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$$ \left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\ \left[{n \atop 0}\right] = \left[{0 \atop n}\right] = 0, \left[{0 \atop 0}\right] = 1 $$

  • Let $T(n, k)$ be A373183, i.e. irregular table $T(n, k)$ ($n\geqslant0, k>0$) read by rows with row polynomials $R(n,x)$ such that

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

  • Let

$$ \ell(n) = \left\lfloor\log_2 n\right\rfloor, \\ \ell(0) = -1 $$

  • Let

$$ f(n) = \ell(n) - \ell(n-2^{\ell(n)}) - 1 $$

  • 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 $b(n, m, k)$ be the family of integer sequences such that

$$ b(n,m,k) = (m+1)^{f(n)}b(n-2^{\ell(n)}, m+1, k) - m^{f(n)+1}b(n-2^{\ell(n)}, m, k-1), \\ b(0, m, k) = [k \leqslant m]\left[{m \atop m-k+1}\right], b(n, m, 0) = 0 $$

Here square bracket denotes Iverson bracket.

I conjecture that

$$ T(n, k) = b(n, 1, \operatorname{wt}(n)-k+2). $$

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))); Vecrev(A/x)
b(n,m,k) = if(k == 0, 0, if(n == 0, if(m<k, 0, abs(stirling(m, m-k+1, 1))), my(L = logint(n, 2), A = n - 1 << L, B = L - if(A == 0, -1, logint(A, 2)) - 1); (m+1)^B*b(A, m+1, k) - m^(B+1)*b(A, m, k-1)))
row2(n) = my(v1); v1 = Vecrev(vector(hammingweight(n)+1, i, b(n,1,i)))
test(n) = row1(n) == row2(n)

Is there a way to prove it?

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    $\begingroup$ It seems correct. See the mathematica code as supplement to the pari/gp code. $\endgroup$
    – 138 Aspen
    Commented Jun 13 at 13:16

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