Let $\sigma(n)$ be the divisor sum of $n$: $$ \sigma(n) = \sum_{d|n} d. $$ I was interested in the parity of $\sigma(n)$ and tried to check whether $\sigma(n)$ is unexpected often even for odd $n$ and vice versa. No result.

Thus I conjectured that $$ \lim_{x\longrightarrow\infty} \dfrac{|\{1:n\leq x,~ Parity(n) \not = Parity(\sigma(n)) \}|}{|\{1:n\leq x, ~Parity(n) = Parity(\sigma(n)) \}|} = 1.$$

I checked this fraction for $x$ up to $100000$ and the conjecture seems to be true. Here are some values:

x        Fraction
---      ---
10,000   1.02799
50,000   1.01264
100,000  1.00898

However I am not able to prove it. Can you prove that $\sigma(n)$ does not prefer any parity?

  • $\begingroup$ I would look at the formula for the sigma function (as referred by @FabianSchn ). It shouldn't be too hard from there. $\endgroup$ – marlasca23 Sep 21 '17 at 22:01
  • $\begingroup$ @marlasca23 Thanks, can you please elaborate how you would proceed? Of course I dont know any reason why $\sigma(n)$ should prefer some parity but how to prove it...? $\endgroup$ – Fabian Schn. Sep 21 '17 at 22:09

$\sigma(n)$ is a multiplicative function such that $$ \sigma(p^k) = 1+p+\ldots+p^k = \frac{p^{k+1}-1}{p-1} $$ for any prime $p$ and any $k\geq 1$. If $p=2$ we have that $\sigma(p^k)$ is certainly odd, while if $p>2$ the parity of $\sigma(p^k)$ only depends on the parity of $k$. In particular $\sigma(n)$ is odd iff $n$ is a number of the form $2^k(2m+1)^2$. The set of these numbers has asymptotic density zero (since in the interval $[1,M]$ there are less than $3\sqrt{M}$ numbers of this kind), hence the parity of $n$ and the parity of $\sigma(n)$ are essentially unrelated (loosely speaking, $\sigma(n)$ is almost surely even), as stated by your claim.

  • 1
    $\begingroup$ Thanks, great answer. $\endgroup$ – Fabian Schn. Sep 21 '17 at 22:39
  • $\begingroup$ @FabianSchn. Now $f(n) = (-1)^{1+\sum_{p^k \| n} \sigma(p^k)}$ is more interesting. It is multiplicative and equally distributed. $\sum_{n=1}^\infty f(n) n^{-s} = \frac{(1-2^{1-s})(1+2^{-s})}{1-2^{-s}}\frac{\zeta(2s)}{\zeta(s)}$ which is analytic for $\Re(s) \ge 1$ so that (with the same method as for the prime number theorem) $\sum_{n \le x} f(n) = o(x)$. The Riemann hypothesis is equivalent to $\sum_{n \le x} f(n) = O(\sqrt{x}\log^2 x)$ $\endgroup$ – reuns Sep 22 '17 at 7:31

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