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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

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

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  • $\begingroup$ I'm confused. First, $\sigma_k()$ is defined but never used. Second, where $2z + 1/z$ comes from? $\endgroup$ – Max Alekseyev Apr 9 '11 at 6:31
  • $\begingroup$ Notation $\sigma_k$ 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
  • $\begingroup$ For clarity comment above is incorporated to question. $\endgroup$ – Luis H Gallardo Apr 10 '11 at 0:45
  • $\begingroup$ @Luis, I just happened to look at this one again today, and I think you should have $I(p^{4k+1}) = 2z$ and $I(m^2) = 1/z$ owing to parity constraints. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 28 '12 at 12:34
  • $\begingroup$ $\sigma(p^{4k+1})$ is even $\endgroup$ – Luis H Gallardo Feb 2 '12 at 19:32
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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 $$\frac{4}{5} = \lim_{k\to\infty} \frac{5^{4k+1}}{\sigma(5^{4k+1})}.$$

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