Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. If $\sigma(y) < 2y$, $y$ is called deficient; if $\sigma(z) > 2z$, $z$ is called abundant.


(1) Can $\sigma(2^r)$ be abundant for some integer $r > 1$?

(2) If so, what conditions on $r$ guarantee abundance?

My Attempt

If $\sigma(2^r) = 2^{r + 1} - 1$ is a Mersenne prime, then it is deficient. Hence we consider the case when $\sigma(2^r) = 2^{r + 1} - 1$ is a composite Mersenne number.

I found this related question in MSE. It essentially states the following proposition:

Proposition A. If $n = {2^{m-1}}(2^m - 1)$, where $m$ is a positive integer such that $2^m - 1$ is composite, then $n$ is abundant.

Suppose to the contrary that $\sigma(2^r) = 2^{r+1} - 1$ is a deficient composite Mersenne number. Since $2^{r - 1}$ is also deficient, then subject to certain constraints on the abundancy index $$I\left(\sigma(2^r)\right)=\dfrac{\sigma\left(\sigma(2^r)\right)}{\sigma(2^r)},$$ (such as a suitable upper bound), then it might be possible to show that $$I(k) = I\left({2^{r-1}}\sigma(2^r)\right) < 2,$$ contradicting Proposition A.


In Retrospect

However, I do realize that this argument is not rigorous, and may in fact be fallacious. This explains my second question.

  • 1
    $\begingroup$ What about $r = 11$? $\sigma(2^{11}) = 2^{12}-1 = 4095 = 3^2\cdot 5 \cdot 7 \cdot 13$, and $\sigma(\sigma(2^{11})) = \sigma(3^2)\sigma(5)\sigma(7)\sigma(13) = 13\cdot 6\cdot 8 \cdot 14 = 8736 > 2\cdot 4095$. $\endgroup$ Jun 1, 2016 at 18:52
  • $\begingroup$ @DanielFischer, yes that is correct. I guess my main question would be: What condition(s) characterize(s) such integers $r$? $\endgroup$ Jun 1, 2016 at 18:57

1 Answer 1


Since integer $r > 1$ implies that $r - 1 \geq 1$, and since $$2 \mid 2^{r - 1}$$ and $$\gcd(2^{r - 1}, \sigma(2^r)) = 1,$$ then a condition on $r$ that guarantees abundance (by ensuring a contradiction against Proposition A) follows from the fact that the abundancy index is weakly multiplicative: $$\dfrac{3}{2} \cdot I\left(\sigma(2^r)\right) \leq I(2^{r-1})I\left(\sigma(2^r)\right) = I\left(2^{r-1}\sigma(2^r)\right) = I(k) < 2$$ so that $$I\left(\sigma(2^r)\right) < \dfrac{4}{3}.$$ This condition on $r$ will ensure a deficient $k = {2^{r-1}}\sigma(2^r)$ thereby contradicting Proposition A.

Consequently, the condition $$I\left(\sigma(2^r)\right) > \dfrac{4}{3}$$ is necessary for $\sigma(2^r)$ to be abundant. (Note that $$I\left(\sigma(2^r)\right) \neq \dfrac{4}{3}$$ since it is assumed that $r \neq 1$.)

Added June 2 2016

Is the condition $$I\left(\sigma(2^r)\right) > \dfrac{4}{3}$$ also sufficient for $\sigma(2^r)$ to be abundant?


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