A289329 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).

1, 3, 1, 26, 13, 1, 330, 215, 30, 1, 5496, 4350, 851, 54, 1, 113160, 104314, 25835, 2365, 85, 1, 2774160, 2895372, 859024, 101505, 5335, 123, 1, 78901200, 91356684, 31357452, 4503709, 308700, 10486, 168, 1, 2554231680, 3230924400, 1252435820, 210546700, 17685969, 788200, 18690, 220, 1

Offset

0,2

Comments

The polynomials p_n(x) are monic.

The coefficient of x^(n-1) in p_n(x) is A022264(n).

For positive integers a, b, we have b!*4^b*p_a(b) = a!*4^a*p_b(a).

The coefficients are conjectured to be positive integers.

Links

Table of n, a(n) for n=0..44.

M. Harned and M. Rittenberg, NIM Counting, Part 2, Girls' Angle Bulletin, Vol. 10, No. 5 (2017), 6-10.

Formula

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)).

Example

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).
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:
(2,2)->(2,1)->(2,0)->(1,0)->(0,0)
(2,2)->(2,1)->(2,0)->(0,0)
(2,2)->(2,1)->(1,1)->(1,0)->(0,0)
(2,2)->(2,1)->(1,1)->(0,1)->(0,0)
(2,2)->(2,1)->(0,1)->(0,0)
(2,2)->(2,0)->(1,0)->(0,0)
(2,2)->(2,0)->(0,0)
(2,2)->(1,2)->(1,1)->(1,0)->(0,0)
(2,2)->(1,2)->(1,1)->(0,1)->(0,0)
(2,2)->(1,2)->(0,2)->(0,1)->(0,0)
(2,2)->(1,2)->(0,2)->(0,0)
(2,2)->(1,2)->(1,0)->(0,0)
(2,2)->(0,2)->(0,1)->(0,0)
(2,2)->(0,2)->(0,0)

Crossrefs

Sequence in context: A363736 A098815 A300457 * A033464 A170924 A173007

Adjacent sequences: A289326 A289327 A289328 * A289330 A289331 A289332

Keyword

nonn,easy,tabl

Author

Jane Wang, Jul 02 2017

Status

approved