Sum-regular $\{0,1\}$-matrices Let $n\in\mathbb{N}$ be a positive integer. We say that an $n\times n$-matrix $A$ with all entries in $\{0,1\}$ is $k$-regular for some $k\in \{0,\ldots,n\}$ if the sum of every row and the sum of every column of $A$ equals $k$. Let $M(n, k)$ be the number of $k$-regular $n\times n$-matrices with all entries in $\{0,1\}$.
It's easy to see that $M(n,1)=n!$. Moreover, a symmetry argument shows that $M(n,k) = M(n, n-k)$ for all $k\in \{0,\ldots,n\}$.
Question. Given $n>1$, is it true that for all $k\in\{0,\ldots, n\}$ we have $M(n,k)\leq M(n, \lfloor\frac{n}{2}\rfloor)$?
 A: I can prove a related result. Perhaps someone can
modify the proof to solve Dominic's problem.
I use multivariate notation such as $x^\alpha=x_1^{\alpha_1}\cdots
x_m^{\alpha_m}$, where $\alpha=(\alpha_1,\dots,\alpha_m)$. Let
$\alpha\in \{0,1,2,\dots\}^m$ and $\beta\in\{0,1,2,\dots\}^n$. Let $f(\alpha,\beta)$ be the
number of $m\times n$ matrices with entries $0,1,2$,
and with each entry equal to 1 colored either red or blue, with row
sum vector $\alpha$ and column sum vector $\beta$.
Theorem.
   $$ f(\alpha,\beta) \leq f((n,n,\dots,n),(m,m,\dots,m)). $$
Proof. Let $g(\alpha,\beta)$ be the
number of $m\times n$ matrices with entries $-2,0,2$,
and with each entry equal to 0 colored either red or blue, with row
sum vector $\alpha$ and column sum vector $\beta$. By dividing each
entry of such a matrix by 2 and then adding 1, it is clear that
 $$ f(\alpha,\beta)=g\left(2\alpha-2(n,\dots,n),
    2\beta-2(m,\dots,m)\right). $$ 
Hence we want to show that
  $$ g(\alpha,\beta) \leq g((0,0,\dots,0),(0,0,\dots,0)). $$
We have for fixed $m,n$ that
  $$ \sum_{\alpha,\beta} g(\alpha,\beta)x^\alpha y^\beta =
      \prod_{r=1}^m\prod_{s=1}^n (x_r^{-1}y_s^{-1}+x_ry_s)^2. $$
Since for any integer $k$ we have $\int_0^{2\pi}e^{ikx}dx = 1$ if $k=
  0$ and otherwise is $0$, it follows that
  $$ g(\alpha,\beta) = \frac{1}{(2\pi)^{m+n}}
    \int_0^{2\pi}\cdots \int_0^{2\pi} e^{-i(\alpha_1 \theta_1+\cdots+
     \alpha_m\theta_m+\beta_1\psi_1+\cdots+\beta_n\psi_n)}\\
     \prod_{r=1}^m\prod_{s=1}^n (e^{-i(\theta_r+\psi_s)}
     +e^{i(\theta_r+\psi_s)})^2\,d\theta\,d\psi. $$
Now $(e^{-i(\theta_r+\psi_s)}+e^{i(\theta_r+\psi_s)})^2$ is a
nonnegative real number. Hence by the triangle inequality,
  \begin{eqnarray*} g(\alpha,\beta) & \leq &
      \frac{1}{(2\pi)^{m+n}}
    \int_0^{2\pi}\cdots \int_0^{2\pi}
     \prod_{r,s=1}^n (e^{-i(\theta_r+\psi_s)}
     +e^{i(\theta_r+\psi_s)})^2\,d\theta\,d\psi\\ & = &
     g((0,\dots,0),(0,\dots,0)).\ \Box \end{eqnarray*}
A: I'm pretty sure this is unknown, though it would be great if I'm wrong.
Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim  e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper of Canfield and McKay.
It is also known to be true for $n\le 20$.
To spell out the asymptotics a bit more, the paper cited above shows that
$$ M(n,k) = (e^{-1/2}+o(1)) \binom {n}{k}^{\!2n} \bigl( \lambda^\lambda(1-\lambda)^{1-\lambda}\bigr)^{n^2}, \qquad(*)$$
as $n\to\infty$, where $\lambda=k/n$ and $cn/\log n\le k\le n-cn/\log n$ for a particular constant $c$. The value $c=1/3$ will do. This expression without the error term is unimodal for $0\le k\le n$ and even with the error term it is unimodal if $n$ is large enough. The same formula holds for $k=o(n^{1/2})$ (McKay and Wang, 2003). A new method of Liebenau and Wormald will (I'm 100% confident) show that $(*)$ is also true for the intermediate ranges of $k$, but it is not published yet. Then we will know that $M(n,k)$ is unimodal in $k$ provided $n$ is large enough.
If we just want to know whether $M(n,k)\le M(n,\lfloor n/2\rfloor)$, as asked, and don't care about unimodality, then what remains asymptotically is to show that
$M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$. This follows from the pairing (configuration) model for random $k$-regular bipartite graphs; namely
$$ M(n,k) \le \frac{ (nk)! }{ (k!)^{2n} }.\qquad(\#)$$
For large $k$, $(\#)$ is a pretty terrible bound, but if I didn't miscalculate it is sufficient to prove that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$.
That completes the proof that $M(n,\lfloor n/2\rfloor)$ is the largest value if $n$ is large enough.
