RE: [tuning-math] Digest Number 1321

Gene,

Any thoughts on what a sufficient condition might be?

Regards,
Yahya

-----Original Message-----
Date: Tue, 21 Jun 2005 02:59:34 -0000
From: "Gene Ward Smith" <gwsmith@...>
Subject: Re: Tonality diamond conjecture

For most primes, it appears the p-diamond comma is p^2/(p^2-1). However,
in some cases (7, 41, 43, 73, 83, 157, 193, 211 ...) it is of the form
(p^2+1)/p^2. I've not found an example where it is of neither form.
All primes where (p^2+1)/p^2 is minimal must be such that p^2+1
factors into primes of size p or less, but this condition is not
sufficient, 47 being the first counterexample.

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--- In tuning-math@yahoogroups.com, "Yahya Abdal-Aziz" <yahya@m...> wrote:

> Gene,
>
> Any thoughts on what a sufficient condition might be?

The question is what will insure that the p-diamond comma is of the form
(n^2+1)/n^2.

For any integer n, we can define oddpart(n) to be n with all the
powers of 2 divided out, so that n = 2^k oddpart(n), with oddpart(n)
odd. We define another function, mind(n), as the least divisor greater
than or equal to sqrt(n). Then a necessary and sufficient condition is
that
mind(oddpart(n^2+1)) < n. This fails, for instance, when n=73, where
the odd part of 47^2+1 factors as 5*13*17, and the product of any two
of the factors is always greater than 47. Or, for instance, the odd
part of
57^2+1 factors as 5^3*13; 5^3 gives 125, greater than 57, and 5^2
leaves us with 5*13=65, greater than 57.

>> Gene,
>>
>> Any thoughts on what a sufficient condition might be?
>
>The question is what will insure that the p-diamond comma is of the form
>(n^2+1)/n^2.

Do you mean n-diamond comma?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> Gene,
> >>
> >> Any thoughts on what a sufficient condition might be?
> >
> >The question is what will insure that the p-diamond comma is of the
form
> >(n^2+1)/n^2.
>
> Do you mean n-diamond comma?

Right, sorry.

(edited)
>The question is what will insure that the n-diamond comma is of the form
>(n^2+1)/n^2.
>
>For any integer n, we can define oddpart(n) to be n with all the
>powers of 2 divided out, so that n = 2^k oddpart(n), with oddpart(n)
>odd. We define another function, mind(n), as the least divisor greater
>than or equal to sqrt(n). Then a necessary and sufficient condition is
>that mind(oddpart(n^2+1)) < n. This fails, for instance, when n=73,
>where the odd part of 47^2+1 factors as 5*13*17, and the product of
>any two of the factors is always greater than 47. Or, for instance,
>the odd part of 57^2+1 factors as 5^3*13; 5^3 gives 125, greater than
>57, and 5^2 leaves us with 5*13=65, greater than 57.

Nice. -Carl