Strange shape of the distribution of the sum of the binomial coefficients ${n\choose r^2}$over squares Update: Initially the question was posted for $a = 1$. Now it has been generalized for any real $a > 0$
What is known about the distribution of the sum of the binomial coefficients over multiple of squares? My experimental data seems to suggest that for a given positive real  $a > 0$
$$
s_{n,a} = 
\sum_{1\leq \lfloor ak^2 \rfloor\leq n}{n\choose \lfloor ak^2 \rfloor}=
{n\choose \lfloor 1^2 a \rfloor} + {n\choose \lfloor 2^2 a \rfloor}  + \cdots + {n\choose \lfloor r^2 a \rfloor} \approx \frac{2^n}{\sqrt{2an}}
$$
Clearly the sum will be dominated by the term closest to the central binomial coefficient which in this case is the square nearest to $n/2$. What I found interesting is the shape of the histogram of the distribution of the ratios of the actual sum to its asymptotic estimate i.e. $\dfrac{s_n \sqrt{2an}}{2^n}$ are similar for all $a$ and look like an acr-sine distribution as mentioned in the comments.
Histogram of distribution for $a = 1$

Question 1: Why does it have an arc-sine like distribution?
Question 2 Where does the spikes occur? E.g. for $a = 1$, the spikes occur roughly at $1 \pm 1/6$.
Related question: What is the sum of the binomial coefficients $n \choose p$ over prime numbers?
 A: This is a part of ongoing research with Nilotpal Kanti Sinha. The answer does not explain fully why we have such distribution, but it will help understanding the distribution of the values of $\frac{s_n{\sqrt{2an}}}{2^n}$. I used $S_{n,a}$ notation to include its dependence on $a$. 
Theorem 1

Let $\beta_n=\beta_{n,a}=\frac n2 - a(\lfloor \sqrt{\frac n{2a}}\rfloor)^2$. Then
  $$
\frac{S_{n,a}\sqrt{2an}}{2^n}=\frac{2\sqrt a}{\sqrt{\pi}} \sum_{k\in\mathbb{Z}} \exp\left\{-2\left(\frac{\beta_n}{\sqrt n}+k\sqrt{2a}\right)^2\right\}+O\left(\frac{\log^3 n}{\sqrt n}\right).
$$

Corollary 1

Let $x_n=x_{n,a}=\beta_n/(\sqrt{2an})$. Then
  $$
\frac{S_{n,a}\sqrt{2an}}{2^n}=1+\sum_{k=1}^{\infty}2 e^{-\frac{\pi^2}{4a} k^2} \cos2\pi k x_n+O\left(\frac{\log^3 n}{\sqrt n}\right). $$

The first harmonic gives a good approximation of the magnitude of the oscillation $1\pm 2e^{-\frac{\pi^2}{4a}}$.
Proof of Theorem 1
Lemma 1

Let $g(n)$ be a function satisfying $|g(n)|\leq 6\log n$. Then as $n\rightarrow\infty$,
  $$
\frac1{2^n}\binom n{\left\lfloor\frac n2 + g(n)\sqrt n\right\rfloor}=\frac2{\sqrt{2\pi n}} e^{-2(g(n))^2}\left(1+O\left(\frac{\log^3 n}{\sqrt n}\right)\right).
$$

Lemma 2[Hoeffding's Inequality]

Let $h>0$ and $B_n=\{k\leq n \ | \ \left|k-\frac n2\right|\geq h\sqrt n\}$. Then
  $$
\frac1{2^n}\sum_{k\in B_n} \binom nk \leq 2e^{-2h^2}.
$$

We apply Lemma 1 and Lemma 2 with $h=5\log n$. Let $g_0(n)=\frac{\beta_n}{\sqrt n}$ and $g_k(n)=\frac{\frac n2-\left\lfloor a\left(\left\lfloor \sqrt{\frac{n}{2a}}\right\rfloor +k\right)^2\right\rfloor}{\sqrt n}$ for $k\in\mathbb{Z}$ and $|k|\leq \frac6a \log n$.
$$
\frac{S_{n,a}\sqrt{2an}}{2^n}=\sum_{|k|\leq \frac6a\log n} \frac{2\sqrt{a}}{\sqrt{\pi}} e^{-2(g_k(n))^2}\left(1+O\left(\frac{\log^3 n}{\sqrt n}\right)\right)+O\left(e^{-(\log n)^2} \right).
$$
Since $g_k(n)=\frac{\beta_n}{\sqrt n }- k\sqrt{2a} + O\left(\frac{k^2}{\sqrt n}\right) = \frac{\beta_n}{\sqrt n}-k\sqrt{2a}+O\left(\frac{\log^2 n}{\sqrt n}\right)$, we have by the mean value theorem,
$$
e^{-2(g_k(n))^2} = e^{-2\left(\frac{\beta_n}{\sqrt n} - k\sqrt{2a}\right)^2} + O\left(\frac{\log^2 n}{\sqrt n} ke^{-ak^2} \right).
$$
Summing these over $k$ and applying Lemma 2 again gives
\begin{align*}
\frac{S_{n,a}\sqrt{2an}}{2^n}&=\sum_{|k|\leq \frac 6a\log n} \frac{2\sqrt a}{\sqrt{\pi}} \exp\left\{-2\left(\frac{\beta_n}{\sqrt n} -k\sqrt{2a}\right)^2\right\}+O\left( \frac{\log^3 n}{\sqrt n}\right)\\
&=\frac{2\sqrt a}{\sqrt{\pi}}\sum_{k\in\mathbb{Z}} \exp\left\{-2\left(\frac{\beta_n}{\sqrt n} +k\sqrt{2a}\right)^2\right\}+O\left( \frac{\log^3 n}{\sqrt n}\right).
\end{align*}
Hence, Theorem 1 follows.
Let $c>0$ and define
$$
f(t)=e^{-\pi\left(\frac{x+t}c\right)^2}.
$$
Then its Fourier transform $\hat{f}(\xi)$ is
$$
\hat{f}(\xi)=\int_{-\infty}^{\infty} f(t)e^{-2\pi i t \xi} \ dt = ce^{2\pi i x\xi}e^{-\pi(c\xi)^2}.
$$
Applying Poisson summation formula, we obtain for any $x\in\mathbb{R}$,
$$
\sum_{k\in \mathbb{Z}} e^{-\pi\left(\frac{x+k}c\right)^2}=c\sum_{k\in\mathbb{Z}} e^{2\pi i xk}e^{-\pi(ck)^2}=c\left(1+\sum_{k=1}^{\infty} 2e^{-\pi(ck)^2} \cos 2\pi kx\right).
$$
Putting $c=\frac{\sqrt{\pi}}{2\sqrt a}$, Corollary 1 follows.
