It is known that Striling numbers of the second kind satisfy the relation $$ \sum\limits_{k=0}^{n}{n \brace k}(x)_k = x^n. $$ where $(x)_n$ is the falling factorials such that $$ (x)_n = x(x-1)(x-2)\dotsc(x-n+1). $$
- Let $\operatorname{wt}(n)$ be A000120 (i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$)).
- Let $T(n,k)$ be an integer coefficients (A358612) such that $$ T(2n+1,k) = kT(n,k) + T(n,k-1), \\ T(2n,k) = kT(n,k) + T(n,k-1) - \frac{1}{k-1}\left(T(2n,k-1)+T(n,k-1)\right), \\ T(n,1) = T(0,1) = T(0,2) = 1. $$
- Let $P(n,k)$ be an integer coefficients (A373183) with row polynomials $R(n,x)$ such that $$ R(2n+1,x) = xR(n,x), \\ R(2n,x) = x(R(n,x+1) - R(n,x)), \\ R(0,x) = x. $$
I conjecture that $$ \frac{1}{x}\sum\limits_{k>0}T(n,k)(x)_k $$$$ \frac{1}{x}\sum\limits_{k=1}^{\operatorname{wt}(n)+2} T(n,k)(x)_k = \sum\limits_{k=1}^{\operatorname{wt}(n)+1}P(n,k)x^k(-1)^{\operatorname{wt}(n)-k+1}. $$
Note that is the same as $R(n,x)$ with ordinary sign alternating which depends on number of terms in the row.$$ T(2^n-1,k) = {n+2 \brace k}, $$ and $$ R(2^n-1,x) = x^{n+1}. $$
Here is the PARI/GP program to check it numerically:
row1(n) = my(A, v1, v2); v1 = [1, 1]; if(n == 0, v1, forstep(i=logint(n, 2), 0, -1, A = bittest(n, i); v2 = vector(#v1+A, i, 0); v2[1] = 1; for(j=2, #v2, v2[j] = j*if(j==#v1+1, 0, v1[j]) + v1[j-1] - if(A, 0, (v2[j-1] + v1[j-1])/(j-1))); v1 = v2); v1)
row2(n) = my(x = 'x, A = x); forstep(i=if(n == 0, -1, logint(n, 2)), 0, -1, A = if(bittest(n, i), x*A, x*(subst(A, x, x+1) - A))); Vecrev(A/x)
row3(n) = my(v1); v1 = row1(n); v1 = Vecrev(sum(i=1, #v1, prod(j=1, i, x-j+1)*v1[i])/x^2)
row4(n) = my(v1); v1 = row2(n); v1 = vector(#v1, i, (-1)^(#v1-i)*v1[i])
test1(n) = row3(n) == row4(n)
Is there a way to prove it?