1

The topic of odd perfect numbers likely needs no introduction.

In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Let $$D(x) = 2x - \sigma(x)$$ denote the deficiency of $x$, and let $$s(x) = \sigma(x) - x$$ denote the sum of aliquot/proper divisors of $x$. Finally, denote the abundancy index of $x$ by $$I(x) = \frac{\sigma(x)}{x}.$$

Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $q$ is prime and $N$ is perfect, we obtain $$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$ so that we get $$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$ Using the identity $$\frac{D(x)}{x} = 2 - I(x)$$ we obtain the bounds $${q^k}\bigg(\frac{q-2}{q-1}\bigg) < D(q^k) \leq {q^{k-1}}\bigg(q-1\bigg)$$ and $$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q}.$$ This implies that $${2q^k n^2}\cdot\bigg(\frac{q-2}{(q-1)(q+1)}\bigg) < D(q^k)D(n^2) < {2q^k n^2}\cdot\bigg(\frac{q-1}{q^2}\bigg).$$ Dividing both sides of the last inequality by $2q^k n^2$, we get $$\frac{q-2}{(q-1)(q+1)} < \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-1}{q^2}.$$ But we know that $$D(q^k)D(n^2)=2s(q^k)s(n^2),$$ which can be verified by a direct, brute-force computation. Thus, the fraction in the middle of the last inequality simplifies to $$\frac{D(q^k)D(n^2)}{2q^k n^2}=\bigg(\frac{s(q^k)}{q^k}\bigg)\bigg(\frac{s(n^2)}{n^2}\bigg)=\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) = 3 - \bigg(I(q^k) + I(n^2)\bigg).$$

We therefore finally have the bounds $$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2) < 3 - \bigg(\frac{q-2}{(q-1)(q+1)}\bigg)$$ which does not improve on the known bounds $$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$ (See this paper for a proof.)

Here are my:

QUESTIONS Is it possible to improve on the bounds for $D(q^k)$ and $D(n^2)$ (where $q^k n^2$ is an odd perfect number with special prime $q$) to hopefully produce stronger bounds for $I(q^k) + I(n^2)$? If so, how could this be done?

Jose Arnaldo Bebita Dris
  • 8,471
  • Note that $$3 - \bigg(\frac{q-1}{q^2}\bigg) \geq \frac{71}{25} = 2.84$$ while $$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) \geq \frac{57}{20} = 2.85.$$ – Jose Arnaldo Bebita Dris Oct 31 '20 at 22:05
  • 1
    FYI, let $f(k):=I(q^k) + I(n^2)$. Then, $f'(k)\lt 0$, so $\lim_{k\to\infty}f(k)\lt f(k)\le f(1)$ which is the known bounds. So, I think (1) if you prove $k\not=1$, then you can get an improved upper bound. (2) if you get an improved upper bound, then you can say $k\not=1$. (3) if you prove that there is an integer $N$ such that $k\le N$, then you can get an improved lower bound. (4) if you get an improved lower bound, then you can say that there is an integer $N$ such that $k\le N$. (I'm not saying that if you can't improve the bounds for $k$, you can't improve the bounds for $f(k)$.) – mathlove Nov 01 '20 at 12:44
  • Thank you for your comments, @mathlove! You are on a roll! =) Would you mind fleshing out your comments as an actual answer, and include additional details if needed, so that I may be able to upvote and accept? =) – Jose Arnaldo Bebita Dris Nov 01 '20 at 12:55

2 Answers2

1

On OP's request, I am converting my comment into an answer.

Let $f(k):=I(q^k) + I(n^2)$.

Then, we have $$f'(k)=\frac{-( q^{2 k + 2}- 4 q^{2 k + 1}+2 q^{k + 1} + 2 q^{2 k} - 1) \log(q)}{(q - 1)q^k (q^{k + 1} - 1)^2}$$ which is negative, so we see that $f(k)$ is decreasing.

It follows that $$\lim_{k\to\infty}f(k)\lt f(k)\le f(1)\tag1$$ which is the known bounds $$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg)\tag2$$

So, I think that this means the following :

  • If you prove that $k\not=1$, then you can get an improved upper bound for $f(k)$.

  • If you get an improved upper bound for $f(k)$, then you can say that $k\not=1$.

  • If you prove that there is an integer $a$ such that $k\le a$, then you can get an improved lower bound for $f(k)$.

  • If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$.

(I'm not saying that if you can't improve the bounds for $k$, you can't improve the bounds for $f(k)$.)

mathlove
  • 151,597
  • Thank you for your answer, @mathlove! I noticed that you used the same variable for the upper bound of $k$ as my variable for the odd perfect number $N$. We do agree, however, that the inequality $$k \leq N = q^k n^2$$ is trivial, do we not? =) – Jose Arnaldo Bebita Dris Nov 02 '20 at 04:56
  • @Arnie Bebita-Dris : Thanks for noticing it. That's not what I meant. I should have used a different variable. I've edited the answer. – mathlove Nov 02 '20 at 05:03
  • My apologies for the bother @mathlove, but I have been trying to prove your third and fourth assertions, and I just realized that I could not reproduce your proofs. Would you mind sharing a sketch of your argument? – Jose Arnaldo Bebita Dris Nov 02 '20 at 06:34
  • Okay @mathlove, just now I used WolframAlpha to get the partial fraction decomposition of $I(q^k) + I(n^2)$ and obtained $$\frac{3q^2 - 4q + 2}{q(q - 1)} - \frac{1}{q^k (q - 1)} + \frac{2(q - 1)}{q(q^{k+1} - 1)}.$$ I am unable to proceed further, since I do not know how to handle the negative summand. – Jose Arnaldo Bebita Dris Nov 02 '20 at 06:58
  • 1
    @Arnie Bebita-Dris : The key is the fact that $f(k)$ is decreasing. For the third, if you prove that there is an integer $a$ such that $k\le a$, then, considering the graph of $y=f(k)$, we see that $f(a)$ is an improved lower bound. For the fourth, if you get an improved lower bound $g(q)$, then from the above fact, we see that there is only one $k=k_0$ such that $f(k)=g(q)$. Then, we have $k\le \lceil k_0\rceil$. – mathlove Nov 02 '20 at 07:07
  • Thank you for your further comments, @mathlove. For your proof of the fourth implication, I noticed you used the variable $q$ (for the special prime?) as an argument for the lower bound $g$. Does this mean that the lower bound obtained depends on $q$? – Jose Arnaldo Bebita Dris Nov 02 '20 at 08:08
  • 1
    @Arnie Bebita-Dris : Yes, that is what I meant. – mathlove Nov 02 '20 at 08:10
0

Let me try to work backwards from $$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$ This can be rewritten as $$\frac{q-1}{q(q+1)} \leq 3 - \bigg(I(q^k) + I(n^2)\bigg) = \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-2}{q(q-1)}.$$ We also have $$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q},$$ which we can rewrite as $$q < \frac{2n^2}{D(n^2)} \leq q + 1.$$ We therefore obtain $$\frac{q-1}{q+1} = \frac{q(q-1)}{q(q+1)} < \bigg(\frac{q-1}{q(q+1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{D(q^k)}{q^k} < \bigg(\frac{q-2}{q(q-1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{(q+1)(q-2)}{q(q-1)},$$ which implies that $$\frac{q-1}{q+1} < 2 - I(q^k) < \frac{(q+1)(q-2)}{q(q-1)},$$ which simplifies to $$\frac{q^2 - q + 2}{q^2 - q} = 2 - \bigg(\frac{(q+1)(q-2)}{q(q-1)}\bigg) < I(q^k) < 2 - \bigg(\frac{q-1}{q+1}\bigg) = \frac{q+3}{q+1}.$$ Note that both bounds do not improve on the currently known $$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}.$$

Jose Arnaldo Bebita Dris
  • 8,471