Let
- $T(n,k)$ be A381299, i.e., irregular triangular array read by rows: $T(n,k)$ is the number of ordered set partitions of $[n]$ with exactly $k$ descents, $n \geqslant 0, 0 \leqslant k \leqslant \binom{n}{2}$.
- $R(n,k,x)$ be the family of polynomials such that $$ R(n,k,x) = x^k\left(R(n,0,x) - \sum\limits_{j=0}^{k-1} R(n-1,j,x)\right), \\ R(n,0,x) = -1 + 2 \sum\limits_{j=0}^{n-1} R(n-1,j,x), \\ R(n,n,x) = 1. $$
I conjecture that for $n>0$ we have $$ \frac{R(n+1,0,x)+1}{4} = \sum\limits_{k=0}^{\binom{n}{2}} T(n,k) x^k. $$
Here is the PARI/GP program to generate $R(n,k,x)$:
upto(n) = {my(A, v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, v3 = v2; A = 2*vecsum(v3)-1; v2[1] = A; for(j=1, i-1, v2[j+1] = x^j*(A -= v3[j])); v2[i+1] = 1; v1[i+1] = v2[1]); v1} Note that this algorithm is pretty simple and very fast.
Is there a way to prove it?