A special case of a well known result of Ingham is that

$$\sum_{n\leq x} d(n)d(n+1)=\frac{6}{\pi^2}x(\log x)^2+O(x\log x)$$

where $d(n)$ is the number of divisors of $n$.

Ingham's results, which are arrived at by a long but elementary calculation, may be combined with a further long and non-elementary (i.e. involving some complex analysis) calculation, but which does not use any properties of non-principal characters or $L$-functions, to arrive at the conclusion that the following average over all $q$ less than $x $ of all non-principal character sums:

$$\frac{1}{x}\sum_{q\leq x}\frac{1}{\phi(q)}\sum_{\chi\neq \chi_0}\left(\sum_{n\leq x}\chi(n)d(n)\right)=O(\log x).$$

In fact, it can be shown that this statement is also sufficient for the truth of Ingham's result.

It seems to me that statements like this would have more direct proofs, perhaps analytical using known or hypothetical properties of $L^2(s,\chi)$, or perhaps using properties of character sums. Is this so?

  • $\begingroup$ Well, one has (1) sharp estimates for the number of n less than x such that d(n) ~ \log^c(x) for reasonable values of c; (2) the large sieve inequality which should upper bound the left-hand side of the character sum with the summand squared (with an additional weight of q, which is harmless on dyadic scales). One can make maximal use of this information by applying the large sieve inequality to the sum restricted to n s.t. d(n) ~ log^c(x) for each value of c. Is the resulting distributional information alone insufficient? (The calculation is a bit too involved for me to check at the moment.) $\endgroup$ – Mark Lewko Mar 15 '17 at 10:21
  • $\begingroup$ Are you missing a power of x from the right hand side? If we replace the characters in the summand with random signs/unimodular numbers the summand would be about $(x \log^3 x )^{1/2}$ and the outer sums are just averages, so this is suggesting much more cancelation than in the random model. Perhaps I'm missing something? $\endgroup$ – Mark Lewko Mar 15 '17 at 21:48
  • $\begingroup$ Hello Mark, thanks - no, the statement is as it appears. Btw I've been thinking about your earlier comment/question but I don't know how to use the distributional information to arrive at the conclusion. I don't see how to use the large sieve here either, but I think this happening because of cancelation between the character sums. I still can't find a direct proof though and, from the lack of comments/answers, I suppose there isn't an obvious one. $\endgroup$ – Kevin Smith Mar 15 '17 at 22:03
  • $\begingroup$ Thanks. The arithmetic large sieve (at least as I was proposing to use it) wont pick up cancelations between moduli so, given my prior comment, that can't work. I'm curious now if there is any elementary argument that gives a poly-log estimate? $\endgroup$ – Mark Lewko Mar 15 '17 at 22:16
  • $\begingroup$ Oh...interchanging the two inner most sums, gives nearly complete sums for fixed n, and is the source of the extreme cancelation. It seems, though, that this quickly reduces matters back to a non-oscillatory number theoretic quantity... $\endgroup$ – Mark Lewko Mar 15 '17 at 22:37

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