Solution
Wythoff's Losing Pairs up to Fifty
Show the problem again
In Wythoff's game two piles are given. A move removes any positive number of counters from one pile, or the same positive number from both. The player who takes the last counter wins. Count the unordered pairs {a, b} with 1 ≤ a ≤ b ≤ 50 such that the player to move loses with optimal play from piles (a, b). Answer with an integer.
Worked solution
The losing positions form the pairs (a_k, b_k) with a_k = ⌊kφ⌋ and b_k = a_k + k = ⌊kφ²⌋, where φ is the golden ratio; equivalently a_k is the smallest positive integer not yet used and b_k = a_k + k. These pairs partition the positive integers. We need b_k ≤ 50. Since a_19 = ⌊19φ⌋ = 30 and b_19 = 49, while a_20 = 32 and b_20 = 52, exactly the pairs k = 1, ..., 19 qualify (and every one has a_k ≥ 1). Piles equal to each other are never losing here. The count is 19, as confirmed by a brute-force evaluation of the game on all piles up to 200.
Source: Wythoff's game (1907); the Beatty-sequence structure is classical. Statement written for AxiomIQ.