MacstersUndead said: I found out that falling factorial coefficients are related to signed Stirling numbers of the first kind, so you can generate those numbers and set up a system of equations which can be represented by a matrix. In your example, we have \(\displaystyle A(n^5 - 10n^4 + 35n^3 - 50n^2 + 24n) + B(n^4 - 6n^3 + 11n^2 - 6n) + C (n^3 - 3n^2 + 2) + D(n^2 - n) + En = n^5\) We can represent this as a linear equation (keeping only the coefficients that are signed Stirling numbers), with a table of signed values similar to Stirling numbers of the first kind - Wikipedia In your example, we have (imagine the vectors as columns). A(1, -10, 35, -50, 24) + B(0,1,-6,11,-6) + C(0,0,1,-3,2) + D(0,0,0,1,-1) + E(0,0,0,0,1) = (1,0,0,0,0) Click to expand... Well, it looks like brute force is out. I had suspected a relationship involving the Stirling numbers but I hadn't had time to investigate very far. Even if I ignored the derivation the explicit formula is going to be too complicated to work with well. The vector idea is a good one! I hadn't thought to look at it like that, though I've seen it used before. (This is what happens to a Physicist!) Thanks for the help! -Dan