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 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. 4
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 Cf. A241522 (4 piles), A241523 (5 piles). Cf. A241717 (first differences). Sequence in context: A160715 A160120 A130665 * A212062 A256926 A101534 Adjacent sequences:  A236302 A236303 A236304 * A236306 A236307 A236308 KEYWORD nonn AUTHOR Tanya Khovanova and Joshua Xiong, Apr 21 2014 STATUS approved

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Last modified March 18 20:22 EDT 2019. Contains 321293 sequences. (Running on oeis4.)