So I'm working on a problem where I get to a point where I have to count the number of solutions to an equation or at least find a decent upper bound to be used in an estimate I need later.
The equation is as follows for a given $n$ and for each $k\in\{1,\dots, n-1\}$ let $S_{k}^{n}:=A^{k}\times B^{n-k}\subset\mathbb{N}^{n}$ denote the set of solutions to the equation
$$\sum_{i=1}^{k}a_{i}=\sum_{j=1}^{n-k}b_{j}$$
where $a_{i},b_{j}\in\{2,2^2,\dots ,2^m,\dots \}$ and the $a_{i}$ and $b_{j}$ are non-decreasing ie $a_{1}\leq a_{2}\leq \dots \leq a_{k}$ and $b_{1}\leq b_{2}\leq \dots \leq b_{n-k}$.
If $(a,b)=(a_{1},\dots,a_{k},b_{1},\dots b_{n-k})$ denotes a point in $S_{k}^{n}$ and $P(a,b)$ denotes the number of unique permutations of the sequence $(a,-b)=(a_{1},\dots,a_{k},-b_{1},\dots -b_{n-k})$ then I'm trying to find a good upper bound on this sum $$\Lambda(S_{k}^{n})=\sum_{(a,b)\in S_{k}^{n}}\left[P(a,b) \left(\prod_{i=1}^{k}\frac{1}{a_{i}} \right)\left(\prod_{j=1}^{n-k}\dfrac{1}{b_{j}}\right)\right].$$
So I see that this is somewhat related to the number of way that a binary number $s$ can be represented by binary digits less then $s$, I think its kinda like a partition of $s$ with only so many summands.
I'm not exactly a pro with thinking about binary digits so any insight or help would be greatly appreciated.
I need this estimate so that I can show that the following sum is finite
$$\sum_{n=1}^{\infty}\dfrac{1}{2^n}\left[\sum_{k=1}^{n}\Lambda(S_{k}^{n})\right]<\infty$$