A241717 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 the largest pile is n.

1, 3, 3, 9, 3, 9, 15, 21, 3, 9, 15, 21, 27, 33, 39, 45, 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171

Offset: 0

Author

Tanya Khovanova and Joshua Xiong, Apr 27 2014

Keywords

• nonn

Comments

This is the finite difference of A236305.

Starting from index 1 all elements are divisible by 3, and can be grouped into sets of size 2^k of an arithmetic progression 6n-3.

It appears that the sum of all terms of the first n rows of triangle gives A000302(n-1), see Example section. - Omar E. Pol, May 01 2015

Examples

			If the largest number is 1, then there should be exactly two piles of size 1 and one empty pile. There are 3 ways to permute this configuration, so a(1)=3.
From _Omar E. Pol_, Feb 26 2015: (Start)
Also written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
1;
3;
3, 9;
3, 9, 15, 21;
3, 9, 15, 21, 27, 33, 39, 45;
3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93;
...
Observation: the first six terms of the right border coincide with the first six terms of A068156.
(End)
From _Omar E. Pol_, Apr 20 2015: (Start)
An illustration of initial terms in the fourth quadrant of the square grid:
---------------------------------------------------------------------------
n   a(n)             Compact diagram
---------------------------------------------------------------------------
.            _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
0    1      |_| |_  |_ _ _  |_ _ _ _ _ _ _  |
1    3      |_ _| | |_ _  | |_ _ _ _ _ _  | |
2    3      | |_ _| |_  | | |_ _ _ _ _  | | |
3    9      |_ _ _ _| | | | |_ _ _ _  | | | |
4    3      | | | |_ _| | | |_ _ _  | | | | |
5    9      | | |_ _ _ _| | |_ _  | | | | | |
6   15      | |_ _ _ _ _ _| |_  | | | | | | |
7   21      |_ _ _ _ _ _ _ _| | | | | | | | |
8    3      | | | | | | | |_ _| | | | | | | |
9    9      | | | | | | |_ _ _ _| | | | | | |
10  15      | | | | | |_ _ _ _ _ _| | | | | |
11  21      | | | | |_ _ _ _ _ _ _ _| | | | |
12  27      | | | |_ _ _ _ _ _ _ _ _ _| | | |
13  33      | | |_ _ _ _ _ _ _ _ _ _ _ _| | |
14  39      | |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |
15  45      |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.
It appears that a(n) is also the number of cells in the n-th region of the diagram, and A236305(n) is also the total number of cells after n-th stage.
(End)
		

Links

• T. Khovanova and J. Xiong, Nim Fractals, arXiv:1405.594291 [math.CO] (2014), p. 6 and J. Int. Seq. 17 (2014) # 14.7.8.

Crossrefs

Cf. A011782, A068156, A236305 (partial sums), A241718 (4 piles), A241731 (5 piles).

Programs

Mathematica

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

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) = 6*c-3.