# [seqfan] Re: sigma_2(n)*sigma_3(n)/sigma(n)

Robert Gerbicz robert.gerbicz at gmail.com
Fri Oct 26 11:01:16 CEST 2018

```First see what is r=sigma_2(p^k)/sigma(p^k), let x=p^k, then
r=(p^(2*k+2)-1)/(p^2-1)*(p-1)/(p^(k+1)-1)=(p^2*x^2-1)/(p^2-1)*(p-1)/(p*x-1)=(p*x+1)/(p+1)

a case: k is even, then x==1 mod (p+1), so (p*x+1)==-1+1==0 mod (p+1). So
here even r is an integer, done.
b case: k is odd, then
sigma_3(p^k)=(p^3*x^3-1)/(p^3-1), and you need to multiple this by r to
p^3*x^3-1==0 mod (p+1)
p^3-1==-2 mod (p+1)
This means, that if p=2 then sigma_3(p^k) is divisible by (p+1), so we
got the theorem.
If p is odd, then (p+1)/2 divides sigma_3(p^k), but we have another
factor=2 in (p*x+1), because p and x is odd.

ps. note that the stronger argument looks like also true: sigma_3(p^k) is
divisible by (p+1) if k is odd, but not needed this.

Frank Adams-watters via SeqFan <seqfan at list.seqfan.eu> ezt írta (időpont:
2018. okt. 26., P, 5:28):

> I am submitting a new sequence, https://oeis.org/draft/A320917, with this
> definition. The problem is to prove that it is always an integer.
>
> This reduces to showing that a(p^k) is a polynomial in p for fixed
> positive integer k, where p is a prime. This is trivially a rational
> function, so we need to show that that rational function, in lowest terms,
> has denominator = 1.
>
> I have verified this up to k = 1024, but I don't see how to continue to an
> actual proof.
>
> Failing that, it would be interesting to see what happens at k = 2^32, so
> k+1 is the the first non-prime Fermat number. Checking this is far beyond
> any brute-force computation I can make, nor do I know any way to compute it
> efficiently. I'm not certain that there is any tie to the Fermat Numbers;
> it's just an intuitive leap.
>