# Tight sublinear estimates for a triple partial binomial summation

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 $$\log_2\Bigg(\underbrace{n^{2\gamma}\mu(\ln n)}_{\substack{\mbox{coming from}\\\mbox{outer two}\\\mbox{summations}}}\binom{n}{n/4}\Bigg) at any $$\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)).

• what does at least $\gamma=\gamma'=1/2$ mean? $\min\{\gamma,\gamma'\}\geq 1/2$? – kodlu Jan 14 at 7:03
• No just for the diagonal case at 1/2. – 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 $$\log_2(\text{the triple sum})-n\gg\ln n.$$

• What is the correct asymptotic? – VS. Jan 26 at 6:08
• @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. – Iosif Pinelis Jan 26 at 13:35