[seqfan] Re: sigma_2(n)*sigma_3(n)/sigma(n)
giovanni.resta at iit.cnr.it
Fri Oct 26 11:42:11 CEST 2018
On 10/26/2018 04:37 AM, Frank Adams-watters via SeqFan wrote:
> 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.
We have sigma_2(p^k)*sigma_3(p^k)/sigma(p^k) =
((p^(2+2k)-1)/(p^2-1) * (p^(3+3k)-1)/(p^3-1)) / ((p^(1+k)-1)/(p-1)) =
((p^(k+1)+1) (p^(3 k+3)-1)) / ((p+1) (p^3-1)),
because p^(2+2k)-1 is the difference of two squares and we can apply
x^2-1 = (x+1)(x-1).
We observe that (p^(3 k+3)-1)) is always divisible by (p^3-1) (it is
Now, if k is even, k+1 is odd, so p^(k+1)+1 is divisible by p+1 giving
1-p+p^2-p^3... +p^k as result, and we are done.
If k is odd, k+1 is even and in general p^(k+1)+1 is not divisible by
p+1 (like p^2+1 is not divisible by p+1).
However, if k is odd, then 3k+3 is even, so p^(3k+3)-1 beside being
divisible by p^3-1 is also divisible by p+1. Since p+1 and p^3-1 have no
common factors, then the ratio (p^(3 k+3)-1)) / ((p+1) (p^3-1)) is
integer and we are done.
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