Let
- $a(n)$ be A000123 (i.e., number of binary partitions: number of partitions of $2n$ into powers of $2$), whose ordinary generating function is $$ \frac{1}{1-x} \prod\limits_{j=0}^{\infty} \frac{1}{{1-x^{2^j}}}. $$ Also we have $$ a(2n+1) = a(2n) + a(n), \\ a(2n) = a(2n-1) + a(n), \\ a(0) = 1. $$
- $$ \ell(n) = \left\lfloor\log_2 n\right\rfloor. $$
- $T(n,k)$ be an integer coefficients defined for $n > 0, 0 \leqslant k \leqslant \ell(n)$ such that for $m > 0, 2^{m-1} \leqslant k < 2^m$ we have $$ \sum\limits_{i=1}^{m} \frac{2^{\frac{i(i+1)}{2}}T(k,i-1)(-1)^{m-i}}{(1-x)^{i+1}} \prod\limits_{j=0}^{\infty} \frac{1}{{1-x^{2^j}}} = \sum\limits_{i=0}^{\infty} a(2^m i + k) x^i. $$
I conjecture that for $n > 0, 0 \leqslant k \leqslant \ell(n), n \ne 2^m, m \geqslant 0$ we have $$ T(n,k) = T(n-1,k) - T\left(\left\lfloor\frac{n}{2}\right\rfloor,k \right). $$ I also conjecture that for $n \geqslant 0$ we have $$ \sum\limits_{k=1}^{2(n+1)} \frac{2^{\frac{k(k+1)}{2}}T(2^{2n+1},k-1)(-1)^{2n+k}}{k} = 0. $$
Here is the PARI/GP program to generate given coefficients:
upto1(n, m) = {my(x = 'x);
2^(m*(m+1)/2)/((1-x)^(m+1)*prod(j=0, l(n), 1 - x^(2^j))) + x*O(x^n)}
row(n) = {my(L = logint(n, 2), M1, M2, M3, v1);
M1 = matrix(L+1, L+1, i, j, (-1)^(L-j+1)*polcoeff(upto1(i-1, j), i-1, x));
M2 = matrix(L+1, 1, i, j, polcoeff(upto1((i-1)<<(L+1) + n, 0), (i-1)<<(L+1) + n, x));
M3 = matsolve(M1, M2);
v1 = vector(L+1, i, M3[i, 1])}
Is there a way to prove it? Is there a way to find a formula for $T(n,k)$ for $n = 2^m, m \geqslant 0$?