A018892 and extensions

[hv at crypt.org]

simplifies to two unit fractions.
       { A018892(n)    if n == 1 (mod 2)
       { d(P^2) (d(Q^2) - 1) / 4    if n == 1 (mod 3)
       { d(P^2) (d(Q^2) + 1) / 4    if n == 2 (mod 3)
satisfied myself of their correctness; while I plan to investigate m/n
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Date: Tue, 31 Jul 2007 14:56:53 +0200
From: Peter Pein <petsie at dordos.net>
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To: Max Alekseyev <maxale at gmail.com>
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Subject: Re: a propos divisors...
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Max Alekseyev schrieb:
> On 7/28/07, Peter Pein <petsie at dordos.net> wrote:
> 
>>> Each here line contains a pair:
>>> n k
>>> such that sigma(3,k)=sigma(3,n) and k=a(n) for r=3 (following your
>>> notations above).
>>>
>>> I will let my program to run for a couple more days to reach 10^8 bound.
> 
> I stopped the program after it reached 7*10^7 as it started to eat too
> much memory.
> The results are attached. Please format them and submit to OEIS as you
> feel appropriate.
> 
> P.S. Eventually I've defended Ph.D. in Computer Science ;)

... if you felt that this has been necessary... ;-)
> 
> Regards,
> Max
> 
> 
Thank you again, Max.

As soon as I've got a little bit more time to spare, I'll enter the values

Peter