A241523 The number of P-positions in the game of Nim with up to 5 piles, allowing for piles of zero, such that the number of objects in each pile does not exceed n.

1, 16, 61, 256, 421, 976, 2101, 4096, 4741, 6736, 10261, 15616, 23221, 33616, 47461, 65536, 68101, 75856, 88981, 107776, 132661, 164176, 202981, 249856, 305701, 371536, 448501, 537856, 640981, 759376, 894661, 1048576, 1058821, 1089616

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. (0,1,1,0,0) is considered different from (1,0,1,0,0).

a(2^n-1) = 2^(4n).

Examples

			If the largest number is not more than 1, then there should be an even number of piles of size 1. We can choose the first four piles to be either 0 or 1, then the last pile is uniquely defined. Thus, a(1)=16.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 9 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A236305 (3 piles), A241522 (4 piles).

Cf. A241731 (first differences).

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}]

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^(4*b) + 10*2^(2*b)*c^2 + 5*c^4.