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Edit [in response to a comment from anon]: Hereinafter, $N$ is a positive integer, $\sigma(N)$ is the sum-of-divisors of $N$, $\omega(N)$ is the number of distinct prime factors of $N$, and $\Omega(N)$ is the number of prime factors of $N$ (counting multiplicities).

Thus, $N$ is a perfect number if $\sigma(N) = 2N$.

A 2013 preprint by Nielsen claims to have proved that $\omega(N) \geq 10$ if $N$ is an odd perfect number. A paper by Ochem and Rao containing inequalities relating $\Omega(N)$ and $\omega(N)$ for $N$ an odd perfect number has been recently accepted in the Mathematics of Computation. The state-of-the-art result for $\Omega(N)$ remains to be Hare's $\Omega(N) \geq 75$.

[Edit - August 29] The state-of-the-art result for $\Omega(N)$ (where $N$ is an odd perfect number) is now Ochem and Rao's $\Omega(N) \geq 101$. [End edit]

[End edit - July 30 2013]

If there exists an $i \in \left[1,\omega(N)\right]$ such that

$$N \leq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}),$$

then $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive)

is ${\it not}$ an odd perfect number. (See Theorem 4.2.5, page 112 in this M.Sc. thesis.)

In particular, suppose $i = 1$. (That is, let $p_1$ be the smallest prime factor of $N$.) Then we have

$${{p_1}^{\alpha_1}}\prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} = N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}} \leq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}) < \frac{9}{4}{{p_1}^{2\alpha_1}},$$

from which it follows that

$$\prod_{i=2}^{\omega(N)}{{p_2}^{\alpha_i}} \leq \prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} < \frac{9}{4}{{p_1}^{\alpha_1}}.$$

But we also have

$${p_2}^{\Omega(N) - {\alpha_1}} = \prod_{i=2}^{\omega(N)}{{p_2}^{\alpha_i}} \leq \prod_{i=2}^{\omega(N)}{{p_i}^{\alpha_i}} < \frac{9}{4}{{p_1}^{\alpha_1}} < \frac{9}{4}{{p_2}^{\alpha_1}} < {{p_2}^{\alpha_1 + 1}},$$

from which we obtain

$$\frac{\Omega(N) - 1}{2} < \alpha_1.$$

Note that we have obtained the result:

"If $$N = {p_1}^{2\alpha_1}{q^k}\prod_{i=2}^{\omega(N) - 1}{{p_i}^{\alpha_i}}$$ is an odd (positive integer) with $\frac{\Omega(N) - 1}{2} < \alpha_1,$ then $N$ is ${\it not}$ perfect."

Taking the contrapositive of the result we have obtained, we have: "If $$N = {p_1}^{2\alpha_1}{q^k}\prod_{i=2}^{\omega(N) - 1}{{p_i}^{\alpha_i}}$$ is an odd perfect number with smallest prime factor $p_1$ and Euler prime $q$, then $\alpha_1 \leq \frac{\Omega(N) - 1}{2}$."

Somebody, please tell me that I ${\it did}$ make a logical error somewhere -- I am finding it increasingly hard to spot my own mistakes these days. =(

Thank you!

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  • $\begingroup$ How does one get $\sigma(p^a)<\frac{3}{2}p^a$ without assuming $p\ne2$? What is "the result we have obtained" that you are trying to take the contrapositive of - can you type it out all as one sentence with logical symbols and all? Also, there are things you shouldn't just leave your readers to work out, namely that $N=\prod p_i^{a_i}$ is $N$'s prime factorization with $p_i$s listed in increasing order, and what $\omega,\sigma,\Omega$ are. $\endgroup$ – anon Jul 30 '13 at 4:04
  • $\begingroup$ Okay thanks anon, for your clarification. Editing my post in response to your comment now. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 30 '13 at 4:39
  • $\begingroup$ Done editing my post to conform to the details that you require @anon. In particular, to answer your first question, I am limiting my $N$ to odd integers only. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 30 '13 at 5:02
  • $\begingroup$ Let $A$ be "$N\leq \frac{3}{2}{p_1}^{\alpha_1}\sigma({p_1}^{\alpha_1})$" and let $B$ be "$N$ is not an odd perfect number" and let $C$ be "$\frac{\Omega(N) - 1}{2} < \alpha_1$". It seems to me that you are saying that since both $A\implies B$ and $A\implies C$ are true, $C\implies B$ is true. $\endgroup$ – mathlove Nov 13 '18 at 6:50
  • $\begingroup$ @mathlove: I think what I currently have are (to use your notation): $$\bigg(A \implies B\bigg) \land \bigg(C \implies B\bigg) \implies \bigg((A \lor C) \implies B\bigg).$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 13 '18 at 6:59
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You've written

(1) If there exists an $i \in \left[1,\omega(N)\right]$ such that $N \leq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i})$, then $N$ (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive) is not an odd perfect number.

(2) If $N \leq \frac{3}{2}{p_1}^{\alpha_1}\sigma({p_1}^{\alpha_1})$ where $p_1$ is the smallest prime factor of $N$, then $\frac{\Omega(N) - 1}{2} < \alpha_1$.

(3) We have obtained the result: "If $N$ is an odd positive integer with $\frac{\Omega(N) - 1}{2} < \alpha_1,$ then $N$ is not perfect."

(4) Taking the contrapositive of the result we have obtained : "If $N$ is an odd perfect number with smallest prime factor $p_1$ and Euler prime $q$, then $\alpha_1 \leq \frac{\Omega(N) - 1}{2}$."

Now, let

$\qquad A$ : $N \leq \frac{3}{2}{p_1}^{\alpha_1}\sigma({p_1}^{\alpha_1})$

$\qquad B$ : $N$ is not an odd perfect number

$\qquad C$ : $\frac{\Omega(N) - 1}{2} < \alpha_1$

From $(1)$, we have $$A\implies B$$

From $(2)$, we have $$A\implies C$$

Now you are claiming in $(3)$ that $$C\implies B$$ However, I don't see any proof for this.

So, I think that you have not proven what you claimed in $(4)$.

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  • $\begingroup$ Thank you for your answer, @mathlove. Can you comment on my answer (below yours) please? =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 13 '18 at 12:47
  • $\begingroup$ @Jose Arnaldo Bebita Dris : It looks correct to me. $\endgroup$ – mathlove Nov 14 '18 at 3:39
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(This is only a partial answer. This approach considers the largest prime factor of $N$ (instead of the smallest) when computing a bound for $\Omega(N)$.)

If $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ ${\it is}$ an odd perfect number (where the $p_i$'s are primes ordered in increasing magnitude and the $\alpha_i$'s are all positive), then

$$N \geq \frac{3}{2}{p_i}^{\alpha_i}\sigma({p_i}^{\alpha_i}).$$

(See Theorem 4.2.5, page 112 in this M.Sc. thesis.)

(Let $r=\omega(N)$.) In particular, suppose $i = r$. (That is, let $p_r$ be the largest prime factor of $N$.) Then we have

$${{p_r}^{\alpha_r}}\prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} = N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}} \geq \frac{3}{2}{p_r}^{\alpha_r}\sigma({p_r}^{\alpha_r}) > \frac{3}{2}{{p_r}^{2\alpha_r}},$$

from which it follows that

$$\prod_{i=1}^{r-1}{{p_{r-1}}^{\alpha_i}} \geq \prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} > \frac{3}{2}{{p_r}^{\alpha_r}}.$$

But we also have

$${p_r}^{\Omega(N) - {\alpha_r}} > {p_{r-1}}^{\Omega(N) - {\alpha_r}} = \prod_{i=1}^{r-1}{{p_{r-1}}^{\alpha_i}} \geq \prod_{i=1}^{r-1}{{p_i}^{\alpha_i}} > \frac{3}{2}{{p_r}^{\alpha_r}} > {p_r}^{\alpha_r},$$

from which we obtain

$$\Omega(N) > 2\alpha_r.$$

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