Question

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

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

Answer 1

Note that $$z = \frac{p^{4k+1}}{\sigma(p^{4k+1})} = \frac{1-p^{-1}}{1-p^{-(4k+2)}} > 1-p^{-1}.$$ Clearly, for a fixed p, the larger is $k$ the closer is $z$ to this lower bound.

Since the lower bound is greater than $z_0$ (for any prime $p\geq 5$) and grows with $p$, it is beneficial to take $p$ as small as possible. The smallest such prime $p\equiv 1\pmod{4}$ is $p=5$.

Therefore, for any prime $p\equiv 1\pmod{4}$ and $k\geq 0$, we have $$z - z_0 = \frac{p^{4k+1}}{\sigma(p^{4k+1})} - \frac{\sqrt{2}}{2} > \frac{4}{5} - \frac{\sqrt{2}}{2} > 0.09$$ where $$\frac{4}{5} = \lim_{k\to\infty} \frac{5^{4k+1}}{\sigma(5^{4k+1})}.$$