A241718 The number of P-positions in the game of Nim with up to 4 piles, allowing for piles of zero, such that the number of objects in the largest pile is n.

1, 7, 13, 43, 25, 79, 133, 211, 49, 151, 253, 379, 457, 607, 757, 931, 97, 295, 493, 715, 889, 1135, 1381, 1651, 1681, 1975, 2269, 2587, 2857, 3199, 3541, 3907, 193, 583, 973, 1387, 1753, 2191, 2629, 3091, 3313, 3799, 4285, 4795, 5257, 5791, 6325

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, Apr 27 2014

Keywords

• nonn

Comments

This is the first difference of A241522.

Examples

			If the largest pile is 2, then there are 6 positions that are permutations of (0,0,2,2) plus 6 positions that are permutations of (1,1,2,2) and one position (2,2,2,2). Therefore, a(2)=13.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO], 2014, p. 8 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A241522, A241717 (3 piles), A241731 (5 piles).

Programs

Mathematica

Table[Length[Select[Flatten[Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0, a}], 2], Max[#] == a &]], {a, 0, 50}]

Formula

If b = floor(log_2(n)) is the number of digits in the binary representation of n and c = n + 1 - 2^b, then a(n) = (12*c-6)*2^b + a(c-1).