# Problem 488: Unbalanced Nim
Alice and Bob have enjoyed playing Nim every day. However, they finally
got bored of playing ordinary three-heap Nim. So, they added an extra
rule: - Must not make two heaps of the same size. The triple (a,b,c)
indicates the size of three heaps. Under this extra rule, (2,4,5) is one
of the losing positions for the next player. To illustrate: - Alice
moves to (2,4,3) - Bob moves to (0,4,3) - Alice moves to (0,2,3) - Bob
moves to (0,2,1) Unlike ordinary three-heap Nim, (0,1,2) and its
permutations are the end states of this game. For an integer N, we
define F(N) as the sum of a+b+c for all the losing positions for the
next player, with 0 < a < b < c < N. For example, F(8) = 42,
because there are 4 losing positions for the next player, (1,3,5),
(1,4,6), (2,3,6) and (2,4,5). We can also verify that F(128) = 496062.
Find the last 9 digits of F(1018).