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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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%I #22 Mar 17 2021 07:57:42

%S 1,6,7,36,13,42,43,216,49,78,55,252,85,258,259,1296,265,294,127,468,

%T 133,330,307,1512,337,510,343,1548,517,1554,1555,7776,1561,1590,559,

%U 1764,421,762,595,2808,601,798,463,1980,637,1842,1819,9072,1849

%N 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.

%C First differences of A237686.

%H T. Khovanova and J. Xiong, <a href="http://arxiv.org/abs/1405.5942">Nim Fractals</a>, arXiv:1405.594291 [math.CO] (2014), p. 16 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Khovanova/khova6.html">J. Int. Seq. 17 (2014) # 14.7.8</a>.

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

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

%e 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 Table[Length[

%t Select[Flatten[

%t Table[{n, k, j, BitXor[n, k, j]}, {n, 0, a}, {k, 0, a}, {j, 0,

%t a}], 2], Total[#] == a &]], {a, 0, 100, 2}]

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

%K nonn

%O 0,2

%A _Tanya Khovanova_ and _Joshua Xiong_, May 02 2014