We have a series of conferences with $4^n-3^n$ attendees. We started with the second in the series for 7 attendees. The caterer provided a 3-top and 4-top table. She said, "We can use two table sizes for each event: a $(2^n)$-top and a $(2^n-1)$-top, which will allow us to configure the exact seating."
The next event will have 37 attendees and 7-top and 8-top tables, where we use 2 8-tops and 3 7-tops. The event number is $n.$
Is there a combinatorial method to determine how many of each size is required?
We are looking for $a,b\in\mathbb{N}$ such that $$ 4^n - 3^n = a2^n + b(2^n-1). $$ Assuming that there is a solution you can find it as follows: Let $k$ and $d$ be such that $$ 0\le d = k2^n -\left(4^n-3^n\right) \le 2^n. $$ If $d\le k$, then your solution is given by $a = k-d$ and $b = d$, as $$ a2^n + b(2^n-1) = (k-d)2^n + d(2^n-1) = k2^n - d = 4^n-3^n. $$ If $d>k$, then there is no solution. In your example we have $$ 5\times2^3 -\left(4^3-3^3\right) = 5\times 8-37 = 3 \le 8. $$ So we have $k=5$ and $d=3$, which give $a=5-3=2$ and $b=3$, as in your solution.