A238759 The number of P-positions in the game of Nim with up to five piles, allowing for piles of zero, such that the total number of objects in all piles is 2n.

1, 10, 15, 100, 65, 150, 175, 1000, 565, 650, 475, 1500, 925, 1750, 1875, 10000, 5565, 5650, 3475, 6500, 3725, 4750, 3875, 15000, 8425, 9250, 6375, 17500, 10625, 18750, 19375, 100000, 55565, 55650, 33475, 56500, 31725, 34750, 23875, 65000

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, May 02 2014

Keywords

• nonn

Comments

First differences of A238147.

Examples

			The P-positions with the total of 4 are permutations of (0,0,0,2,2) and (0,1,1,1,1). Therefore, a(2)=15.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 17 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A238147 (partial sums), A048883 (3 piles), A237711 (4 piles), A241523, A241731.

Programs

Mathematica

Table[Length[
  Select[Flatten[
    Table[{n, k, j, i, BitXor[n, k, j, i]}, {n, 0, a}, {k, 0, a}, {j,
      0, a}, {i, 0, a}], 3], Total[#] == a &]], {a, 0, 90, 2}]
(* Second program: *)
(* b = A238147 *) b[n_] := b[n] = Which[n <= 1, {1, 11}[[n+1]], OddQ[n], 11 b[(n-1)/2] + 5 b[(n-1)/2 - 1], EvenQ[n], b[(n-2)/2 + 1] + 15 b[(n-2)/2]];
Join[{1}, Differences[Array[b, 40, 0]]] (* Jean-François Alcover, Dec 14 2018 *)

Formula

a(2n+1) = 10*a(n), a(2n+2) = a(n+1) + 5*a(n).