Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$

Question

Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).

Let $a(n,m)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)=m$. In other words, $a(n,m)$ is the $n$-th number with binary weight equals $m$.

I conjecture that $$a(1,2^{n-1}+n)+\sum\limits_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)=(2^{n+1}+1)2^{2^{n-1}+n-1}-1$$

I also conjecture that numbers of the form $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ have only $2$ partitions into parts with binary weight equals $2^{n-1}+n$, exactly the number itself and the sum above.

Is there a way to prove it?

Answer 1

Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have $$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$

Then the sum in question can be easily computed: \begin{split} & a(1,2^{n-1}+n)+\sum_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)\\ &= \sum_{i=0}^{2^{n-1}+1} a(i+1,2^{n-1}+n) - (2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-1}) \\ &=(2^{n-1}+1)(2^{2^{n-1}+n+1} - 1) - 2^{n-1}(2^{2^{n-1}+2}-1) + 2^{2^{n-1}+n-1} \\ &= (2^{n+1}+1)2^{2^{n-1}+n-1} - 1. \end{split}