Let
- $$ \ell(n) = \left\lfloor\log_2 n\right\rfloor. $$
- $\operatorname{wt}(n)$ be A000120, i.e., number of ones in the binary expansion of $n$. Here $$ \operatorname{wt}(2n+1) = \operatorname{wt}(n) + 1, \\ \operatorname{wt}(2n) = \operatorname{wt}(n), \\ \operatorname{wt}(0) = 0. $$
- $T(n,k)$ be A078121, i.e., an integer coefficients such that $$ T(n,k) = \begin{cases} 1 & \text{if } k = 0 \\ \displaystyle{ \sum\limits_{i=k-1}^{n-1} T(n-1,i) T(i,k-1) } & \text{otherwise} \end{cases} $$
- $a(n)$ be A000123, i.e., number of binary partitions: number of partitions of $2n$ into powers of $2$. Here $$ a(2n+1) = a(2n) + a(n), \\ a(2n) = a(2n-1) + a(n), \\ a(0) = 1. $$
- $Q(n,k)$ be A272020, i.e., irregular triangle read by rows: strictly decreasing sequences of positive numbers given in lexicographic order.
- $R(n,k)$ be A066099, i.e., triangle read by rows, in which row $n$ lists the compositions of $n$ in reverse lexicographic order.
- $b(n)$ be an integer sequence such that $b(n) = \nu_1$ where we start with vector $\nu$ of fixed length $\ell(n)+2$ with elements $\nu_i = 1$ (that is, $ \nu = \{1,1,\dotsc,1\} $), reserve $t$ as an empty vector of fixed length $\ell(n)+2$ and for $i$ from $1$ to $\operatorname{wt}(n)$, for $j$ from $1$ to $Q(n,i)-R(n,i)+1$ apply $t := \nu$ (at the beginning of each cycle for $i$) and also apply $$ \nu_j := \sum\limits_{k=0}^{j+R(n,i)-1} T(j+R(n,i)-1,k) t_{k+1}. $$
I conjecture that $ b(n) = a(n) $.
Is there a way to prove it?