A question about odd perfect numbers

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!

• 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. – anon Jul 30 '13 at 4:04
• Okay thanks anon, for your clarification. Editing my post in response to your comment now. – Jose Arnaldo Bebita-Dris Jul 30 '13 at 4:39
• 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. – Jose Arnaldo Bebita-Dris Jul 30 '13 at 5:02
• 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. – mathlove Nov 13 '18 at 6:50
• @mathlove: I think what I currently have are (to use your notation): $$\bigg(A \implies B\bigg) \land \bigg(C \implies B\bigg) \implies \bigg((A \lor C) \implies B\bigg).$$ – Jose Arnaldo Bebita-Dris Nov 13 '18 at 6:59

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 $$(4)$$.

• Thank you for your answer, @mathlove. Can you comment on my answer (below yours) please? =) – Jose Arnaldo Bebita-Dris Nov 13 '18 at 12:47
• @Jose Arnaldo Bebita Dris : It looks correct to me. – mathlove Nov 14 '18 at 3:39

(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

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