A241522 -id:A241522 - cp's OEIS frontend

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

Tanya Khovanova and Joshua Xiong, May 02 2014

nonn

First differences of A237686.

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

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

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

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}]

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

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.

Original entry on oeis.org

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

Tanya Khovanova and Joshua Xiong, Apr 21 2014

nonn

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

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

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.

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

Cf. A241717 (first differences).

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

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

A237686 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 doesn't exceed 2n.

Original entry on oeis.org

1, 7, 14, 50, 63, 105, 148, 364, 413, 491, 546, 798, 883, 1141, 1400, 2696, 2961, 3255, 3382, 3850, 3983, 4313, 4620, 6132, 6469, 6979, 7322, 8870, 9387, 10941, 12496, 20272, 21833, 23423, 23982, 25746, 26167, 26929, 27524, 30332, 30933

Offset: 0

Tanya Khovanova and Joshua Xiong, May 02 2014

nonn

Partial sums of A237711.

			There is a position (0,0,0,0) with a total of zero. There are 6 positions with a total of 2 that are permutations of (0,0,1,1). Therefore, a(1)=7.

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

Cf. A237711 (first differences), A130665 (3 piles), A238147 (5 piles), A241522, A241718.

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}]

a(2n+1) = 7a(n) + a(n-1), a(2n+2) = a(n+1) + 7a(n).

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.

Original entry on oeis.org

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

Tanya Khovanova and Joshua Xiong, Apr 24 2014

nonn

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).

			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.

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

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

Cf. A241731 (first differences).

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}]

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.

A241718 The number of P-positions in the game of Nim with up to 4 piles, allowing for piles of zero, such that the number of objects in the largest pile is n.

Original entry on oeis.org

1, 7, 13, 43, 25, 79, 133, 211, 49, 151, 253, 379, 457, 607, 757, 931, 97, 295, 493, 715, 889, 1135, 1381, 1651, 1681, 1975, 2269, 2587, 2857, 3199, 3541, 3907, 193, 583, 973, 1387, 1753, 2191, 2629, 3091, 3313, 3799, 4285, 4795, 5257, 5791, 6325

Offset: 0

Tanya Khovanova and Joshua Xiong, Apr 27 2014

nonn

This is the first difference of A241522.

			If the largest pile is 2, then there are 6 positions that are permutations of (0,0,2,2) plus 6 positions that are permutations of (1,1,2,2) and one position (2,2,2,2). Therefore, a(2)=13.

T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO], 2014, p. 8 and J. Int. Seq. 17 (2014) # 14.7.8.

Cf. A241522, A241717 (3 piles), A241731 (5 piles).

Table[Length[Select[Flatten[Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0, a}], 2], Max[#] == a &]], {a, 0, 50}]

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) = (12*c-6)*2^b + a(c-1).

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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.

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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.

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A237686 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 doesn't exceed 2n.

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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.

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A241718 The number of P-positions in the game of Nim with up to 4 piles, allowing for piles of zero, such that the number of objects in the largest pile is n.

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