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

Robert Gerbicz robert.gerbicz at gmail.com
Fri Oct 26 11:20:49 CEST 2018

```"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."
Yes, this is trivial: sigma_3(p^k)=1+p^3+...+p^(3*k)==1-1+1-1+...+1-1==0
mod (p+1) if k is odd.
This also shortens the above proof.

Robert Gerbicz <robert.gerbicz at gmail.com> ezt írta (időpont: 2018. okt.
26., P, 11:01):

> 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
> get your term, and here
>    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.
>>
>> Franklin T. Adams-Watters
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>

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