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%I #14 Sep 03 2023 15:02:11
%S 1,3,1,26,13,1,330,215,30,1,5496,4350,851,54,1,113160,104314,25835,
%T 2365,85,1,2774160,2895372,859024,101505,5335,123,1,78901200,91356684,
%U 31357452,4503709,308700,10486,168,1,2554231680,3230924400,1252435820,210546700,17685969,788200,18690,220,1
%N Triangle of coefficients of polynomials p_n(x) (exponents in increasing order) where 2^(x-n-1)p_n(x)/n! counts the ways a game of Nim with two piles of sizes x and n can be played out (x > 0, n >= 0).
%C The polynomials p_n(x) are monic.
%C The coefficient of x^(n-1) in p_n(x) is A022264(n).
%C For positive integers a, b, we have b!*4^b*p_a(b) = a!*4^a*p_b(a).
%C The coefficients are conjectured to be positive integers.
%H M. Harned and M. Rittenberg, <a href="http://girlsangle.org/page/bulletin-archive/GABv10n05E.pdf">NIM Counting, Part 2, Girls' Angle Bulletin</a>, Vol. 10, No. 5 (2017), 6-10.
%F p_n(x) = n!*Sum_{k=0..n} (4^(n-k-1)*(binomial(n-1,k+1) + 4*binomial(n-1,k) + 4*binomial(n-1,k-1))*binomial(x-1,k)).
%e p_0(x) = 1 since a Nim game with one stack with x tokens can be played out in 2^(x-1) ways (since each way is equivalent to placing dividers into some of the x-1 spaces between tokens).
%e p_2(2) = 26 + 13*2 + 2^2 = 56 since there are 2^(2-2-1)*56/2! = 14 ways to play out a Nim game that starts with two stacks with two tokens in each stack:
%e (2,2)->(2,1)->(2,0)->(1,0)->(0,0)
%e (2,2)->(2,1)->(2,0)->(0,0)
%e (2,2)->(2,1)->(1,1)->(1,0)->(0,0)
%e (2,2)->(2,1)->(1,1)->(0,1)->(0,0)
%e (2,2)->(2,1)->(0,1)->(0,0)
%e (2,2)->(2,0)->(1,0)->(0,0)
%e (2,2)->(2,0)->(0,0)
%e (2,2)->(1,2)->(1,1)->(1,0)->(0,0)
%e (2,2)->(1,2)->(1,1)->(0,1)->(0,0)
%e (2,2)->(1,2)->(0,2)->(0,1)->(0,0)
%e (2,2)->(1,2)->(0,2)->(0,0)
%e (2,2)->(1,2)->(1,0)->(0,0)
%e (2,2)->(0,2)->(0,1)->(0,0)
%e (2,2)->(0,2)->(0,0)
%K nonn,easy,tabl
%O 0,2
%A _Jane Wang_, Jul 02 2017