A023194 numbers n such that sum of divisors of n is prime.

[Gabe Cunningham]

 From the fact that the sigma (the sum-of-divisors function) is
multiplicative, we can derive that the sigma(n) is even except when n
is a square or twice a square. If n = 2(2k+1)^2, that is, n is twice
an odd square, then sigma(n) = 3*sigma((2k+1)^2). If n = 2(2k)^2, that
is, n is twice an even square, then sigma(n) is only prime if n is a
power of 2; otherwise we have sigma(n) = sigma(8*2^m) * sigma(k/2^m)
for some positive integer m. So the only possible candidates for
values of n other than squares such that sigma(n) is prime are odd
powers of 2. But sigma(2^(2m+1)) = 2^(2m+2)-1 = (2^(m+1)+1) * (2^(m+1)
- 1), which is only prime when m=0, that is, when n=2. So 2 is the
only non-square n such that sigma(n) is prime.

- Gabe

(apologies to zak for the duplicate.)

Quoting zak seidov <zakseidov at yahoo.com>:

> Are all terms in A023194,
> except the first one,
> the full squares?
> Is there any non-square number n, except 2,
> such that sum of divisors of n is prime?
> Cf. A107926 (just submitted),
> Zak
> 
> 
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