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$$ R(2n+1,x) = xR(n,x), \\ R(2n,x) = x(R(n,x+1) - R(n, x)), \\ R(0, x) = x $$

  • Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here

$$ \operatorname{wt}(2n+1) = \operatorname{wt}(n) + 1, \\ \operatorname{wt}(2n) = \operatorname{wt}(n), \\ \operatorname{wt}(0) = 0 $$

  • Let $T(n,k)$ be an integer coefficients (A358612) such that

$$ T(2n+1, k) = kT(n, k) + T(n, k-1), \\ T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\ T(n, 1) = T(0, 2) = 1 $$

I conjecture that

$$ R(n,x)=\sum\limits_{k=1}^{\operatorname{wt}(n)+1} x^k(-1)^{\operatorname{wt}(n)-k+1}\sum\limits_{i=1}^{\operatorname{wt}(n)-k+3} s(\operatorname{wt}(n)-i+3, k+1)T(n, \operatorname{wt}(n)-i+3). $$

Here is the PARI/GP program to check it numerically:

row1(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))); A
row2(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)
row3(n) = my(x = 'x, v1); v1 = row2(n); my(A = #v1 - 2); sum(k=1, A+1, x^k*(-1)^(A-k+1)*sum(i=1, A-k+3, stirling(A-i+3, k+1, 1)*v1[A-i+3]))
test(n) = row1(n) == row3(n)

Is there a way to prove it?

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