Note that there's no need to actually perform N-M+1 exponentiations and N-M additions. Two exponentiations, a subtraction, and a division will give you the sam

Note that there's no need to actually perform N-M+1 exponentiations and N-M additions. Two exponentiations, a subtraction, and a division will give you the same answer. Sum[A^i,{i,1,N}] = (A^(N+1) - 1) / (A-1). Sum[A^i,{i,M,N}] = (A^(N+1) - A^M) / (A-1). Consider A=10, M=6, N=2. Sum = (10,000,000 - 100) / 9 = (9,999,900) / 9 = 1,111,100, which is easily checked mentally. Yes, I've stared at permutations for far too long, but I can't have been the first to notice the pattern. It must be in plenty of textbooks. Another cute combinatorics question: using an alphabet of size A, generate the Nth largest M-character sequence where no two adjacent characters are equal. A simple count-and-check algorithm takes O(N) time, but there's a simple O(M) algorithm (that is, constant time if the number of characters is fixed), assuming constant-time basic arithmetic operations.