I have a sequence (well a sequence of sequences) that I can't figure out how to generate directly, despite the fact I can generate it recursively
Specifically, I can write the elements of a matrix of polynomial coefficients, $F^{(b)}$ in terms of some sequence $A$ as
$$ F^{(b)} = \frac{1}{2^{(b+(i-j))}} {b-(i-j) \choose j} A^{(b)}_{(i-j)} $$
I'm sure there's something notable here about how most terms depend on $(i-j)$ but I don't know what exactly that is
The problem is I don't know how to generate $A^{(b)}$ directly, here it is as a matrix ($b=0...5$ down the rows)
$$ A = \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3 & -3 & 0 & 0 & 0 \\ 1 & 6 & -9 & 3 & 0 & 0 \\ 1 & 10 & -15 & -15 & 75 & 0 \\ 1 & 15 & -15 & -120 & 585 & -855 \\ \end{array} \right) $$
Now, I can also generate $F^{(b)}$ through a recursion, but it's nasty, starting from the generating equation
$$ \sum_{j=0}^{b} f^{(b)}_{k j} a^j + (-1)^{a} G = \sum_{i=0}^{a} {(-1)}^{(a-i)} \left(\sum_{j=0}^b \left(f_{k j}^{(b-1)} + (i-(k-1)) f^{(b-1)}_{k-1 j}\right) i^j \right) $$
where I ignore $G$ for now, I can turn this into a sequence of matrix transformations to get
$$ F^{(b)} = \left((I - D_1 R)F^{(b-1)} + D_1 F^{(b-1)} D_{-1}\right) S_2 \left[2^{-i}\right] S_1 (P^{-1}) S_2 \left[2^{i}\right] S_1 P $$
where $D_1$ is the matrix with ones on the first subdiagonal (shifts rows down), $D_{-1}$ has ones on the first superdiagonal, $R_{i j} = i \delta_{i j}$, $S_1$ is the matrix of Stirling numbers of the first kind, $S_2$ is the matrix of Stirling numbers of the second kind, $P$ is the matrix of binomial coefficients, and the two square-bracket matrices have the given element on the diagonal
There is obviously a lot of simplification that will happen here, just given the sheer number of similarity transforms happening, but I can't seem to figure out how to make that happen
Turns out I didn't really need this, but for the heck of it, here's a recursive generator just to show how nasty it is
$$ A^{(b)} = \sum_{j=0}^{k} C_j^{(b)} (b-k+1)^{[k-j]} $$
where $C_j^{(b)}$ is an auxiliary sequence defined by
$$ C_j^{(b)} = C_j^{(b-1)} + \sum_{w=0}^{j-1} (-1)^{j-w+1} C_w^{(b-1)} \left((b-j)^{[j-w]} + (-1)^{\delta_{w, j-1} } 2 (j-1) (b-j+1)^{[j-1-w]}\right) $$
which is obviously not particularly nice but it works