A237686 The number of P-positions in the game of Nim with up to four piles, allowing for piles of zero, such that the total number of objects in all piles doesn't exceed 2n.

1, 7, 14, 50, 63, 105, 148, 364, 413, 491, 546, 798, 883, 1141, 1400, 2696, 2961, 3255, 3382, 3850, 3983, 4313, 4620, 6132, 6469, 6979, 7322, 8870, 9387, 10941, 12496, 20272, 21833, 23423, 23982, 25746, 26167, 26929, 27524, 30332, 30933

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, May 02 2014

Keywords

• nonn

Comments

Partial sums of A237711.

Examples

			There is a position (0,0,0,0) with a total of zero. There are 6 positions with a total of 2 that are permutations of (0,0,1,1). Therefore, a(1)=7.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 16 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A237711 (first differences), A130665 (3 piles), A238147 (5 piles), A241522, A241718.

Programs

Mathematica

Table[Length[
  Select[Flatten[
    Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0,
      a}], 2], Total[#] <= a &]], {a, 0, 100, 2}]

Formula

a(2n+1) = 7a(n) + a(n-1), a(2n+2) = a(n+1) + 7a(n).