ? ?13 ? ?40
> > > > ? ? 20 ? ?10 ? ?5
>
> > > > There are 15 odds and 14 sets of evens; distributed as follows;
>
> > > > ? ? ? ? ? ? ?e ? ? ? ? ? ? ? ? ? ? ? ?N_e
> > > > ?Number of Evens ? ? ? ? ? ? ? ? ? Number
> > > > ? ? ? ? ? in Set ? ? ? ? ? ? ? ? ?of Sets
> > > > ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?8
> > > > ? ? ? ? ? ? 2 ? ? ? ? ? ? ? ? ? ? ?1
> > > > ? ? ? ? ? ? 3 ? ? ? ? ? ? ? ? ? ? ? ? 4
> > > > ? ? ? ? ? ? 4 ? ? ? ? ? ? ? ? ? ? ? ? 0
> > > > ? ? ? ? ? ? 5 ? ? ? ? ? ? ? ? ? ? ? ? 0
> > > > ? ? ? ? ? ? 6 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?1
>
> > > > The ratio of two successive odds; ie O_i and O_k is
>
> > > > ? ? ? ? ?(O_k) / (O_i) ?= ?3.0 / (2^e) ? [rounded to the nearest
> > > > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? odd number]
>
> > > This is only an approximation, not exact.
>
> > > The exact relationship is:
>
> > > (0_k)/(0_i) = (9*(0_i) + 3.0) / ((2**e)**2*(0_k) - 2**e)
>
> > > as demonstated here:
>
> > > bill ? ? me ? ? ? ?actual
> > > 1.500000 ? ? ? ?1.501961 ? ? ? ?1.501961
> > > 1.500000 ? ? ? ?1.501305 ? ? ? ?1.501305
> > > 1.500000 ? ? ? ?1.500870 ? ? ? ?1.500870
> > > 1.500000 ? ? ? ?1.500579 ? ? ? ?1.500579
> > > 1.500000 ? ? ? ?1.500386 ? ? ? ?1.500386
> > > 1.500000 ? ? ? ?1.500257 ? ? ? ?1.500257
> > > 1.500000 ? ? ? ?1.500172 ? ? ? ?1.500172
> > > 0.046875 ? ? ? ?0.046879 ? ? ? ?0.046879
> > > 0.375000 ? ? ? ?0.375610 ? ? ? ?0.375610
> > > 0.375000 ? ? ? ?0.376623 ? ? ? ?0.376623
> > > 0.375000 ? ? ? ?0.379310 ? ? ? ?0.379310
> > > 1.500000 ? ? ? ?1.545455 ? ? ? ?1.545455
> > > 0.750000 ? ? ? ?0.764706 ? ? ? ?0.764706
> > > 0.375000 ? ? ? ?0.384615 ? ? ? ?0.384615
>
> > > > Then (O_o) / (O_1) is the product of the individual ratios, or
>
> > > > ? ? ? ? (O_o) / (O_1) ?= ? PROD { [ 3.0 / (2^e) ] ^ (N_e) } .
>
> > > > For the sample sequence;
>
> > > > (O_o) / (O_1) ?= ?1.5^8 * 0.75^1 * 0.0375^4 * 0.046875^1
> > > > ? ? ? ? ? ? ? ?= ?25.63 * 0.75 ? * 0.0198 ? * 0.046875
> > > > ? ? ? ? ? ? ? ?= ?0.017817948013544
> > > > ? 5 / 255 ? ? ?= ?0.019607843137254902
>
> > > Which clearly illustrates that you have only an approximation.
>
> > > Watch what happens when we take the product of the true ratios.
>
> > > bill ? ?0.0178179480135
> > > me ? ? ?0.0196078431373
> > > actual ?0.0196078431373
>
> > > No loss of precision whatsoever.
>
> > > > This information lets us speculate as to what distributions
> > > > of the evens are necessary for a Collatz sequence to loop,
>
> > > Only partly to do with the distribution of evens. Loop requirements
> > > are well established and depend solely on the factorization of the
> > > Crossover Point:
>
> > > ? CP = Z/(X-Y)
>
> > > X is 2**sum(even_sets)
> > > Y is 3**count(even_sets)
> > > Z is based on evens distribution, but includes factors of 3:
>
> > > ? even_sets = [1,2,3,4]
>
> > > ? Z = 3**0*2**(1+2+3) + 3**1*2**(1+2) + 3**2*2**(1) + 3**3*2**(0)
>
> > > > oscillate in a finite range,
>
> > > What does that mean? Oscillating in a finite range is impossible.
> > > There is a finite quantity of numbers and no number in that range
> > > can appear twice (otherwise you have a loop cycle). So every
> > > successor must be different and so you would eventually run out
> > > of distinct numbers (as your range is finite). If you expanded the
> > > range, you would still eventually run out of numbers which can only
> > > be accomodated by an infinite number of range expansions.
>
> > > Thus, your "oscillating in a finite range" is EITHER a loop cycle
> > > OR a run to infinity. There can't be anything in between.
>
> > > > or to expand indefinitely.
> > > > These can be compared to the ?expected distribution.
>
> > > > The expected distribution is,
>
> > > > ? ? ? ? ? ? e ? ? ? ? ? ? ? ? ? ? ? ? N_e
> > > > ? ? ? ? ? ? 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?M/2
> > > > ? ? ? ? ? ? 2 ? ? ? ? ? ? ? ? ? ? ? ? M/4
> > > > ? ? ? ? ? ? 3 ? ? ? ? ? ? ? ? ? ? ? ? M/8
> > > > ? ? ? ? ? ? | ? ? ? ? ? ? ? ? ? ? ? ? ?|
> > > > ? ? ? ? ? ? x ? ? ? ? ? ? ? ? ? ? ? ? M/(2^x)
>
> > > > M = total number of intervals between successive odd numbers.
>
> > > > For M = 128, ?the expected distribution gives,
>
> > > > ?(O_o) / (O_1) ?= ?4.1470896E-32
>
> > > > For M = 16,384, ? the expected distribution gives,
>
> > > > ? (O_o) / (O_1) ?= ?0.369378054E -2050
>
> > > > This will more than cancel a CS that starts with a seed ?of a (GOOGOLPLEX) ^ 2.
>
> > > I'll have to get back to you on this part.
>
> > > > Another that mitigates against an infinite CS is 2^n.
>
> > > > Hypothesis: There is a CS that oscillates in a finite range, but does
> > > > not loop..
>
> > > Not possible for reasons given above.
>
> > > > Every time the sequence goes through the range, it must generate a new
> > > > set of even numbers (unless it loops) ?Eventually all of the evens are
> > > > exhausted and the CS must leave the range.
>
> > > So why even bring it up?
>
> > > > Hypothesis: There is a CS that oscillates in a constantly expanding
> > > > finite range, but does not loop..
>
> > > No such thing as "a constantly expanding finite range". The only
> > > choices are finite range (loop cycle) or infinite range.
>
> > > > This is the same as saying, "The CS will expand to infinitely or
> > > > collapse to zero."
>
> > > Don't you mean 1?
>
> > The purpose of this effort is to suggest a set of criteria
> > necessary to reach a specific goals; ie "looping" or
> > "constant expansion". These criteria are designed so that
> > their probabalities can be estimated.
>
> > If there is a loop cycle, then ?(O_o) / (O_1) = 1.0000000.
> > Even
> > with accurate ratios, this is very hard to achieve. ?But my
> > approximations give a hint as to what evens distributions are
> > necessary. ?One approximate distribution is A 1's and B 2's,
> > where
> > ? ? ?0.75^(A + B) = 1 / (2 ^ A). ?This reduces to
> > ? ? ?(A + B)/A = 2.40942083965320901 or
> > ? ? ? ? B/A ? ?= 1.40942083965320901
>
> Since B is 2 evens, you should be mre concerned
> about the ratio
>
> (2*B+A)/M
>
> which has to be > log(3)/log(2) if you want your
> loop to be in the positive domain.
>
>
>
> > A = 1000 and B = 1409 are very accurate (but not exact)
> > solutions.
>
> Which, unfortunately, is slightly less than log(3)/log(2).
>
> > The expected distribution for M = 2409 is appx 1205 1's, appx 602 2's
> > and, appx 603 3's and larger.
>
> A geometric distribution in other words.
>
> > Intuitively, the problems with other appropriate distributions
> > are equally troublesome.
>
> > But there is still hope. There is an almost infinite number
> > of candidate seeds.
> > If the CS expands infinitely, then (O_o) / (O_1) > 1 )for
> > all seeds. ?Here, the approximations give a satisfactory
> > picture of the evens distributions necessary to accomplish
> > this goal.
>
> Well, you won't get expansion to infinity with a
> geometric distribution as they are convergent.
>
>
>
> > A sustained ratio of 5 1's for every 4 2's with no e's larger
> > than 2 will achieve astounding growth in the CS. ?For 150 1's
> > and 150 2's; ?
>
> But that's not 5 1's to 4 2's. Shouldn't it be 150 & 120?
>
> > (O_o) / (O_1) ?= 199,853,840,664. ?
Not yet. How can I access it?
mensa...@aol.com
>> explained?
> Not yet. How can I access it?
Title: Blueprint for Failure:
How to Construct a Counterexample
to the Collatz Conjecture
Author: Paul Stadfeld
Full paper at S.A.T.O. Volume 5.3. (2006)
<http://home.zonnet.nl/galien8>
mensa...@aol.com
Did you delete your last post?
Anyway, you mentioned how big the numbers would
be in a counterexample of 275000 elements, so
here's my take on that:
Well, yeah, a sequence that diverged for 275000
iterations would end up at a large number, but such
a number won't be a counterexample because to be
a loop cycle, the Sequence Vector (in this case
[1,1,1...<275000 1's>...1,1,1]) must be "prime",
i.e., not partitionable into identicle sub-sequences.
More interesting is this particular value of C for
3n+C:
1541095
from which I get a Sequence Vector
>>> sv = cf.build_sv(19,19,1000000,1541095)
that has
>>> len(sv)
123296
odd numbers and
>>> sum(sv)
246588
even numbers. The evens distribution has
set size : number of sets
>>> hist
{1: 61247,
2: 31347,
3: 15288,
4: 7778,
5: 3879,
6: 1869,
7: 971,
8: 457,
9: 236,
10: 119,
11: 45,
12: 24,
13: 16,
14: 8,
15: 5,
16: 4,
17: 1,
18: 2}
which is a pretty good match to a Negative
Binomial (Geometric) Distribution, just as
we would expect.
All Sequence Vectors appear infinitely many
times on the Collatz graph. The first occurence
of this particular one has a Hailstone having
>>> print cf.gmpy.numdigits(a0)
58828
decimal digits. The Seed that generates it
has
>>> print cf.gmpy.numdigits(g[1])
74231
decimal digits. Obviously, the Hailstone/Seed
ratio must be quite small:
5.7233608...<53812 digits>...022437364648645e-15404
That is, of course, in 3n+1. And you would get
similar results for other typically huge Sequence
Vectors. What makes this one special is that in
the system where I first found it, 3n+1541095,
the Sequence Vector is a loop cycle!
bill
On Nov 9, 12:00 am, "mensana...@aol.com" <mensana...@aol.com> wrote:
> Did you delete your last post?
Yes.
I am trying to read "Blueprint...." , but I am having some
difficulty. I will start a new thread to discuss my comments
and questions.
Bill J