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I'm a grad student in mathematics and I've been working with a very gifted high school student (likely the smartest high school student I've ever met) on problems he's brought up and some competition math problems. This student has developed an interest in perfect numbers and the question regarding existence of odd perfect numbers. He has come up with a conjecture about odd perfect numbers, but I have not studied number theory and hence am not necessarily aware of well-known results of the field. So, here we are.

His idea:

Suppose $N \in \mathbb{N}$, with prime decomposition $N = p_1^{q_1}\cdots p_n^{q_n}$ Define $\tilde{N} = p_1\cdots p_n$.

Conjecture: If $N$ is an odd perfect number, then the sum of reciprocals of all factors of $\tilde{N}$ (excluding 1, including $\tilde{N}$) is less than 1.

Q. Does this conjecture appear to be equivalent to something that has already been established? If this conjecture is true, does it appear to have any obvious implications?

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    $\begingroup$ On how many odd perfect numbers have you guys tested the conjecture? ;) $\endgroup$ Commented Dec 13, 2017 at 23:34
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    $\begingroup$ On every odd perfect number we could find :) $\endgroup$ Commented Dec 13, 2017 at 23:38
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    $\begingroup$ The conjecture is equivalent to something that is well-known. If $\sigma(n)$ denotes the sum of the divisors of $n$, then the sum of the reciprocals of the divisors of $n$ is $\sigma(n)/n$. (Proof: Exercise.) We have that $n$ is perfect if and only if $\sigma(n) = 2n$, so if $n$ is perfect, the sum of the reciprocals of the divisors of $n$ (including $1$ and $n$ itself) is exactly $2$. $\endgroup$ Commented Dec 14, 2017 at 0:00
  • $\begingroup$ Does your student claim that the sum is $\le 1$ or $\lt 1$? $\endgroup$ Commented Dec 14, 2017 at 0:08
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    $\begingroup$ @user8734617 To clarify Jeremy Rouse's comment, $N$ is not square free, by known facts about odd perfect numbers en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers So $\tilde N | N$, properly. So the terms in the OP's sum are a proper subset in the terms in Jeremy's sum. The OP's sum is $<1$. $\endgroup$ Commented Dec 14, 2017 at 0:17

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The earliest result appears to be $$\frac{1}{2}<\sum_{p\mid N}\frac{1}{p}<2 \log \left (\frac{\pi}{2} \right ),$$

as proved by Perisastri in 1958. Other authors have sharpened this result (eg D. Suryanarayana's "On odd perfect numbers. II", Proceedings of the American Mathematical Society, Vol. 14, No. 6 (Dec. 1963), pp. 896-904). For example, for a perfect number $N\equiv 0 \mod 3$, the sum over primes is $$\sum_{p\mid N}\frac{1}{p}=\frac{1}{3}+ \cdots+\frac{1}{p'},$$ see Ribenboim, New Book of Prime Records, p. 101, for a discussion and other references.

  • Perisastri, M., A note on odd perfect numbers, Math. Student 26, 179-181 (1959). ZBL0096.02601.

(note that MathSciNet records the date as 1958, not 1959)

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    $\begingroup$ Something must be wrong in your identity : $p$ is a dummy variable in the lhs, while it is not in the rhs. $\endgroup$ Commented Dec 15, 2017 at 19:40
  • $\begingroup$ Here the number $p$ is prime. $\endgroup$ Commented Mar 25, 2023 at 13:39
  • $\begingroup$ @DenisSerre I made the only sensible edit that I can think of, to remove that notation clash $\endgroup$ Commented Nov 18, 2024 at 22:31
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The conjecture is true. Since $$\sigma(N)=\prod\left(\frac{p^{e+1}-1}{p-1}\right)$$ and $$\sigma(\tilde{N})=\prod(p+1)$$, it follows that $$\frac{\sigma(N)}{N}=\prod\left(\frac{p^{e+1}-1}{p^e(p-1)}\right)$$ and $$\frac{\sigma(\tilde{N})}{N}=\prod\left(\frac{p+1}{p}\right)$$, so $$\frac{\sigma(\tilde{N})}{\sigma(N)}=\prod\left(\frac{p^{e+2}-p^e}{p^{e+2}-p}\right)\le1$$ with equality iff $N$ is squarefree, where $$N=\prod{p^{e}}$$ and $$N=\prod{p}$$.

All products run over distinct prime factors $p$ of $N$ and $\tilde{N}$ in the equations above.

Since $$\frac{\sigma(N)}{N}=2$$, it follows that $$\frac{\sigma(\tilde{N})}{\tilde{N}}\le2$$ with equality iff $N$ is squarefree. Since $N$ is odd, we have that $\sigma(N)$ isn't divisible by $4$. Since $N$ is squarefree, $$\sigma(N)=\prod(p+1)$$. The product has only even factors, and the product of two or more even integers is divisible by $4$. Hence $N$ is prime and $$2N=N+1$$. This gives $N=1$, a contradiction. Hence $$\frac{\sigma(\tilde{N})}{\tilde{N}}\lt2$$, so $$\frac{\sigma(\tilde{N})}{\tilde{N}}-1\lt1$$, and the desired statement follows.

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