A241522 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 each pile does not exceed n.

1, 8, 21, 64, 89, 168, 301, 512, 561, 712, 965, 1344, 1801, 2408, 3165, 4096, 4193, 4488, 4981, 5696, 6585, 7720, 9101, 10752, 12433, 14408, 16677, 19264, 22121, 25320, 28861, 32768, 32961, 33544, 34517, 35904, 37657, 39848, 42477, 45568, 48881, 52680, 56965, 61760, 67017

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, Apr 24 2014

Keywords

• nonn

Comments

P-positions in the game of Nim are tuples of numbers with a Nim-Sum equal to zero. (0,1,1,0) is considered different from (1,0,1,0).

Partial sums of A241718.

Examples

			If the largest number is 1, then there should be an even number of piles of size 1. Thus, a(1)=8.
		

Links

• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 42-43.
• 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. A236305 (3 piles), A241523 (5 piles).

Cf. A241718 (first differences).

Programs

Mathematica

Table[Length[Select[Flatten[Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0, a}], 2], #[[4]] <= 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) = 2^(3*b) + 6*c^2*2^b + a(c-1).

a(2^n-1) = 2^(3*n).