Cleaner form for transformation involving Stirling numbers

Question

I have a specific transformation involving treating the Stirling numbers of the first kind as a matrix, given by, for some positive $b$

$$ C^{(b)} = (S_1^{(b)})^{T} \left(J_b\ \left(\frac{1}{2^{i}}{{b}\choose{i}}\right) \right) S_1^{(b)} $$

where $S_1^{(b)}$ is a $b+1 \times b+1$ matrix of Stirling numbers of the first kind, $J_b$ is the exchange matrix (matrix with 1s along the anti-diagonal), and the remaining matrix is simply a diagonal matrix with binomial coefficients divided by powers of two along the diagonal

schematically, this looks like, for example for $b=2$,

$$ C^{(2)} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & \frac{1}{2^{2}}{{2}\choose{2}} \\ 0 & \frac{1}{2^{1}}{{2}\choose{1}} & 0 \\ \frac{1}{2^{0}}{{2}\choose{0}} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \\ \end{bmatrix} $$

and the specific matrix elements are given by

$$ c^{(b)}_{k l} = \begin{cases} \frac{1}{2^b} s(b, l) & k = 0 \\ s(b, k) & l = 0 \\ \sum_{w=l}^{b-k} \frac{1}{2^w} {{b}\choose{w}} s(b-w, k) s(w, l) & else \end{cases} $$

Given the connections between changes of basis to rising/falling factorials, Stirling numbers, and binomial coefficients this feels like it has some deeper meaning, but this is of course complicated by the fact that we can't read this as a plain change of basis since $S_1^{(b)}$ is not unitary

For anyone who wants it, here's some Mathematica code to generate this matrix

cMat[b_] :=
 With[
  {
   S = Table[StirlingS1[k, w], {k, 0, b}, {w, 0, b}],
   J = Array[KroneckerDelta[#, b - #2] &, {b, b} + 1, 0],
   W = DiagonalMatrix[Table[Binomial[b, w]/2^w, {w, 0, b}]]
   },
  Transpose[S] . (J . W) . S
  ]
cMat[3] // MatrixForm

$$ \left( \begin{array}{cccc} 0 & \frac{1}{4} & -\frac{3}{8} & \frac{1}{8} \\ 2 & -\frac{9}{4} & \frac{3}{4} & 0 \\ -3 & \frac{3}{2} & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right) $$

Answer 1

Combinatorial interpretation:

Let $\Sigma_b$ represent the collection of colored permutations of the following kind: for a permutation $w$ of $\{1, \ldots, b\}$, its cycles are marked either "colored" or "uncolored", and if $x$ is an entry in a colored cycle then $x$ is colored either red or blue.

For example, when $b = 2$, there are 14 colored permutations in $\Sigma_2$: five have underlying permutation $(12)$ (one in which the cycle is uncolored, four in which the cycle is colored) and nine have underlying permutation $(1)(2)$.

The rescaled matrix entry $2^b \cdot |c^{(b)}_{kl}|$ counts the permutations in $\Sigma_b$ that have $k$ colored and $l$ uncolored cycles, while the checkerboard sign pattern just tracks the parity of those permutations.

Generatingfunctionology:

You ask about connections with falling/rising factorials; to me that's really a question about the generating function for these numbers. For a fixed $b$, we can define the following bivariate generating function: $$ F_b(x, y) = \sum_{k = 0}^b \sum_{\ell = 0}^b c^{(b)}_{k, \ell} x^k y^{\ell}. $$ One could compute formulas for this either from the combinatorial interpretation or directly from the formula you gave: \begin{align*} F_b(x, y) & = \sum_{k = 0}^b \sum_{\ell = 0}^b \sum_{w = \ell}^{b - k} \frac{1}{2^w} \binom{b}{w} s(b - w, k)s(w, \ell) x^k y^{\ell} \\ & = \sum_{w = 0}^b \frac{1}{2^w} \binom{b}{w} \sum_{k = 0}^{b - w} s(b - w, k) x^k \sum_{\ell = 0}^w s(w, \ell) y^\ell \\ & = \sum_{w = 0}^b \frac{1}{2^w} \binom{b}{w} (x)_{b - w} (y)_{w}.\end{align*} There are lots of fun ways that one can rewrite this; for example, it is equal to $$b! \cdot \sum_{w = 0}^b \frac{1}{2^w} \binom{x}{b - w} \binom{y}{w}.$$ If it weren't for the power of $2$, this would be Vandermonde's identity, but as it is it does not factor further in general. We can also compute the natural trivariate generating function $$ F(x, y, z) = \sum_{b \geq 0} F_b(x, y) \frac{z^b}{b!}, $$ and from the previous equation we have \begin{align*} F(x, y, z) & = \sum_{b \geq 0}\sum_{w = 0}^b \frac{1}{2^w} \binom{x}{b - w} \binom{y}{w} z^b \\ & = \sum_{w \geq 0} \frac{z^2}{2^w}\binom{y}{w} \sum_{b \geq w} \binom{x}{b - w} z^{b - w} \\ & = \sum_{w \geq 0} \left(\frac{z}{2}\right)^w \binom{y}{w} \cdot (1 + z)^x \\ & = \left(1 + \frac{z}{2}\right)^y (1 + z)^x, \end{align*} which is rather nice.