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Not an answer but a simple heuristic argument: if you set $r=(ab)_p$, the OP's sum is equal to $$\sum_{1\le r\le p-1}\dfrac{1}{r}\sum_{1\le a\le p-1}\dfrac{1}{a}(ra^{-1})_p$$ This proves immediately t …

It is known since Gauss that $$r_3(n)=12\dfrac{h(D)}{w(D)/2}(1-(D/2))\sum_{d\mid f}\mu(d)(D/d)\sigma(f/d)\;,$$ where $-n=D(2^vf)^2$, $D$ a fundamental discriminant, $v\ge-1$, $h(D)$ is the class numbe …

The asymptotics of $W_k(x)$ for fixed real $x$ and $|k|\to\infty$ are given by $$W_k(x)=2k\pi i-\log(2k\pi i)+\log(x)+\dfrac{\log(2k\pi i)-\log(x)}{2k\pi i}+O(\log(|k|)^2/k^2)$$ (see for instance arXiv …

Since $\sum_{n\ge0}J_n(t)^2=1/2$ and $\sum_{n\ge2t}|J_n(t)|$ is negligible, applying Cauchy-Schwartz gives $S(t)=O(t^{1/2})$. Experimentally, I would guess that $S(t)=O(\log(t)^{5/2})$ or so. Also, in …

By induction, it is immediate that $a_{m,n}\le 2\min(m,n)$. In addition, when $m$ is fixed and $n>m$ we have $a_{m,n}=2m-P_m(n)/2^n$, where $P_m(x)$ is a polynomial of degree $m$, so $2m$ is an expone …

Partial answer (but the general case should be similar): using Stirling+Euler-MacLaurin+Glaisher's constant we have for $a=1$: $$\prod_{1\le j\le n}\binom{n}{j}\sim n^{-(n/2+1/3)}e^{n^2/2+n(1-\log(2\p …

Assuming the GRH for L-functions attached to modular forms, if $D$ is a cuspidal eigenform of weight $k$ with rational coefficients such as $\Delta$ and $\Lambda(s)=(2\pi)^{-s}\Gamma(s)L(s)$ the compl …

Not an answer but a conjectural answer for the value of $C(m)$ supported by extensive numerical evidence: $$C(m)=\dfrac{1}{m-1}\sum_{2\le k\le m-1}\binom{m-1}{k}\zeta(k)+\sum_{2\le k\le m-2}\dfrac{1}{ …