As usual $\sigma_k(n)$ denotes the sum of the $k$-th powers of the positive divisors of an integer $n.$ Note that $k$ is also an integer so that it may be negative.

There are no odd perfect numbers known but there are many numbers $N$ such that $$ \sigma(N) \equiv 2 \pmod{4} $$ i.e.; many numbers of the form:

$$ N = p^{4k+1}m^2 $$ with $\gcd(p,m)=1$ and $p$ a prime number with $p \equiv 1 \pmod{4}.$

Since $$ 2\sqrt{2} $$ equals the minimum of the $2z+1/z$ when $0 < z \leq 1,$ attained for $$ z=z_0 = \frac{\sqrt{2}}{2}, $$

by putting

$$ z =\frac{p^{4k+1}}{\sigma(p^{4k+1})} $$

It may have some interest the

Question: For which numbers $k$ and prime numbers $p \equiv 1 \pmod{4}$
the $z$ above is `close`

to $z_0$, say appears as a convergent in the continued fraction of $z_0.$

question inspired by some MO posts of Arnie Dris:

a) In his notation I am asking here when the sum below is `minimal`

$$ I(p^{4k+1}) + I(m^2) = \sigma_{-1} (p^{4k+1}) + \sigma_{-1}(m^2) $$

b) Arnie wanted $2N=\sigma(N)$ so that in his case

$$ I(p^{4k+1}) = 1/z $$

and

$$ I(m^2)= 2z $$

so that

$$ I(p^{4k+1})+I(m^2)=2z+1/z $$

`$\sigma_k$`

is used in line 4, with $k=1$, in line $13$ with $k=1,$ and twice in line $-1$ with $k=-1.$ Arnie wanted $2N=\sigma(N)$ so that in his case $$ I(p^{4k+1}) = 1/z $$ and $$ I(m^2)= 2z $$ so that $$ I(p^{4k+1})+I(m^2)=2z+1/z $$ $\endgroup$ – Luis H Gallardo Apr 9 '11 at 21:46