A237711 The number of P-positions in the game of Nim with up to four piles, allowing for piles of zero, such that the total number of objects in all piles is 2n.

1, 6, 7, 36, 13, 42, 43, 216, 49, 78, 55, 252, 85, 258, 259, 1296, 265, 294, 127, 468, 133, 330, 307, 1512, 337, 510, 343, 1548, 517, 1554, 1555, 7776, 1561, 1590, 559, 1764, 421, 762, 595, 2808, 601, 798, 463, 1980, 637, 1842, 1819, 9072, 1849

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, May 02 2014

Keywords

• nonn

Comments

First differences of A237686.

Examples

			The P-positions with the total of 4 are permutations of (0,0,2,2) and (1,1,1,1). Therefore, a(2)=7.
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 16 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A237686 (partial sums), A048883 (3 piles), A238759 (5 piles), A241522, A241718.

Programs

Mathematica

Table[Length[
  Select[Flatten[
    Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0,
      a}], 2], Total[#] == a &]], {a, 0, 100, 2}]

Formula

a(2n+1) = 6a(n), a(2n+2) = a(n+1) + a(n).

G.f.: Product_{k>=0} (1 + 6*x^(2^k) + x^(2^(k+1))). - Ilya Gutkovskiy, Mar 16 2021