Let $p\in(0,1)$ be any constant (if that helps or is necessary, $p=1/2$ is what I care the most about), and fix $C > 0$.
What can we say about the total variation distance between $\operatorname{Binomial}(n,p)$ and $\operatorname{Binomial}(n+C\lfloor\sqrt{n}\rfloor, p)$? That is, how does the quantity $$ \delta_{p,C}(n)\stackrel{\rm def}{=}\frac{1}{2}\sum_{k=0}^{n+C\lfloor\sqrt{n}\rfloor}\left\lvert \binom{n}{k} p^k(1-p)^{n-k}-\binom{n+C\lfloor\sqrt{n}\rfloor}{k} p^k(1-p)^{n+C\lfloor\sqrt{n}\rfloor-k} \right\rvert $$ behaves, when $n\to\infty$?
From simulations, it seems to converge to some value $\ell=\ell(p,C)\in(0,1)$; which I suppose would make sense "in hindsight" as the mode is shifted by roughly $\sqrt{n}$, while a constant fraction of the probability weight is roughly uniformly spread in an interval of size $\approx\sqrt{n}$ centered around the mode.
However, I do not trust my coding skills enough to be convinced by simulations, and in any case experiments are no proof. And trying to expand and analyze $\delta_{p,C}$ by massaging or poking the expression above just has lead, so far, to cumbersome and ugly-looking expressions.
Is there a simple (hopefully not brute-force-computation) argument to study the asymptotic behavior of $\delta_{p,C}$ when $n\to\infty$?
(if this helps in a first time, $C=1$ is fine, and restricting to $n=k^2$ for some $k$ to avoid messy floors is definitely OK).
Closing this question: I will not give a full answer, but basically there is at least a sort-of-straightforward approach to give rather good bounds on this quantity. Namely,
Another option is to use known approximations of the tails of the Binomials (for their CDF's $F_1,F_2$), and simply see what value of $x$ maximizes (the approximation of) $\lvert F_1(x)-F_2(x)\rvert$.
Finally, a third option (messier) to get a lower bound is to tackle the quantity above by restricting the sum to the "indices that matter," that is within a constant factor of standard deviations from the means (as the Binomials are really tighly concentrated); and then to lower bound the remaining summands by some ather horrendous Taylor approximation of the binomial coefficients (writing $k=\alpha\sqrt{n}$ for say $\alpha\in[-1/2,1/2]$), since that's the range of indices that are left)