Attempt 2 Let x&y = 2^k = (100..00)2 with k zeroes. Clearly, k+1-th bit should be 1 in both x and y (a&b=1 <=> a=b=1). And clearly x and y should have no other common bits. Let's find the number of pairs for each k. First, take only those numbers with k+1-th bit set. Consider one of them as a candidate for x. If it has ones in some other positions, then y should have zeroes in those positions. This translates into following: let mask = !(x) + 2^k and find all other numbers with k+1-st bit set, that are submasks of mask. For each k, you can find this number with dp over submasks.