A236305 The number of P-positions in the game of Nim with up to 3 piles, allowing for piles of zero, such that the number of objects in each pile does not exceed n.

1, 4, 7, 16, 19, 28, 43, 64, 67, 76, 91, 112, 139, 172, 211, 256, 259, 268, 283, 304, 331, 364, 403, 448, 499, 556, 619, 688, 763, 844, 931, 1024, 1027, 1036, 1051, 1072, 1099, 1132, 1171, 1216, 1267, 1324, 1387, 1456, 1531, 1612, 1699

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, Apr 21 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) is considered different from (1,0,1) and (1,1,0).

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

Partial sums of A241717.

This sequence seems to be A256534(n+1)/4. - Thomas Baruchel, May 15 2018

Examples

			If the largest number is 1, then there should be an even number of piles of size 1. Thus, a(1)=4.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1024
• 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, p. 37.
• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 7 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A241522 (4 piles), A241523 (5 piles).

Cf. A241717 (first differences).

Programs

Mathematica

Table[Length[Select[Flatten[Table[{n, k, BitXor[n, k]}, {n, 0, a}, {k, 0, a}], 1], #[[3]] <= a &]], {a, 0, 100}]

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^(2*b) + 3*c^2.

a(n) = 4^floor(log(n)/log(2)) + 3*(n mod 2^floor(log(n)/log(2)))^2 (conjectured). - Thomas Baruchel, May 15 2018