A238147 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 doesn't exceed 2n.

1, 11, 26, 126, 191, 341, 516, 1516, 2081, 2731, 3206, 4706, 5631, 7381, 9256, 19256, 24821, 30471, 33946, 40446, 44171, 48921, 52796, 67796, 76221, 85471, 91846, 109346, 119971, 138721, 158096, 258096, 313661, 369311

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, May 02 2014

Keywords

• nonn

Comments

Partial sums of A238759.

Examples

			There is 1 position (0,0,0,0,0) with a total of zero. There are 10 positions with a total of 2 that are permutations of (0,0,0,1,1). Therefore, a(1)=11.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 18 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A238759 (first differences), A130665 (3 piles), A237686 (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], #[[5]] <= a &]], {a, 0, 35}]
(* Second program: *)
a[n_] := a[n] = Which[n <= 1, {1, 11}[[n+1]], OddQ[n], 11 a[(n-1)/2] + 5 a[(n-1)/2 - 1], EvenQ[n], a[(n-2)/2 + 1] + 15*a[(n-2)/2]];
Array[a, 34, 0] (* Jean-François Alcover, Dec 14 2018 *)

Formula

a(2n+1) = 11a(n) + 5a(n-1), a(2n+2) = a(n+1) + 15a(n).