9
$\begingroup$

Let $\sigma_0(n)$ be the divisor counting function: $$\sigma_0(n) = \sum_{d \vert n} 1.$$ I ran some numerical experiments that showed when $p$ is prime, the function $\sigma_0(n)$ is equidistributed mod $p$. That is, for any residue class $a \mod p$, $$\lim_{X \to \infty} \dfrac{ \vert \{ n<X: \sigma_0(n) \equiv a \mod p \} \vert }{X} = \dfrac{1}{p}.$$

Is this fact correct? If so, could anyone sketch a proof / provide a reference for a proof?

$\endgroup$
2
  • 14
    $\begingroup$ Seems unlikely because the product formula suggests that a residue of 0 mod p should not be as likely as nonzero residues. For starters if p=2 the only way to get an odd value is if n is a square, so asymptotically a=0 happens 100% of the time. I could believe that all nonzero residues mod p are equally likely. $\endgroup$ Apr 7, 2023 at 21:43
  • 5
    $\begingroup$ Anurag gave an excellent answer but it's worth pointing out that a more general theorem on the distribution of $\sigma_0(n)$ in coprime residue classes was given by Narkiewicz in: On distribution of values of multiplicative functions in residue classes, Acta Arithmetica 7 (1967), vol. 12, issue 3, pp. 269--279. $\endgroup$ Apr 8, 2023 at 16:12

1 Answer 1

17
$\begingroup$

$\newcommand{\Y}{\mathfrak{X}_p(X)}$I haven't checked all the details on the application of Selberg–Delange below, so it's possible something I am saying is nonsense, but even after accounting for the correction noted by Noam Elkies in the comments, it appears to be the case that $\sigma_0(n)$ is not, in general, equidistributed among the non-zero congruence classes mod $p$ unless $2$ is a primitive root modulo $p$ (see A001122 on OEIS). In fact, we have that $$\lim_{X \to \infty}\biggl(\frac{1}{\Y}\sum_{n\leqslant X} 1_{\sigma_0(n) \equiv a \mod p}\biggr) = \frac{1}{p-1}\biggl(1+\frac{1}{\delta_p}\sum_{\substack{\chi \neq \chi_0\\\chi(2) = 1}} \overline{\chi}(a)G_{\chi}(1)\biggr),$$ where the sum runs over all nontrivial Dirichlet character mod $p$ with $\chi(2) = 1$, $$\Y = \sum_{n\leqslant X} 1_{p \nmid \sigma_0(n)}, \qquad \delta_p = \lim_{p \to \infty} \frac{\Y}{X},$$ are the counting function and density of the set of $n$ with $p \nmid \sigma_0(n)$ and $G_\chi$ is defined below. Note if $n$ is squarefree, then $\sigma_0(n) = 2^{\omega(n)}$, where $\omega(n)$ is the prime-divisor counting function, and hence $p \nmid \sigma_0(n)$ for $p > 2$. Since the squarefrees have positive density, this implies that $\Y \gg X$ and $\delta_p > 0$.

When $2$ is a primitive root modulo $p$, the only character with $\chi(2) = 1$ is the trivial one, so this would imply equidistribution in that case. But, for example, if you take $p = 7$ or $p = 17$, and consult the table of values of Dirichlet characters mod $p$, then you'll find that the only characters which have $\chi(2) = 1$ are the trivial character and the quadratic character; from this it follows that $\sigma_0(n)$ is biased towards being a quadratic residue over being a nonresidue. The number of characters that will occur in this sum for a given prime $p$ is $1$ less than A001917 on OEIS.

Roughly speaking, this is because $\sigma_0(p) = 2$, and hence the characters which have $\chi(2) = 1$ give a main term contribution. Here's a sketch:

Guided by Weyl's criterion for $(\mathbb{Z}/p\mathbb{Z})^\times$, suppose $(a,p)=1$, and note that orthogonality of Dirichlet characters reads $$1_{n \equiv a \mod p} = \frac{1}{p-1}\sum_{\chi} \overline{\chi}(a) \chi(n),$$ where the sum is over all Dirichlet characters mod $p$. Thus, $$\frac{1}{\Y}\sum_{n\leqslant X} 1_{\sigma_0(n) \equiv a \mod p} = \frac{1}{p-1} + \frac{1}{p-1}\sum_{\chi\neq \chi_0} \overline{\chi}(a)\left( \frac{1}{\Y} \sum_{n\leqslant X} \chi(\sigma_0(n)) \right), \tag{$\star$}\label{star}$$ where we have separated the contribution of the trivial character. It thus suffices to study $$M_\chi(X) = \sum_{n\leqslant X} \chi(\sigma_0(n)), $$ which is a mean-value of multiplicative function. The characters which satisfy $M_\chi(X) = o(X)$ do not contribute a main term to \eqref{star} since $\Y \gg X$, while the characters with $M_\chi(X) \asymp X$ do.

Standard multiplicative number theory techniques apply here. To put this into effect, define the Dirichlet series $$F_\chi(s) = \sum_{n\geqslant 1} \frac{\chi(\sigma_0(n))}{n^s},$$ so that an application of Perron's formula gives that $$\sum_{n\leqslant X} \chi(\sigma_0(n)) = \frac{1}{2\pi i} \int_{2-i\infty}^{2+i\infty} F_\chi(s) X^s \frac{ds}{s}.$$ Investigating the Euler product of $F_\chi$, we find $$F_\chi(s) = \prod_p \biggl(\sum_{k=0}^\infty \frac{\chi(\sigma_0(p^k))}{p^{ks}}\biggr) = \prod_p \biggl(\sum_{k=0}^\infty \frac{\chi(k+1)}{p^{ks}}\biggr) = \zeta(s)^{\chi(2)} G_\chi(s),$$ where $G_\chi(s)$ is convergent in $\Re(s) > 1/2$. An application of the Selberg-Delange method (see Chapter 5 of Tenenbaum's "Introduction to analytic and probabilistic number theory", in particular Theorem 5.2 with $z = \chi(2)$, $N = 1$ and $F = F_\chi$) should then give that $$ \sum_{n\leqslant X} \chi(\sigma_0(n)) = \frac{X (\log X)^{\chi(2) - 1}}{\Gamma(\chi(2))} \biggl(G_\chi(1) + O\Big(\frac{1}{\log X}\Big)\biggr),$$ from which it is clear that if $\chi(2) \neq 1$, then the expression on the right is $o(X)$ while, if $\chi(2) = 1$, then we get something of size $\asymp X$. Putting this back into \eqref{star}, and concentrating on the main terms proves the claim.

$\endgroup$
2
  • 1
    $\begingroup$ thanks so much - this is a wonderful answer! I'll have to go through your argument in more detail, but I appreciate your help :) $\endgroup$ Apr 8, 2023 at 2:17
  • $\begingroup$ Glad it was useful! I found an error already, I forgot to actually take into account the fact that the sums won't have terms when $p \nmid \sigma_0(n)$. I've fixed it now. Hopefully, even if there's an error in the application of Selberg-Delange, this points you in the right direction. $\endgroup$ Apr 8, 2023 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.