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Let $\sigma_0(n)$ be the divisor counting function $$\sigma_0(n) = \sum_{d \vert n} 1.$$ I'm interested in the convolution sum $$ S(n) := \sum_{k=1}^{n-1} \sigma_0(k) \sigma_0(n-k)$$ I ran some quick numerical experiments and it looks like for odd primes $p$, the convolution sum $S(n)$ is equidistributed* mod $p$: $$\lim_{X \to \infty} \dfrac{\lvert \{ n<X: S(n) \equiv a \mod p \} \rvert}{X} = \dfrac{1}{p}.$$ Is this true? And if so, could anyone sketch a proof / provide a reference?

*This is a bit of a lie. In fact, the data seems to suggest that certain residue classes occur more often than others, see this question. But I've stated my observation in this (incorrect) way to avoid making the question too long. In any case, I'd like to learn of any techniques I could use to get a handle of $S(n)$ mod $p$.

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  • $\begingroup$ You probably want to use additive characters mod š¯‘¯ here instead of multiplicative characters. Do you mind saying a bit about your motivation for asking this question? Is it just curiosity, or did it arise from some other considerations? $\endgroup$
    – Anurag Sahay
    Commented Apr 8, 2023 at 22:37
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    $\begingroup$ Sure, here's my motivation: I'm studying a product of two Eisenstein series. The Fourier coefficients of Eisenstein series are divisor functions, and it turns out the Fourier coefficients of the product of Eisenstein series is a convolution of divisor functions. I'm interested in how the Fourier coefficients of this product of Eisenstein series are distributed mod $p$, which leads me to ask how this convolution of divisor sums is distributed mod $p$. $\endgroup$
    – Adithya Chakravarthy
    Commented Apr 8, 2023 at 22:44
  • $\begingroup$ This definition apparently involves $\sigma_0(0)$. Is that right? $\endgroup$
    – colt_browning
    Commented Apr 9, 2023 at 13:51
  • $\begingroup$ @colt_browning Nope, I don't think it involves $\sigma_0(0)$. The summation over $k$ in the definition of $S(n)$ is for $k=1,\dots,n$. $\endgroup$
    – Adithya Chakravarthy
    Commented Apr 9, 2023 at 13:53
  • $\begingroup$ @AdithyaChakravarthy Yes, and the 2nd factor is $\sigma_0(n-k)$, which becomes $\sigma_0(0)$ for $k=n$. $\endgroup$
    – colt_browning
    Commented Apr 9, 2023 at 13:57

2 Answers 2

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Too long for a comment

Maybe this helps though it has $\sigma(n)=\sigma_1(n)$ not $\sigma_0(n)$. It may be that a similar identity may hold for $\sigma_0(n).$ In a long paper available here we find

the convolution sum of the divisor function: $\sum_{k=1}^{n-1} \sigma_1(k)\sigma_1(n-k),$ which was presented by Besge 1. After that, convolution sums for such divisor functions have become the subject of interest to many mathematicians.

According to this paper the sum evaluated by Besge satisfies $$ \sum_{k=1}^{n-1} \sigma_1(k)\sigma_1(n-k)=\frac{5}{12}\sigma_3(n)+\left(\frac{1}{12}-\frac{n}{2} \right)\sigma_1(n) $$

There is much more in that first paper I linked to that may be relevant.

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EDITED ANSWER. I'm now quite sure that this is true for every odd prime $p$. Using the list available on the OEIS (A055507), it appears that when you perform a $\chi^2$ test, for each prime $p=3,5,7,13,17,19,23,29$ and $31$, the finite sequence $\{S(n): n\le10000\}$ is uniformly distributed on $\{0,1,\dots,p-1\}$. The $p$-values are always largely above $.05$. In particular there is a quite strong statistical evidence that there are no congruence classes that appear significantly more often than others.

However, the same analysis for the prime $11$ surprisingly returns a $p$-value $p=0.005$, but as, H. Cohen remarked (see comment below), this apparent non-uniformity disappears, when one considers $n\le 10^7$.

enter image description here

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  • $\begingroup$ I performed furhter analyses and I found that equidistribution holds for all $p\le109$ (the greatest I tested), except the case $p=97$ for which the $p$-value is .008. $\endgroup$
    – G. Melfi
    Commented Oct 24, 2023 at 15:15
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    $\begingroup$ You are correct up to 10000: a=5 mod 11 occurs much less than the others (826 compared to an average of 909, and a=9 and a=10 also are small). However if you go further (I went to $10^7$) this disappears. Law of small numbers I guess. $\endgroup$
    – Henri Cohen
    Commented Oct 24, 2023 at 18:05
  • $\begingroup$ Quite interesting how it takes a while for equidistribution to appear. I still have very little idea how to prove this, but thanks for the data. $\endgroup$
    – Adithya Chakravarthy
    Commented Oct 25, 2023 at 15:11

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