Is there a closed-form or combinatorial proof for the 'multisections' of the binomial series $\sum_{k=0}^{n-1}\binom{n}{k}$?

Question

Throughout, let $m,n\in\mathbb{N}^+$. As is well-known, $$ \sum_{k=0}^{r-1}\binom{r}{k}=2^r-1 $$I want to understand the multisections of this sum, namely for $m=2,3,4,5\ldots, 0\le j<m$ I want to evaluate $$ S_{m,j}(n) = \sum _{k=0}^{n-1} \binom{m n}{k m+j} $$Numerical results suggest that $S_{m,j}(n)$ is roughly but not exactly equally distributed in $j$ for fixed $m,n$. I computed the results for $m=2,3,4,6$ (I did $5$ as well but it was much more opaque):

\begin{array}{c|cccccc}\\ m\setminus j& 0 & 1 & 2 & 3 & 4 & 5\\ \hline 2 & 2^{2 n-1}-1 & 2^{2 n-1} & & & & \\ 3 & \frac{2 (-1)^n}{3}+\frac{8^n}{3}-1 & \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3} & \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3} & & & \\ 4 & (-1)^n 2^{2 n-1}+4^{2 n-1}-1 & 4^{2 n-1} & (-1)^{n+1} 2^{2 n-1}+4^{2 n-1} & 4^{2 n-1} & \text{} & \text{} \\ 6 & \frac{1}{3} 2^{6 n-1}+(-1)^n 3^{3 n-1}-\frac{2}{3} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^n 3^{3 n-1}-\frac{1}{6} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^{n+1} 3^{3 n-1}-\frac{1}{6} & \frac{1}{3} 2^{6 n-1}+(-1)^{n+1} 3^{3 n-1}+\frac{1}{3} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^{n+1} 3^{3 n-1}-\frac{1}{6} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^n 3^{3 n-1}-\frac{1}{6} \\ \end{array}

For instance, this table says $S_{3,1}(n) = \sum _{k=0}^{n-1} \binom{3 n}{3k+1} = \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3}$. Clearly we should have $\sum_{j=0}^{m-1}S_{m,j}(n)=2^{mn}-1$ but past that I'm not sure what the multisections should look like. Perhaps there is a combinatorial proof for these sums?

Answer 1

With the adjustment mentioned in the comments (adding 1 to $S_{m,0}(n)$), we have that $S_{m,j}(n)$ is the number of ways to color $mn$ distinguishable objects black and white so that the number of black objects is congruent to $j \pmod m.$

It turns out there's no particular reason to restrict attention to the case where the number of objects is divisible by $m$, so let's talk instead about $\tilde S_{m,j}(n)$, defined to be the number of ways to color $n$ distinguishable objects black and white so that the number of black objects is congruent to $j \pmod m$. So we have $$S_{m,j}(n) = \tilde S_{m,j}(mn).$$

If we're fixing $m$ there's a nice recursive relationship between these: $$\tilde S_{m,j}(n+1) = \tilde S_{m,j}(n) + \tilde S_{m,j-1}(n).$$

That's because given a coloring of $n+1$ balls black or white with $j \pmod m$ black balls, either the last ball is white and the coloring of the first $n$ balls has $j \pmod m$ black balls, or the last ball is black and the coloring of the first $n$ balls has $j-1 \pmod m$ black balls. (Here, $j-1$ is evaluated modulo $m$, of course.)

For a fixed $m$, the set of these observations for all $j$ combines into one matrix equation:

$$ \pmatrix {\tilde S_{m,0}(n+1) \\ \tilde S_{m,1}(n+1) \\ \vdots \\ \tilde S_{m,m-1}(n+1)} = \pmatrix {1 & 0 & 0 & \dots & 0 & 1 \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots &1 & 1 } \pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)} $$

So,

$$ \pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)} = \pmatrix {1 & 0 & 0 & \dots & 0 & 1 \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots &1 & 1 } ^ n \pmatrix {1 \\ 0 \\ \vdots \\ 0} $$

So diagonalizing that matrix should give you explicit formulas for each $\tilde S_{m, j}(n)$ as a linear combination of exponential terms in $n$ where the bases of the exponent are the eigenvalues of the matrix.

Diagonalizing it is pretty easy, if $M$ is that matrix and $R$ is the rotation matrix

$$R = \pmatrix {0 & 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots &1 & 0 }, $$ we have $M = I + R$ and the eigenvalues of $R$ are precisely the $m$'th roots of unity. That is, the eigenvalues of $M$ take the form $\lambda_0 = 2, \lambda_1 = 1+\zeta, \lambda_2 = 1+\zeta^2, ..., \lambda_{m-1} = 1+\zeta^{m-1}$ where $\zeta$ is a primitive $m$'th root of unity.

So, explicitly we expect for each $j$ there should be some coefficients $C_{j, k}$ for which

$$\tilde S_{m,j}(n) = C_{j,0} 2^n + C_{j, 1} (1+\zeta)^n + \dots + C_{j, m-1} (1+\zeta^{m-1})^n$$

In your case, we have

$$S_{m,j}(n) = C_{j,0} 2^{mn} + C_{j, 1} (1+\zeta)^{mn} + \dots + C_{j, m-1} (1+\zeta^{m-1})^{mn}.$$

The thing that makes things nice in the examples you're looking at ($m = 2, 3, 4, 6$) is the fact that the $m$'th power of one plus an $m$'th root of unity is nice for those things. That is for a primitive $m$'th root of unity $\zeta$ with $m$ in that set, $(1+\zeta^k)^m$ is always just an integer. This is not true for general $m$.

If you like, equivalently, the matrix $M^m$ has particularly nice eigenvalues for those values of $m$.

Uniformity

You mentioned that these are roughly uniform. That's because the eigenvector of $M$ corresponding to the largest (in absolute value) eigenvalue $2$ is the uniform vector $(1, \dots, 1)$. All the other exponential terms are relatively small.

Random walks

Another interpretation of this is that $\tilde S_{m, j}(n)$ the number of walks of length $n$ on $\mathbb Z / m \mathbb Z$ which land on $j$, where each step in the walk either leaves you where you started or moves you one step up in $\mathbb Z / m \mathbb Z$. Or, if you normalize, $\tilde S_{m, j}(n) / 2^n$ is the probability that a random walk in $\mathbb Z / m \mathbb Z$ ends up at $j$ after $n$ steps, where the steps in the random walk are chosen uniformly from those two options.

Then the normalized vector

$$ 2^{-n} \pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)} $$

just keeps track of the probability distribution of your location in $\mathbb Z / m \mathbb Z$ after $n$ steps of this random walk, and the normalized matrix $M/2$ is the transition matrix for this random walk.

This should also make it fairly clear that the distribution should approach a uniform one, since there's nothing in the random walk that tends to keep you in near one element of $\mathbb Z / m \mathbb Z$ over another.

More combinatorial approach

For the nice values of $m$ in the question, $M^m$ diagonalizes over the rational numbers. With enough effort, it should be possible to make this observation into a more combinatorial argument for the identities that you're talking about. (There's lots of things you could mean by "combinatorial", e.g. I consider the above to already be fairly combinatorial, especially in terms of the "random walk" interpretation above.) You can be explicit about the diagonalization, and write down some combinatorial interpretations of the individual terms in the diagonalization. You might need to do messy things like putting negative terms on one side of an equation and positive terms on another, and construct bijections recursively and hope that those bijections turn out to be something nice. I don't think the outcome of this process is likely to be worth the effort. It might be easier to work with $M^{2m}$ instead, since then the eigenvalues will all be positive.

The combinatorial things you put together this way can only exist and be meaningful for these special values of $m$, so I don't think they'll give that much insight over the more abstract approach. I think the more abstract approach (with a concrete interpretation in terms of random walks) is "the right way to think about it".

Edit: One exciting form of argument that might exist here: For those values $m$, it might be relevant that $\mathbb Q[\zeta]$ is a degree 2 extension of $\mathbb Q$, and that there's a corresponding rank 2 lattice of algebraic integers. Maybe the walks in $\mathbb Z / m \mathbb Z$ have some interesting combinatorial relationship to this lattice (and walks in this lattice?).