(Preamble: I first thought that this question might be more appropriate for MSE. However, I posted it here nonetheless in the hope that someone with that brilliant idea can help with answering my questions at the end. If this is not appropriate for MO, please let me know and I will be more than happy to migrate the question over to MSE.)

Let $\sigma(x)$ be the sum of the divisors of $x$. We say that $X$ is almost perfect if $\sigma(X) = 2X - 1$.

Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) show that, if $M \neq 2^t$ is an even almost perfect number, then $M$ takes the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite.

Recently, Antalan and Dris have been able to obtain the following bounds: $$\dfrac{\sigma(2^r)}{b} < 1 < \dfrac{\sigma(b)}{b} < \dfrac{\sigma(b^2)}{b^2} < \dfrac{4}{3} < \dfrac{3}{2} \leq \dfrac{\sigma(2^r)}{2^r} < 2 < \dfrac{\sigma(b)}{2^r}.$$ The key ingredient in the proof is to observe that $b^2$ is deficient, so that we can write $$\sigma(b^2) = 2b^2 - D$$ where $D = 2b^2 - \sigma(b^2)$ is the deficiency of $b^2$. Notice that we also have $$D = b^2 - \dfrac{b^2 - 1}{\sigma(2^r)},$$ from which we obtain the upper bound for the abundancy index of $b^2$: $$I(b^2) = \dfrac{\sigma(b^2)}{b^2} < \dfrac{4}{3},$$ by considering $r \geq 1$.

Note that, if $r > 1$, then $$\sigma(2^r) = 2^{r + 1} - 1 \geq 7,$$ so that $$D \geq b^2 - \left(\dfrac{b^2 - 1}{7}\right) = \dfrac{6b^2 + 1}{7}.$$ Consequently, $$I(b^2) = \dfrac{\sigma(b^2)}{b^2} = 2 - \dfrac{D}{b^2} \leq 2 - \left(\dfrac{6b^2 + 1}{7b^2}\right) = \dfrac{8b^2 - 1}{7b^2} = \dfrac{8}{7} - \dfrac{1}{7}\cdot\dfrac{1}{b^2} < \dfrac{8}{7}.$$

By the contrapositive, if $I(b^2) > 8/7$, then $r = 1$. (Note that it is not possible to have $I(b^2) = 8/7$.) We want to show results in the other direction.

Suppose $r = 1$. Then $$I(b^2) = \dfrac{\sigma(b^2)}{b^2} = 2 - \dfrac{D}{b^2} = 1 + \left(\dfrac{b^2 - 1}{b^2}\right)\cdot\left(\dfrac{1}{\sigma(2^1)}\right) = \dfrac{4b^2 - 1}{3b^2}.$$ Since $3 \nmid b$ (by work of Antalan) and $b$ is an odd composite, then $b \geq 5 \cdot 7 = 35$, so that $$I(b^2) = \dfrac{4}{3} + \dfrac{1}{3}\cdot\left(\dfrac{-1}{b^2}\right) \geq \dfrac{1633}{1225} \approx 1.333 > 1.\overline{142857} = \dfrac{8}{7}.$$

Consequently, we have the improved bounds $$\dfrac{8}{7} < \dfrac{\sigma(b^2)}{b^2} < \dfrac{4}{3} < \dfrac{3}{2} = \dfrac{\sigma(2^1)}{2^1} = \dfrac{\sigma(2^r)}{2^r},$$ if $r = 1$, and $$1 < \dfrac{\sigma(b^2)}{b^2} < \dfrac{8}{7} < \dfrac{7}{4} = \dfrac{\sigma(2^2)}{2^2} \leq \dfrac{\sigma(2^r)}{2^r},$$ if $r > 1$.

Now, we want to somehow obtain a contradiction by assuming $r = 1$, then try using a criterion by Dris:

$$N \hspace{0.05in} \text{is almost perfect} \iff \dfrac{2N}{N + 1} \leq \dfrac{\sigma(N)}{N} < \dfrac{2N + 1}{N + 1}.$$

Update - October 01 2016 (11:57 PM) From a recent answer to this MSE post, MSE user Erick Wong asserts that: "This leaves the sufficient condition as the only potential object of study. But the strength of the sufficiency statement grows with the size of the interval. To make the claim stronger we'd need to widen the bounds, not make them tighter." Dagal and myself think that, for purposes of arriving at a contradiction, we must use tighter bounds for $I(N)=\sigma(N)/N$. Are we indeed misguided, to use Erick's term?

Alas, here is where I get stuck.

Added May 14 2016

Taking off from the method in this MSE post, it is easy to show that $$r = 1 \implies \dfrac{8}{7} < I(b^2) < \dfrac{4}{3} \implies 3 \nmid b$$ $$r > 1 \implies I(b^2) < \dfrac{8}{7} \implies 7 \nmid b.$$

Note that the contrapositive of the second implication is that $$7 \mid b \implies r = 1.$$

I am not too sure, though, how these further restrictions might help with trying to obtain a contradiction under the assumption that $r = 1$.


(1) To what extent can the arguments in this post be optimized to yield potentially improved numerical bounds for $I(b^2) = \sigma(b^2)/b^2$?

(2) Is the sought contradiction for proving $r > 1$ achievable under the framework presented in this post?

(3) What (hopefully constructive) suggestions can you give for improving the partitioned bounds for the abundancy index $I(b^2)$ for specified values of $r$?


Perhaps you might consider using the fact, pointed out in an answer to one of your other questions that $\sigma(M)$ and $M$ are relatively prime, to impose further restrictions on $b$. Although this may not answer any of your questions on achieving $r>1$ or bounds on $I(b^2)$, it may restrict the $b$ that you need to consider in searching for even almost perfect numbers of the form mentioned in your question. For example, if $5$ divides $b$, then if a prime $p\equiv 1 \bmod 5$, then $p^2$ won't be the highest power of $p$ dividing $b$ (because then $5$ divides $\sigma(p^4)$ and $\sigma(M)$, contradicting the fact that $gcd(M,\sigma(M))=1$). However, these types of considerations probably won't improve on existing restrictions on $b$ appearing in almost perfect $M$ of the form $2^rb^2$, since in an answer to yet another of your questions, it is already pointed out that the number of almost perfect $N<x$ is at most $x^{1/4+\epsilon}$, for any $\epsilon>0$ and $x>x_0(\epsilon)$, and in fact it is pointed out that the density ought to be logarithmic.

Added later: Here's a way to sharpen the lower bound on $I(b^2)$ when $r=1$. From the proof of Lemma 2.3 in the linked paper of Antalan and Dris, we have $$ (2^{r+1}-1)\sigma(b^2)=2^{r+1}b^2-1. $$ Then $\sigma(b^2)\equiv 1 \bmod 2^{r+1}$. When $r=1$, $\sigma(b^2)\equiv 1 \bmod 4$. Then the sum of exponents of the $1\bmod 4$ primes in the factorization of $b$ is even. Also, $b$ is not a prime power for any $r \geq 1$ because if $b=p^e$ then $\sigma(b^2)\equiv 1\bmod p$ whereas the equation above gives $\sigma(b^2)\equiv -(2^{r+1}-1)^{-1}\bmod p$. Therefore when $r=1$, $b\geq 5^27$ and $I(b^2)\geq \frac{4}{3}-\frac{1}{3(175^2)}$.

  • 1
    $\begingroup$ user85586, indeed if $M = {2^r}{b^2}$ is an almost perfect number which is not a power of two (with $r \geq 1$), then I know that $\gcd\left(b^2, \sigma(b^2)\right) = 1$ (first proved by Dagal and Ymas), which implies by Greening's Theorem that $b^2$ is solitary. Additionally, $b^2$ satisfies the divisibility constraint $\left(\sigma(b^2) - b^2\right) \mid \left(b^2 - 1\right)$ (see OEIS sequence A059046). $\endgroup$ – Jose Arnaldo Bebita Dris May 14 '16 at 5:27
  • $\begingroup$ So now I have the following $\gcd$ constraints: $$\gcd\left(2^r, \sigma(2^r)\right) = 1$$ $$\gcd\left(b^2, \sigma(b^2)\right) = 1$$ $$\gcd\left({2^r}{b^2}, \sigma(2^r)\sigma(b^2)\right) = 1.$$ Lastly, since we know that $$\dfrac{\sigma(2^r)}{b^2} < 1,$$ then we are sure at this point that $$b^2 \nmid \sigma(2^r).$$ $\endgroup$ – Jose Arnaldo Bebita Dris May 14 '16 at 5:34
  • $\begingroup$ But we DO KNOW that $$\sigma(2^r) \mid \left(b^2 - 1\right).$$ $\endgroup$ – Jose Arnaldo Bebita Dris May 14 '16 at 5:35
  • $\begingroup$ The most basic equation to remember is of course $$\left(\sigma(b^2) - b^2\right)\cdot{\sigma(2^r)} = \left(b^2 - 1\right).$$ $\endgroup$ – Jose Arnaldo Bebita Dris May 14 '16 at 5:41
  • $\begingroup$ I also have $\gcd(2^r, b^2) = 1$. $\endgroup$ – Jose Arnaldo Bebita Dris Jun 1 '16 at 19:33

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