Is there tight estimates for the following logarithmic summation ($\gamma,\gamma'\in(0,1)$ and $\mu,\mu'>0$)

$$\log_2\Bigg(\sum_{t=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum_{\ell=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum_{k=\frac t2-n^{\gamma'}\sqrt{\mu'\ln n}}^{\frac t2+n^{\gamma'}\sqrt{\mu'\ln n}}\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)?$$

I am hoping it might give $n - f(n)$ bound where $f(n)$ is $\omega(\ln n)$ or at least $\Omega(1)$ for diagonal case of $\gamma=\gamma'=\frac12$ at some $\mu,\mu'>0$.

Relevant problem is in Tight estimates for binomial summation (and perhaps an upper bound possible might be $$\log_2\Bigg(\underbrace{n^{2\gamma}\mu(\ln n)}_{\substack{\mbox{coming from}\\\mbox{outer two}\\\mbox{summations}}}\binom{n}{n/4}\Bigg)<n H(\frac14+\epsilon)<0.82n$$ at any $\epsilon>0$ since $t/2=n/4$ might be close to the value that yields the bound (together with loose upper bound from Vandermonde's identity for inner sum)).

  • $\begingroup$ what does at least $\gamma=\gamma'=1/2$ mean? $\min\{\gamma,\gamma'\}\geq 1/2$? $\endgroup$ – kodlu Jan 14 at 7:03
  • $\begingroup$ No just for the diagonal case at 1/2. $\endgroup$ – VS. Jan 14 at 8:08

This conjecture does not hold even in the case $\gamma=\gamma'=1/2$.

Indeed, consider the values of $\ell,t,k$ such that $$|\ell-n/2|\ll\sqrt n,\ |t-n/2|\ll\sqrt n,\ |k-t/2|\ll\sqrt n,$$ where $A\ll B$ or, equivalently, $B\gg A$ means that $|A|\le CB$ for some universal real constant $C>0$; as usual, $A\asymp B$ means that $A\ll B\ll A$. By what was shown in this answer, $$\binom\ell k\asymp\frac{2^\ell}{\sqrt\ell}\,e^{-u^2/2},$$ where $$u:=\frac{k-\ell/2}{\sqrt{\ell/2}}\ll1, $$ so that $$\binom\ell k\asymp\frac{2^\ell}{\sqrt n}.$$ Similarly, $$\binom{n-\ell}{t-k}\asymp\frac{2^{n-\ell}}{\sqrt n},$$ whence $$\binom\ell k\binom{n-\ell}{t-k}\asymp\frac{2^n}n.$$ Hence (in the case $\gamma=\gamma'=1/2$), your big triple sum is $\gg \dfrac{2^n}n\,n^{3/2}$ and hence $$\log_2(\text{the triple sum})-n\gg\ln n.$$

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  • $\begingroup$ What is the correct asymptotic? $\endgroup$ – VS. Jan 26 at 6:08
  • $\begingroup$ @VS. : I think the correct asymptotics for your $f(n)$ is $\asymp-\ln n$ if $\gamma=\gamma'=1/2$; otherwise, it will much depend on $\gamma$ and $\gamma'$. However, please ask any additional questions in separate posts. $\endgroup$ – Iosif Pinelis Jan 26 at 13:35

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