gaps in coprimes of primorials

[Jamie M]

Hi,

The gaps in coprimes of primorials is a mirror image ie:


primorial 30 coprimes:

1,7,11,13,17,19,23,29

That sequence of coprimes has the gaps:

6,4,2,4,2,4,6

which looks the same even if someone with dyslexia writes it
(ie a mirror image)


Also tested again for the next primorial 210 coprimes:

1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,
121,127,131,137,139,143,149,151,157,163,167,169,173,179,181,187,191,193,197,199,209,

The list of gaps for the above is:

10,2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10

Again that looks like a mirror image.

So what is the reason for this!??!

cheers,
Jamie

[Jamie M]

This also implies that you can find the second half of coprimes of a
primorial if you know all the coprimes less than the primorial
magnitude/2 (ie for primorial 210 it is 210/2 = 105.

Also if you know just half of the coprimes, you can check the
corresponding position and fill it in, so you can know a random
any half of coprimes and fill in the rest, ie if you know 13 is coprime
for 210, then also you can know that 210-13 is coprime etc.

cheers,
Jamie

[Jamie M]

On 3/27/2016 3:00 PM, Jamie M wrote:

>
> This also implies that you can find the second half of coprimes of a
> primorial if you know all the coprimes less than the primorial
> magnitude/2 (ie for primorial 210 it is 210/2 = 105.
>
> Also if you know just half of the coprimes, you can check the
> corresponding position and fill it in, so you can know a random
> any half of coprimes and fill in the rest, ie if you know 13 is coprime
> for 210, then also you can know that 210-13 is coprime etc.
>
> cheers,
> Jamie
>

The mirrored gap sequence seems to work for any number not just
primorials.


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[Jamie M]

If you take the coprimes and organize them into two columns, ie for
coprimes of 30:


1 29
7 23
11 19
13 17

the sum of the two columns for each row is always 30, ie 1+29, 7+23,
11+19, 13+17


For any natural number n these columns can be generated with
the numbers coprimes adding to n.

Coprimes organized into two columns for different numbers n greater than
2, showing unique all unique products:

n=3
column1 column2 product
1 2 2

n=4
1 3 3

n=5
1 4 4
2 3

n=6
1 5 5

n=7
1 6 6
2 5 10
3 4 12

n=8
1 7 7
3 5 15

n=9
1 8 8
2 7 14
4 5 20

n=10
1 9 9
3 7 21

n=11
1 10 10
2 9 18
3 8 24
4 7 28
5 6 30

n=12
1 11 11
5 7 35

n=13
1 12 12
2 11 22
3 10 30
4 9 36
5 8 40
6 7 42

n=14
1 13 13
3 11 33
5 9 45

n=15
1 14 14
2 13 26
4 11 44
7 8 56

n=16
1 15 15
3 13 39
5 11 55
7 9 63


n=17
1 16 16
2 15 30
3 14 42
4 13 52
5 12 60
6 11 66
7 10 70
8 9 72


cheers,
Jamie



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[Jamie M]

On 3/27/2016 3:47 PM, Jamie M wrote:
> On 3/27/2016 3:20 PM, Jamie M wrote:
>> On 3/27/2016 3:00 PM, Jamie M wrote:
>>
>>>
>>> This also implies that you can find the second half of coprimes of a
>>> primorial if you know all the coprimes less than the primorial
>>> magnitude/2 (ie for primorial 210 it is 210/2 = 105.
>>>
>>> Also if you know just half of the coprimes, you can check the
>>> corresponding position and fill it in, so you can know a random
>>> any half of coprimes and fill in the rest, ie if you know 13 is coprime
>>> for 210, then also you can know that 210-13 is coprime etc.
>>>
>>> cheers,
>>> Jamie
>>>
>>
>> The mirrored gap sequence seems to work for any number not just
>> primorials.
>

Here's a picture of a spreadsheet showing an interesting pattern for
primorials coprimes when organized into two columns so that all
pairs of coprimes add to the primorial size (30 and 210 in the
spreadsheet)

http://i.imgur.com/FNKP7gt.png

The difference (gap) in consecutive products of the coprime pairs
(column E) has greatest common divisor 12 for both primorial 30 and
primorial 210 which is interesting, so that all gaps in the products
of the pairs of coprimes are multiples of 12. Any idea why that is?

cheers,
Jamie

of 12

[James Waldby]

On Mon, 28 Mar 2016 00:56:46 -0700, Jamie M wrote:
> Here's a picture of a spreadsheet showing an interesting pattern for
> primorials coprimes when organized into two columns so that all
> pairs of coprimes add to the primorial size (30 and 210 in the
> spreadsheet)
> http://i.imgur.com/FNKP7gt.png
>
> The difference (gap) in consecutive products of the coprime pairs
> (column E) has greatest common divisor 12 for both primorial 30 and
> primorial 210 which is interesting, so that all gaps in the products
> of the pairs of coprimes are multiples of 12. Any idea why that is?

Let v stand for a primorial of 6 or more (eg 6, 30, 210...)

The difference d between any two "products of coprime pairs" can be
expressed as d = q*(v-q) - p*(v-p), where none of those four factors
are divisible by 2 or 3. We want to show that d mod 12 is 0, which is
equivalent to showing that d is a multiple of 3 and a multiple of 4.

Expanding, we have d = q^2-p^2 + v*(q-p). Because q-p is even and v
is a multiple of 6, v*(q-p) is a multiple of 12 and need not be
considered further. That is, d mod 12 == q^2 - p^2 == (q+p)*(q-p).

Because q+p and q-p are both even, d mod 12 is a multiple of 4.

Now write p = 3u+a, where a is 1 or 2, and q = 3w+b, similarly.
Consider the four (b,a) cases:
(1,1): q-p = 3w+1-3u-1 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3
(1,2): q+p = 3w+1+3u+2 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3
(2,1): q+p = 3w+2+3u+1 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3
(2,2): q-p = 3w+2-3u-2 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3
Thus, d mod 12 is a multiple of 3.

As a multiple of both 3 and 4, d mod 12 is 0, as we wanted to show.

--
jiw

[James Waldby]

On Wed, 30 Mar 2016 21:03:21 +0000, James Waldby wrote:

> On Mon, 28 Mar 2016 00:56:46 -0700, Jamie M wrote:
>> Here's a picture of a spreadsheet showing an interesting pattern for
>> primorials coprimes when organized into two columns so that all
>> pairs of coprimes add to the primorial size (30 and 210 in the
>> spreadsheet)
>> http://i.imgur.com/FNKP7gt.png
>>
>> The difference (gap) in consecutive products of the coprime pairs
>> (column E) has greatest common divisor 12 for both primorial 30 and
>> primorial 210 which is interesting, so that all gaps in the products
>> of the pairs of coprimes are multiples of 12. Any idea why that is?
>
> Let v stand for a primorial of 6 or more (eg 6, 30, 210...)
>
> The difference d between any two "products of coprime pairs" can be
> expressed as d = q*(v-q) - p*(v-p), where none of those four factors
> are divisible by 2 or 3. We want to show that d mod 12 is 0, which is
> equivalent to showing that d is a multiple of 3 and a multiple of 4.

...

> As a multiple of both 3 and 4, d mod 12 is 0, as we wanted to show.

I now see that Leon Aigret already posted about this a couple of days
ago. Oh well.

--
jiw

[Jamie M]

Well your description seems quite concise and I need all the help I
can get so thanks! :)

What I'm interested to find out about is if I can get a formula to
find the sequence of coprimes for a primorial.

ie for primorial 30 the coprimes:

1,7,11,13,17,19,23,29

EulerPhi(30) is the count of coprimes but I'd like to get the
formula to find the actual coprimes. Any idea?

I want to find the coprimes of very large primorials efficiently..

cheers,
Jamie

[James Waldby]

On Wed, 30 Mar 2016 16:42:18 -0700, Jamie M wrote:
...

> What I'm interested to find out about is if I can get a formula to
> find the sequence of coprimes for a primorial.
>
> ie for primorial 30 the coprimes: 1,7,11,13,17,19,23,29
>
> EulerPhi(30) is the count of coprimes but I'd like to get the
> formula to find the actual coprimes. Any idea?
>
> I want to find the coprimes of very large primorials efficiently..

I don't know of a formula. It's possible that the following process
is efficient enough to be worthwhile; or there may be a better
algorithm; I don't know.

Let v = n# for some large n, so v is a large primorial.

First initialize an empty map, L, with at least v bits in it.

For each prime p less than v, set bit p in L.

Go through L in ascending order, and if bit b is set, do Propagate(b).

Propagate(b): Let m=b. While m<v, do { PPass(b); m = m*b }

PPass(b): Let c=m. While c*b < v, do { Set bit c*b. Advance c to
next 1-bit in L. }

What the process outlined above should do (after it sets bits for
primes) is to set bits for multiples of powers of primes for all
primes larger than n and less than a fraction of v, for all
not-too-big powers and primes.

Total work in PPass() probably is like O(w*v*sum(1/p)) where the
sum is over primes p less than sqrt(v), and where w represents
the average work of a single "Advance c to next 1-bit in L" step.

--
jiw

[James Waldby]

On Thu, 31 Mar 2016 05:22:20 +0000, James Waldby wrote:

> On Wed, 30 Mar 2016 16:42:18 -0700, Jamie M wrote:
> ...
>> What I'm interested to find out about is if I can get a formula to
>> find the sequence of coprimes for a primorial.
>>
>> ie for primorial 30 the coprimes: 1,7,11,13,17,19,23,29
>>
>> EulerPhi(30) is the count of coprimes but I'd like to get the
>> formula to find the actual coprimes. Any idea?
>>
>> I want to find the coprimes of very large primorials efficiently..

I snipped the algorithm that I presented earlier, which first sets
a 1-bit for each prime between n and v, then sets bits for various
multiples of set bits. Instead, the following rather-simpler and
probably-faster method turns off bits for multiples of 2, 3,... n:

Let v = n# for some large n, so v is a large primorial.

Initialize a map L with v bits in it, each set to 1.

For each prime p in {2, 3, ... n}, do PPass(p).

PPass(p): While k*p < v, do { Clear bit k*p. Add 1 to k. }

Each bit that's still set after the above process corresponds
to a number co-prime to v.

--
jiw

[Jamie M]

Thanks, right now I just generate a big list of primes and
then check all numbers below the primorial to see if they
are coprime with the primorial, not as fast maybe but I ran
it up to primorial 223092870 now giving this result:

primorial 223092870 coprime equation list gap counts: (36495360 total)

gap count
{[2, 7952175]}
{[4, 7952175]}
{[6, 10169950]}
{[8, 2918020]}
{[10, 3401790]}
{[12, 2255792]}
{[14, 871318]}
{[16, 362376]}
{[18, 396872]}
{[20, 61560]}
{[22, 88614]}
{[24, 48868]}
{[26, 7682]}
{[28, 5664]}
{[30, 2164]}
{[32, 72]}
{[34, 198]}
{[36, 56]}
{[38, 2]}
{[40, 12]}

There should be more ways to optimize finding the coprimes
of primorials, as for one, they are a palindrome, so only
the first half need to be found to fill in the locations
of the second half, but I think there are more ways to
figure out where they are too.

cheers,
Jamie

[Jamie M]

The reason the prime number jumping champions are the primorials is:

First there are two ways to look at this question.

1. the simple answer:

primorials are special numbers since they have the same coprime gaps
as the natural numbers have primes gaps, ie they are more synchronized
with the prime numbers for their "coprime factors" than any other
numbers so that means that they have more possible ways for prime gaps
to consecutively add up to their value as a primorial, this is why the
prime gap jumping champions are primorials.


2. the more complicated answer (work in progress):

By looking at the different ways that all possible prime gaps
can be constructed out of smaller prime gaps, it can be seen that the
primorial numbers have the most possible ways to be constructed
out of the smaller prime gaps.

One there is a list of all these permutations, then for each prime gap
2,4,6,8 etc.. in each permutation, multiply the prime gap by a factor
proportional to it's occurrence, ie if gap 6 occurs often the factor
will be larger, and if gap 20 occurs less often its factor will be
smaller. What you can see from this (probably) is that the primorial
gaps tend to have a larger ratio, or maybe more of, the gaps that
occur more often ie gaps 2,4,6. I'm not sure about this though.


Lots of work still required on all this, just an update...

cheers,
Jamie

[James Waldby]

On Wed, 30 Mar 2016 23:27:21 -0700, Jamie M wrote:
> On 3/30/2016 10:22 PM, James Waldby wrote:
>> On Wed, 30 Mar 2016 16:42:18 -0700, Jamie M wrote:
>> ...
>>> What I'm interested to find out about is if I can get a formula to
>>> find the sequence of coprimes for a primorial.
>>>
>>> ie for primorial 30 the coprimes: 1,7,11,13,17,19,23,29
>>>
>>> EulerPhi(30) is the count of coprimes but I'd like to get the
>>> formula to find the actual coprimes. Any idea?
>>>
>>> I want to find the coprimes of very large primorials efficiently..
>>
>> I don't know of a formula. It's possible that the following process

[snip algorithm that sets bits in a map]

>>
>
> Thanks, right now I just generate a big list of primes and
> then check all numbers below the primorial to see if they
> are coprime with the primorial, not as fast maybe but I ran
> it up to primorial 223092870 now giving this result:

For the purpose of listing gaps between consecutive coprimes, that
method seems reasonable enough, except there's no need to make a big
list of primes, and you might be able to speed up the coprimeness
check by the method illustrated in a demo program included below.
It shows a way of using reciprocals to test if n is a multiple of p
in {5, 7, 11, 13, 17, 19, 23}. On my computer the method runs about
twice as fast as testing if n%p is zero for some p in that set. The
program prints out the following result, where the first line is for
the reciprocals method and the second line is for the modulus method.

ncr = 36495360 for tmax = 223092870 in 0.309669 seconds

nco = 36495360 for tmax = 223092870 in 0.588192 seconds

> primorial 223092870 coprime equation list gap counts: (36495360 total)
>
> gap count
> {[2, 7952175]}
> {[4, 7952175]}

[snip rest of list]

>
> There should be more ways to optimize finding the coprimes
> of primorials, as for one, they are a palindrome, so only

Rather than saying "they are a palindrome" you should say
they are located in symmetric pairs.

> the first half need to be found to fill in the locations
> of the second half, but I think there are more ways to
> figure out where they are too.

Here's the demo program, in C:

/* Re: Coprimes of primorials
Copyright 2016 James Waldby. Offered without warranty
under GPL v3 terms as at http://www.gnu.org/licenses/gpl.html
*/
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
//================================================
double ttime(double base) {
struct timeval tod;
gettimeofday(&tod, NULL);
return tod.tv_sec + tod.tv_usec/1e6 - base;
}
//================================================
int main(int argc, char *argv[]) {
double t0, t1;
long long B32 = 1LL<<32;
unsigned int n, i, nco=0, ncr=0;
unsigned int tmax = 223092870;
const unsigned int primes[] = {5, 7, 11, 13, 17, 19, 23 };
enum { nprimes = sizeof primes / sizeof primes[0] };

// Use of these "reciprocals" depends on overflow from 32 bit vars
unsigned int recips[nprimes];
for (i=0; i<nprimes; ++i) {
recips[i] = (unsigned int)((B32+primes[i]-1)/primes[i]);
}
t0=ttime(0);
for (n=5; n<tmax; n += 4) {
for (i=0; i < nprimes; ++i)
if (n*recips[i] < recips[i]) break;
if (i==nprimes) ++ncr;
n += 2;
for (i=0; i < nprimes; ++i)
if (n*recips[i] < recips[i]) break;
if (i==nprimes) ++ncr;
}
t1=ttime(t0);
printf ("ncr = %d for tmax = %d in %9.6f seconds\n", ncr, tmax, t1);

t0=ttime(0);
for (n=5; n<tmax; n += 4) {
for (i=0; i < nprimes; ++i)
if (n%primes[i] == 0) break;
if (i==nprimes) ++nco;
n += 2;
for (i=0; i < nprimes; ++i)
if (n%primes[i] == 0) break;
if (i==nprimes) ++nco;
}
t1=ttime(t0);
printf ("nco = %d for tmax = %d in %9.6f seconds\n", nco, tmax, t1);
return 0;
}

--
jiw

[Jamie M]

Cool thanks for sharing! :)

cheers,
Jamie

[Jamie M]

Hi,

I think there is a formula to do this too, based on a thing I noticed
before that was answered with a way to find the primorial coprimes:

I found this pattern:

Primorial primefactors coprimes
2 2
6 2x3 2
30 6x5 8 ((5-1)*2)
210 30x7 48 ((7-1)*8)
2310 210x11 480 ((11-1)*48)

There is the primes-1 sequence in there for the coprimes formula,
ie to get EulerPhi(2310)=480 it is (11-1)*48

Question: how do you explain coprimes being
(nextprimefactor-1)*(coprimes of next smallest primorial)?


And got a reply for a formula from a poster in the thread
"what is this pattern in the primes" on sci.electronics.design:


//reply answer

Write the numbers from 1 to your current primorial. If that's 6, then
you have

1 2 3 4 5 6

Now remove the numbers that are divisible by the factors

1 X X X 5 X

Now consider your next primorial, which is given by multiplying the
current by the next prime, p.

Create p - 1 new lines by adding your current primorial to each number,
successively:

1 X X X 5 X
7 X X X 11 X
13 X X X 17 X
19 X X X 23 X
25 X X X 29 X

Now, this contains all the numbers from 1 to the new primorial, with the
multiples of the current primorial's factors already removed.

Since the previous primorial is coprime with the new factor p, each
column that hasn't already been removed contains exactly one number that
is divisible by p. We can rearrange the columns to put that number last,
then remove the entirety of the last row.

1 X X X 11 X
7 X X X 17 X
13 X X X 23 X
19 X X X 29 X

So the count of numbers remaining (being the coprime count for the next
primorial) is (p-1) * the count of numbers in the first row (being the
coprime count for the current primorial).

Sylvia.

//end of reply answer



So I think it is really easy to find the list of coprimes for a given
primorial using that?! :)

cheers,
Jamie

[Jamie M]

Hi,

I found an interesting pattern, that allows an easy way to find all
the coprimes of primorial 2310, given primorial 210 coprimes.

It is probably generalized for all primorials, so could be used
recursively to find a primorials coprimes, by incrementing
through primorials 6,30,210,2310, etc up to the primorial of
interest, to find that primorials coprimes without doing much
math.

So the description of how it works:

take numbers 1 to 210 and arrange into a single row of
210 columns, then label the columns in that row that are
coprimes of 210.

primorial 210 coprimes (count 48):
1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,
121,127,131,137,139,143,149,151,157,163,167,169,173,179,181,187,191,193,197,199,209,

Next extend the single row into a total of 11 rows with row1 counting
from 1 to 210, row2 from 211 to 420, row3 from 421 to 630 etc all the
way to 2310 at the end of the last row11. This makes a table of 210
columns and 11 rows (2310 numbers total)

Now all possible coprimes for primorial 2310 (a total of 480 coprimes)
will be in the 48 columns that were previously labelled as coprimes
of primorial 210. There are 11 rows and 48 columns, so that is 11
more numbers than primorial 2310 has coprimes, and accordingly each
column has a single extra number that isn't a coprime of primorial 2310.

There is a pattern to find these numbers that aren't coprimes too,
I didn't figure out where exactly the pattern is coming from but
it is there.

The pattern to find the single number in each of the 48 columns there
is a formula to find which row the number will be in, from row1 to row
11. You just need to know the location of a single number that isn't
coprime and then you can find all the rest.

Here is the pattern! :D

Given the location of a single number that isn't coprime, in column1,
which is number 2101 in row11, check the gap to the next column
that has coprimes, ie column11, that is a gap of 10 columns (11-1),
so the number that isn't coprime with 2310 in column11 is in 11-10=row1.
11-10 comes from the column1 row 11 minus column11 row 1.

That one is a bit hard to see because of the numbers, but the next
one is easier to see:

The next column that has coprimes is column13 and it has a gap of 13-11
=2 from the last column that has coprimes, so the row position of the
number that isn't a coprime is the last columns row that isn't coprime
minus 2. The previous column row location that isn't coprime is row1,
so minus 2 rows takes the next row to row 10, the second last row.

Ok that one is hard to see too since there is a row overflow in that
example, but this next example is easier to see!

The next column that has coprimes is column17 and it has a gap of
17-13=4 from the last column that has coprimes. The previous column's
row location that isn't coprime is row 10, so 10-4=6 means that the
column17 row location of the number that isn't coprime is row6, and
that is number 1067.

There is a single number that isn't coprime in each of the columns,
and after finding them all the table can be turned back into a single
list of numbers to be used to construct a longer table for the next
bigger primorial and the process repeated of removing one non-coprime
from each column that has coprimes.

I didn't test this pattern for other primorials so I am sure it may
vary somewhat, for primorial210 with 30 columns I didn't figure out
the pattern if there even is one.

Also if a fixed width table is used rather than increasing the width
for bigger primorials, there seems to be an interesting pattern of
how to remove non coprimes too, as multiples of prime numbers in
the coprime rows, for primes of size of the largest prime factor for
the primorial and up the size required for the whole table.

cheers,
Jamie

[Jamie M]

On 4/4/2016 8:32 AM, Jamie M wrote:

> Here is the pattern! :D
>

This might help figure out the pattern for these
numbers that are removed from the primorial 2310
columns of coprimes.

The table referenced has 210 columns and 11 rows
organized like this:

1 2 3 4 5 6 7 ... 210
211 ... 420
...
2101 ... 2310


Out of the 210 columns, there are 48 columns in the above table that
have coprimes of the primorial 2310, and each of the 48 columns has a
single number that isn't coprime to primorial 2310.

numbers in the 48 columns that aren't coprimes of 2310:

column1:
remove 2101 (11*191)

column11:
remove 11 (11*1)

column13:
remove 1903 (11*173)

column17:
remove 1067 (11*97)

column19:
remove 649 (11*59)

colum23:
remove 2123 (11*193)

column29:
remove 869 (11*79)

column31:
remove 451 (11*41)

column37:
remove1507 (11*137)

column41:
remove 671 (11*61)

column43:
remove253 (11*23)

column47:
remove 1727 (11*157)

column53:
remove 473 (11*43)

column59:
remove 1529 (11*139)

column61:
remove 1111 (11*101)

column67:
remove 2167 (11*197)

column71:
remove 1331 (11*121)

column73:
remove 913 (11*83)

column79:
remove 1969 (11*179)

column83:
remove 1133 (11*103)

column89:
remove 2189 (11*199)

..
etc up to column 209:
remove 209 (11*19)


I am pretty sure that these multiples of 11, are just the coprimes
of primorial 210. Ie there are 46 primes up to 210, and 48 coprimes
of 210, so since there are 48 columns, it isn't just 11*primes.

There is a pattern (see my last post) for finding these numbers to
remove, if there is a gap of x columns ie column 83 to column 89,
then there is a row offset of the same gap x between the two
numbers to remove.

cheers,
Jamie

[Rohit Naringrekar]

Hi Jamie,

So nice to see interest in this fascinating sequence.

I have been independently exploring this sequence for several months now, and I cannot believe the features and rich properties this sequence manifests.
I keep observing new patterns every time I inspect it closer or from different view points.

I have elaborated my findings on my website on these pages. Feel free to peruse.

www.rohitn.com/numbers/numbers_CoprimeGaps.aspx
www.rohitn.com/numbers/numbers_ClaimsToProve.aspx

Thanks,
Rohit

[James Waldby]

Rohit Naringrekar <rohitn...@gmail.com> wrote:
...
> www.rohitn.com/numbers/numbers_CoprimeGaps.aspx
> www.rohitn.com/numbers/numbers_ClaimsToProve.aspx
...

Regarding symmetry of (n,h)=1 and (n,n-h)=1 about (n+1)/2, we can show
more generally that (n,h) = (n,n-h), as follows:
Suppose (n,h)=j and (n,n-h)=k. From this, there are u,v and x,y
such that n=ju, h=ju and n=kx, n-h=ky. Since n-h = j(u-v), we
have (n,n-h) >= j, ie, k >= j. Since h = n-(n-h) = k(x-y), we also
have j = (n,h) >= k. k >= j and j >= k implies k = j, as needed.
[While (n,h) = (n,n-h) holds for all n, in next paragraph n is odd.]

This allows proving the first bullet point on the first of those
pages. For some integer d we have ((n-1)/2, n) = ((n+1)/2, n) = d
so there are a, b such that (n+1)/2 = da and (n-1)/2 = db. Now
1 = (n+1)/2 - (n-1)/2 = da - db = d(a-b) which implies d = 1, ie,
that (n+1)/2 and (n-1)/2 are coprime to n. From symmetry and the
fact that (n+1)/2 and (n-1)/2 are coprime to n, we see that there
are an odd number of entries in your lists of coprime run lengths.

Regarding claim 1 on the second of those pages, "Between every 2 natural
numbers, there exists a real number ... such that the reciprocal of the
number is the fractional part of the number", this is equivalent to
saying that given an n, there is an r in (n,n+1) with 1/r = r-n, which
is 1 = r^2 - rn or r^2 - rn - 1 = 0. Apply quadratic formula and take
the positive root, so s = (n + sqrt(n^2 + 4))/2. Clearly, s > n.
Now n+1 > s if 2n+2 > n + sqrt(n^2 + 4) or if n+2 > sqrt(n^2 + 4)
or if (n+2)^2 > n^2 + 4 or if 2n > 0, which is true for all n > 0.

--
jiw

[Rohit Naringrekar]

Hi James,

Thanks for your inputs and proof. The symmetry is trivial to prove. Good clarity of thought with (n, r, n+1) .

Will look forward to any other comments you may have.

Thanks,
Rohit