OFFSET

0,2

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

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.

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

EXAMPLE

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

MATHEMATICA

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

CROSSREFS

KEYWORD

nonn

AUTHOR

Tanya Khovanova and Joshua Xiong, Apr 21 2014

STATUS

approved