Is there a tight bound on following binomial summations involving squares on arithmetic progressions?

Question

The summations of interest is following:

$$\sum_{i=0}^{\lfloor\sqrt n\rfloor}\binom{n}{i^2}$$ $$\sum_{i\in\{a,q+a,2q+a\dots,\lfloor\sqrt n\rfloor\}}\binom{n}{i^2}$$ where $q<n$ and $a\in\{0,1,\dots,q-1\}.$

Is there a tight asymptotic bound on these?

Answer 1

As stated under the OEIS entry $\texttt{A003099}$ mentioned in comments, the quantity $$f(n)=\frac{\sqrt n}{2^n}\sum_{k\geqslant 0}\binom{n}{k^2}$$ stays bounded, in $(\approx 0.587,\approx 0.827)$ for large $n$. More precisely, one obtains

For any $x\in\mathbb{R}$ we have $\displaystyle\color{blue}{\lim_{n\to\infty}f\big(\lfloor2(n+x)^2\rfloor\big)=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-4(m-x)^2}}$.

The idea is basically to put $k=n+m$ above, and use the following consequence $$\lim_{n\to\infty}\frac{\sqrt{2n(n+a)+b}}{2^{2n(n+a)+b}}\binom{2n(n+a)+b}{(n+m)^2}=\sqrt\frac2\pi e^{-(2m-a)^2}$$ of Stirling's formula. Thus, the exact values of the bounds are \begin{align*}\limsup_{n\to\infty}f(n)&=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-(2m)^2}&={0.827112271364145742192103543257}\cdots\\ \liminf_{n\to\infty}f(n)&=\sqrt\frac2\pi\sum_{m\in\mathbb{Z}}e^{-(2m-1)^2}&={0.587247586271786487501375771221}\cdots\end{align*}

I'm sure that the second (more general) sum in the question can be handled similarly.