Does anyone know of the mean value of two Ramanujan Sums when summed over the square of integers?

In my research on the Landau problem regarding nearly square primes, I have run into the mean value of a product of Ramanujan Sums: \begin{equation*} \lim\limits_{N \to \infty} \frac{1}{N} \sum\limits_{n =1}^{N} c_r\left( n^2 \right) c_s\left( n^2 \right) \end{equation*}

Does anyone know of a listing of properties of Ramanujan Sums similar to the this one, but with summations taken over the squares of positive numbers instead summation over the integers?

The closest I can find is the function, $E_G$ found in Sums of products of Ramanujan sums by Toth. I am interested in the particular case where $g_1(n) = n^2$ though and not general case case represented by $E_G$ for and arbitrary set of functions $g_i(n)$

  • $\begingroup$ I have now calculated the mean value in general, see the Added section. $\endgroup$ – GH from MO Sep 14 '15 at 22:04

Here is a partial answer (Added: completed below). Assuming $(r,s)=1$ and denoting $q=rs$, we have $$ \sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) = \sum_{n=1}^N c_q\left( n^2 \right) = \sum_{n=1}^N \sum_{d\mid(q,n^2)}\mu\left(\frac{q}{d}\right)d = \sum_{d\mid q}\mu\left(\frac{q}{d}\right)d\sum_{\substack{{1\leq n\leq N}\\{d\mid n^2}}} 1. $$ Now let $f(d)$ be the number of residue classes modulo $d$ whose square is zero modulo $d$. Then $$ \sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) = \sum_{d\mid q}\mu\left(\frac{q}{d}\right)d f(d)\left(\frac{N}{d}+O(1)\right) = N \sum_{d\mid q}\mu\left(\frac{q}{d}\right) f(d) + O_q(1),$$ whence $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N c_r\left( n^2 \right) c_s\left( n^2 \right) = \sum_{d\mid q}\mu\left(\frac{q}{d}\right) f(d).$$ The right hand side is the multiplicative convolution $g:=\mu\ast f$ evaluated at $q$, in particular it is multiplicative in $q$. Hence it suffices to determine $g$ at prime powers $p^k$, which is straightforward: $$ g(p^k) = f(p^k)-f(p^{k-1}) = p^{\lfloor\frac{k}{2}\rfloor}-p^{\lfloor\frac{k-1}{2}\rfloor}.$$ We infer that the sought mean value equals $$ g(q) = \begin{cases}\phi(\sqrt{q}),&q=\square\\0,&q\neq\square\end{cases} \tag{$\ast$}$$ Note that under our initial assumptions, $q$ is a square if and only if both $r$ and $s$ are squares.

Added. In the general case, i.e. without the coprimality condition on $r$ and $s$, we can proceed as follows. The function $n\mapsto c_r(n^2)c_s(n^2)$ is periodic mod $[r,s]$, hence the mean value exists and equals $$ h(r,s):=\frac{1}{[r,s]}\sum_{n=1}^{[r,s]}c_r(n^2)c_s(n^2). $$ This function satisfies the multiplicativity relation $$ h(rr',ss') = h(r,s)h(r',s'),\qquad (rs,r's')=1, $$ as follows from the Chinese Remainder Theorem and the invariance property $$ c_q(m)=c_q(km),\qquad (k,q)=1. $$ Therefore, $$ h(r,s) = \prod_{p\mid rs}h(p^{v_p(r)},p^{v_p(s)}), $$ so that it suffices to evaluate $h(p^k,p^l)$ for any prime $p$ and any exponents $k,l\geq 0$. This is straightforward, given the explicit formulae for $c_{p^k}(m)$ here. In particular, for $l>k\geq 0$ we obtain $$ h(p^k,p^l)=\phi(p^k)h(1,p^l) =\begin{cases}\phi(p^k)\phi(p^{l/2}),&\text{$l$ even}\\0,&\text{$l$ odd}\end{cases}$$ which generalizes the explicit formula ($\ast$) above.

  • $\begingroup$ how do you make the jump: $\sum\limits_{n =1}^{N} c_r\left( n^2 \right) c_s\left( n^2 \right) = \sum\limits_{n =1}^{N} c_q\left( n^2 \right)$ when $r \ne s$ $\endgroup$ – John Washburn Sep 14 '15 at 18:26
  • $\begingroup$ I assume $(r,s)=1$, not just $r\neq s$. Then, $c_r(m)c_s(m)=c_{rs}(m)$ for all $m$. $\endgroup$ – GH from MO Sep 14 '15 at 18:28
  • $\begingroup$ I am not guaranteed (r,s)=1. The outer summations which I left off are: $r=1 \to \infty$ and $s=1 \to \infty$ $\endgroup$ – John Washburn Sep 14 '15 at 18:36
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    $\begingroup$ Computationally, I got the same result, \begin{align*} \sum\limits_{n =1}^{N} c_r\left( n^2 \right) c_s\left( n^2 \right) &= \begin{cases} \phi\left( r \right) & r = s \\ 0 & r \ne s \end{cases} \end{align*} but now I can prove it. Thanks. $\endgroup$ – John Washburn Sep 14 '15 at 18:40
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    $\begingroup$ @JohnWashburn: I don't agree that the mean is zero when $(r,s)>1$. For example, if $r=p$ and $s=p^2$, where $p$ is a prime, then the mean value is $(p-1)^2$. This is because $c_p(n^2)c_{p^2}(n^2)$ equals zero for $p\nmid n$ and equals $\phi(p)\phi(p^2)$ when $p\mid n$. $\endgroup$ – GH from MO Sep 14 '15 at 20:44

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