Simplest way to generate integer coefficients with row sums equal to the terms of an arbitrary given sequence

Question

• Let $f(n)$ be an arbitrary function.
• 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 $a(n)$ be an integer sequence such that $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m}f(m-j)a(2^j k), \\ a(0) = 1. $$
• Let $T(n, k)$ be an integer coefficients such that $$ T(n,k) = \sum\limits_{i=0}^{2^n-1}[\operatorname{wt}(i)=k]a(i). $$ Here square bracket denotes Iverson bracket.

Here is the explanation how to make row sums equal to the terms of an arbitrary given sequence:

• Let $b(n)$ be an arbitrary sequence with $b(0)=1$.

To calculate $f(n)$ for it, use the following: $$ s(n,m) = \sum\limits_{j=0}^{2^n-1} a(2^m j), \\ s(n,m) = s(n+1, m-1) - \sum\limits_{k=0}^{m-1}f(m-k-1)s(n,k), \\ s(n,0) = b(n), s(0,m) = 1, \\ f(n) = s(1, n) - 1 - \sum\limits_{k=0}^{n-1}f(k). $$

I am interested in the simplest way to generate $T(n,k)$. I am also interested in representing the sequence $T(n+k,n)$ for fixed $k$ and variable $n$ as a polynomial $P_k(n)$ of degree $k$. Here we are need condition $f(0)=1$ to make $T(n,n)=1$ and make $P_k(n)$ an ordinary polynomial (without having to multiply it by a fixed number to the power of $n$).

Answer 1

I don't have any answer, but here are a few terms to be sure it is clear and to show some regularity in the constant or linear terms. Writing $f_n := f(n)$ and taking $f_0=1$, it looks like

$$\begin{array}{r,l}P_0(n)=&1\\ P_1(n) \overset{?}=& \binom{n+1}{1}+\binom{n+1}{2}\;\;f_1\\ P_2(n) \overset{?}=& \binom{n+2}{2}+\binom{n+2}{3}\left(f_2+2f_1\right) &+&\dots\\ P_3(n) \overset{?}=& \binom{n+3}{3} +\binom{n+3}{4}\left(f_3 +2f_2 +3f_1\right) &+&\dots\\ P_4(n) \overset{?}=& \binom{n+4}{4} +\binom{n +4}{5}\left(f_4 +2f_3 +3f_2 +4f_1\right) &+&\dots\\ P_5(n) \overset{?}=& \binom{n+5}{5} +\binom{n +5}{6}\left(f_5 +2f_4 +3f_3 +4f_2 +5f_1\right) &+&\dots \end{array}$$ $$\begin{array}{r,l} P_2(n) \overset{?}=&\dots&+&2\binom{n+2}{4}\;\;f_1^2\\ P_3(n) \overset{?}=&\dots&+&5\binom{n +3}{5}\left(f_2f_1 +f_1^2\right) &+&\dots\\ P_4(n) \overset{?}=&\dots&+&3\binom{n +4}{6}\left(f_2^2 +2f_3f_1 +3f_1^2 +4f_2f1\right) &+&\dots\\ P_5(n) \overset{?}=&\dots&+&7\binom{n +5}{7}\left(f_3f_2+f_2^2 +f_4f_1 +2f_3f_1 +2f_1^2+3f_2f_1\right) &+&\dots \end{array}$$ $$\begin{array}{r,l} P_3(n) \overset{?}=&\dots&+&\;\;5\binom{n +3}{6}\;\;f_1^3\\ P_4(n) \overset{?}=&\dots&+&\;\;7\binom{n +4}{7}\left(2f_1^3+3f_2f1^2\right) &+&\dots\\ P_5(n) \overset{?}=&\dots&+&28\binom{n +5}{8}\left(f_2^2f_1 +f_3f_1^2 +f_1^3 +2f_2f_1^2\right) &+&\dots \end{array}$$ $$\begin{array}{r,l} P_4(n) \overset{?}=&\dots&+&14\binom{n +4}{8}\;\;f_1^4\\ P_5(n) \overset{?}=&\dots&+&42\binom{n +5}{9}\left(f_1^4+2f_2f_1^3\right) &+&42\binom{n +5}{10}\;\;f_1^5 \end{array}$$