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$\mu=1+\epsilon$ where $\epsilon>0$ holds.

1.Is there a good bound for $$T=\frac{\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}^2}{2^n}?$$

This quantity can be interpreted as $$\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}\mathbb P(\frac n2+i)$$ where $\mathbb P(\frac n2+i)$ is under bionmial distribution and thus has probability $\frac{\binom{n}{\frac n2 +i}}{2^n}$ which is 'trucated expected value of $\binom{n}{\frac n2 +i}$'.

Computing few values suggests $\log_2T< n-\log_2\sqrt{\mu n\ln n}$ at $\mu\rightarrow1^+$.

For example at $n=1000$ to $50000$ gives such margin.

  1. How large can $\mu$ be for this $\log_2T< n-\log_2\sqrt{\mu n\ln n}$ bound to hold up?

Naively I can get $$<\frac{2^n}{\sqrt{n\pi/2}}(1-o(1))$$ by using $$2^nT<\binom{n}{\frac n2}\sum_{i=-\sqrt{\mu n\ln n}}^{\sqrt{\mu n\ln n}}\binom{n}{\frac n2 +i}.$$

I also know we can prove $$\binom{n}{\frac n2 +i}\asymp\frac{2^{nH(\frac12+\frac in)}}{\sqrt{n\pi/2}}$$ approximation.

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    $\begingroup$ I don't know if $\mu$ is something special or just a parameter, but it is odd that increasing $\mu$ makes your sums larger and your conjectured answers smaller. $\endgroup$ – Brendan McKay Jan 18 at 10:44
  • $\begingroup$ $\mu=1+\epsilon$ at $\epsilon>0$. That is correct. Perhaps then the relation is flawed. On the $\log_2$ scale I see $\ll n-\log_2 f(n)$ where $f(n)$ seems to be $\Omega(\sqrt{\mu n\ln n})$ when $\mu=1.001$. $\endgroup$ – VS. Jan 18 at 10:57
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    $\begingroup$ The normal approximation of the binomials will give you accurate values. $\endgroup$ – Brendan McKay Jan 18 at 11:12
  • $\begingroup$ can you explain the n-log_2f(n) and what f(n) should i expect? $\endgroup$ – VS. Jan 18 at 11:15
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Your main conjecture is not quite correct.

Indeed, for each natural $j$, let $B_j$ be a random variable (r.v.) with the binomial distribution with parameters $j$ and $1/2$, and let $C_j$ be an independent copy of $B_j$. Let also $u:=\sqrt{\mu n\ln n}$. Then for all even natural $n$ $$T/2^n=U_n:=\sum_{k\colon\,|k-n/2|<u}P(B_n=k,C_n=n-k) =\sum_{k=0}^nP(B_n=k,C_n=n-k)-R_n=P(B_{2n}=n)-R_n, $$ where $$R_n:=\sum_{k\colon\, n/2\ge|k-n/2|\ge u}P(B_n=k,C_n=n-k)\le P(B_n\ge n/2+u)^2 \le e^{-4u^2/n}=1/n^{4\mu^2}, $$ where, in turn, the latter inequality is an instance of an exponential Hoeffding inequality. On the other hand, by Stirling's formula, $$P(B_{2n}=n)\sim1/\sqrt{\pi n}$$ (the asymptotics everywhere here are as $n\to\infty$). So, $$T_n\sim2^n/\sqrt{\pi n}$$ and hence $$\log_2T=n-\log_2\sqrt{(1+o(1))\pi n}.$$

So, the inequality $\log_2T<n-\log_2\sqrt{\mu n\ln n}$ does not hold for any real $\mu\ge1$ if $n$ is large enough.

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