A241523 -id:A241523 - cp's OEIS frontend

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

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

A241522 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 each pile does not exceed n.

Original entry on oeis.org

1, 8, 21, 64, 89, 168, 301, 512, 561, 712, 965, 1344, 1801, 2408, 3165, 4096, 4193, 4488, 4981, 5696, 6585, 7720, 9101, 10752, 12433, 14408, 16677, 19264, 22121, 25320, 28861, 32768, 32961, 33544, 34517, 35904, 37657, 39848, 42477, 45568, 48881, 52680, 56965, 61760, 67017

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) is considered different from (1,0,1,0).

Partial sums of A241718.

			If the largest number is 1, then there should be an even number of piles of size 1. Thus, a(1)=8.

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, pp. 42-43.
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. A236305 (3 piles), A241523 (5 piles).

Cf. A241718 (first differences).

Table[Length[Select[Flatten[Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0, a}], 2], #[[4]] <= 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) = 2^(3*b) + 6*c^2*2^b + a(c-1).

a(2^n-1) = 2^(3*n).

A241731 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 the largest pile is n.

Original entry on oeis.org

1, 15, 45, 195, 165, 555, 1125, 1995, 645, 1995, 3525, 5355, 7605, 10395, 13845, 18075, 2565, 7755, 13125, 18795, 24885, 31515, 38805, 46875, 55845, 65835, 76965, 89355, 103125, 118395, 135285, 153915, 10245, 30795, 51525, 72555, 94005, 115995

Offset: 0

Tanya Khovanova and Joshua Xiong, Apr 27 2014

nonn

This is the finite difference of A241523.

			If the largest pile is 1, then there are 10 positions that are permutations of (0,0,0,1,1) plus 5 positions that are permutations of (0,1,1,1,1). Therefore, a(1)=15.

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. A241523, A241717 (3 piles), A241718 (4 piles).

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], 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)= 10*2^(2*b)*(2*c-1) + 20*c^3 - 30*c^2 + 20*c - 5.

A238147 The number of P-positions in the game of Nim with up to five 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, 11, 26, 126, 191, 341, 516, 1516, 2081, 2731, 3206, 4706, 5631, 7381, 9256, 19256, 24821, 30471, 33946, 40446, 44171, 48921, 52796, 67796, 76221, 85471, 91846, 109346, 119971, 138721, 158096, 258096, 313661, 369311

Offset: 0

Tanya Khovanova and Joshua Xiong, May 02 2014

nonn

Partial sums of A238759.

			There is 1 position (0,0,0,0,0) with a total of zero. There are 10 positions with a total of 2 that are permutations of (0,0,0,1,1). Therefore, a(1)=11.

T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 18 and J. Int. Seq. 17 (2014) # 14.7.8.

Cf. A238759 (first differences), A130665 (3 piles), A237686 (4 piles), A241523, A241731.

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}]
(* Second program: *)
a[n_] := a[n] = Which[n <= 1, {1, 11}[[n+1]], OddQ[n], 11 a[(n-1)/2] + 5 a[(n-1)/2 - 1], EvenQ[n], a[(n-2)/2 + 1] + 15*a[(n-2)/2]];
Array[a, 34, 0] (* Jean-François Alcover, Dec 14 2018 *)

a(2n+1) = 11a(n) + 5a(n-1), a(2n+2) = a(n+1) + 15a(n).

A238759 The number of P-positions in the game of Nim with up to five piles, allowing for piles of zero, such that the total number of objects in all piles is 2n.

Original entry on oeis.org

1, 10, 15, 100, 65, 150, 175, 1000, 565, 650, 475, 1500, 925, 1750, 1875, 10000, 5565, 5650, 3475, 6500, 3725, 4750, 3875, 15000, 8425, 9250, 6375, 17500, 10625, 18750, 19375, 100000, 55565, 55650, 33475, 56500, 31725, 34750, 23875, 65000

Offset: 0

Tanya Khovanova and Joshua Xiong, May 02 2014

nonn

First differences of A238147.

			The P-positions with the total of 4 are permutations of (0,0,0,2,2) and (0,1,1,1,1). Therefore, a(2)=15.

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

Cf. A238147 (partial sums), A048883 (3 piles), A237711 (4 piles), A241523, A241731.

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], Total[#] == a &]], {a, 0, 90, 2}]
(* Second program: *)
(* b = A238147 *) b[n_] := b[n] = Which[n <= 1, {1, 11}[[n+1]], OddQ[n], 11 b[(n-1)/2] + 5 b[(n-1)/2 - 1], EvenQ[n], b[(n-2)/2 + 1] + 15 b[(n-2)/2]];
Join[{1}, Differences[Array[b, 40, 0]]] (* Jean-François Alcover, Dec 14 2018 *)

a(2n+1) = 10*a(n), a(2n+2) = a(n+1) + 5*a(n).

cp's OEIS Frontend

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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A241522 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 each pile does not exceed n.

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A241731 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 the largest pile is n.

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A238147 The number of P-positions in the game of Nim with up to five piles, allowing for piles of zero, such that the total number of objects in all piles doesn't exceed 2n.

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A238759 The number of P-positions in the game of Nim with up to five piles, allowing for piles of zero, such that the total number of objects in all piles is 2n.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula