Sum of divisors

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#1 h Apr 18, 2025, 2:47 PM

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Does there exist $M>0$ , such that $\forall m>M$ , there exists an integer $n$ satisfying $\sigma(n)=m$ ?
$\sigma(n)=$ the sum of all positive divisors of $n$ .

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#2 h Apr 27, 2025, 10:17 PM

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No. For example, take $m=q$ as a large prime. Then by multiplicativity of $\sigma(n)$ , the only chance for $\sigma(n)=q$ would be that $n=p^k$ is a prime power, so $q=p^k+p^{k-1}+\dots+p+1$ for some (even) $k \ge 2$ .
But it is easy to see that there are only roughly $\sqrt{X}$ numbers of the form $x^k+x^{k-1}+\dots+x+1$ up to $X$ , while of course there are roughly $X/log(X)$ many choices for $q$ , so "most" $q$ are not of the form $\sigma(n)$ .

In fact, a slight modification of this argument shows that "most" numbers are not values of $\sigma(n)$ in the sense that the set of such values has density zero in the natural numbers.

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