Let $q(n)$ be A007814, i.e., number of trailing zeros in the binary representation of $n$. Here $$q(2n+1)=0, q(2n)=q(n)+1$$
Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$). Here $$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1,\operatorname{wt}(2n)=\operatorname{wt}(n),\operatorname{wt}(0)=0$$
Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
Let $f(n)$ be A054429, i.e., simple self-inverse permutation of natural numbers: list each block of $2^n$ numbers (from $2^n$ to $2^{n+1} - 1$) in reverse order. Here $$f(n)=3\cdot2^{\ell(n)}-n-1$$
Let $$g(n,m)=2(f(n+2^{\ell(n)+1})+2^{\ell(n)+2}(2^m-1))$$
Let $h(n)$ be A059893, i.e., reverse the order of all but the most significant bit in binary expansion of $n$: if $n = 1ab\cdots yz$ then $h(n) = 1zy\cdots ba$.
Let $a(n)$ be A347205. Here $$a(2n+1)=a(n), a(2n)=a(n)+a(n-2^{q(n)}), a(0)=1$$
Let $T(n,k)$ be an integer coefficients described in my previous question.
I conjecture that $$a(g(n,m))=\frac{1}{(\operatorname{wt}(n)+1)!}\sum\limits_{k=1}^{\operatorname{wt}(n)+2}(m+1)^{k-1}T(h(n),\operatorname{wt}(n)-k+3)$$
Is there a way to prove it?
