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A339634 Number of final Tic-Tac-Toe positions on a (2*n+1) X 3 board that result in a tie. 3

%I #62 Mar 28 2021 22:46:17

%S 3,16,30,72,178,444,1114,2808,7098,17984,45656,116106,295718,754226,

%T 1926060,4924188,12602416,32284214,82777240,212415744,545495716,

%U 1401849594,3604921774,9275890122,23881602058,61518226734,158548607640,408814563524,1054590179342

%N Number of final Tic-Tac-Toe positions on a (2*n+1) X 3 board that result in a tie.

%C The number of (2*n+1) X 3 0,1-matrices with 3*n+2 1's and 3*n+1 0's and no consecutive horizontal, vertical, nor diagonal triples of 111 or 000.

%H Doron Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/Combo20/ClassProjects.html">Class Projects for Combinatorics Fall 2020 (Rutgers University)</a>; <a href="/A339634/a339634.pdf">Local copy</a> [Pdf file only, no active links]

%H Doron Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/Combo20/projects/ComboProject5.txt">Initial Maple Package for Project 5 Combinatorics Fall 2020 (Rutgers University)</a>; <a href="/A339634/a339634.txt">Local copy</a>

%H Doron Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/math454_20.html">Math 454, Section 02 (Combinatorics) Fall 2020 (Rutgers University)</a>; <a href="/A339634/a339634_1.pdf">Local copy</a> [Pdf file only, no active links]

%F a(n) = [x^(2*n+1)*t^ceiling(3*(2*n+1)/2)] (4*t^17*x^11 + 4*t^16*x^11 + 8*t^16*x^10 + 12*t^15*x^10 + 6*t^15*x^9 + 8*t^14*x^10 + 8*t^14*x^9 + 8*t^13*x^9 - 2*t^13*x^8 + 6*t^12*x^9 - 16*t^12*x^8 - 2*t^11*x^8 - 26*t^11*x^7 - 26*t^10*x^7 - 19*t^10*x^6 - 38*t^9*x^6 - 7*t^9*x^5 - 19*t^8*x^6 - 13*t^8*x^5 - 13*t^7*x^5 - t^7*x^4 - 7*t^6*x^5 + 10*t^6*x^4 - t^5*x^4 + 16*t^5*x^3 + 16*t^4*x^3 + 9*t^4*x^2 + 18*t^3*x^2 + 9*t^2*x^2) / (t^12*x^8 + t^11*x^7 + t^10*x^7 + t^9*x^6 - 2*t^6*x^4 - t^5*x^3 - t^4*x^3 - t^3*x^2 + 1) for n >= 1.

%e For n = 0 it is a 3 X 1 matrix, and there are 3 arrangements of 2 1's and a single 0 such that there are no 3-streaks of 1's nor 0's in the matrix.

%e For n = 1 it is the classic 3 X 3 Tic-Tac-Toe board, having 1's as X's and 0's as O's.

%p # Maple program adapted from OddTTT3(N) in Project 5 of Doron Zeilberger's Combinatorics Class Fall 2020 (Rutgers University).

%p A339634List:=proc(n) local f,i,t,x,N:

%p f:=(4*t^17*x^11 + 4*t^16*x^11 + 8*t^16*x^10 + 12*t^15*x^10 + 6*t^15*x^9 + 8*t^14*x^10 + 8*t^14*x^9 + 8*t^13*x^9 - 2*t^13*x^8 + 6*t^12*x^9 - 16*t^12*x^8 - 2*t^11*x^8 - 26*t^11*x^7 - 26*t^10*x^7 - 19*t^10*x^6 - 38*t^9*x^6 - 7*t^9*x^5 - 19*t^8*x^6 - 13*t^8*x^5 - 13*t^7*x^5 - t^7*x^4 - 7*t^6*x^5 + 10*t^6*x^4 - t^5*x^4 + 16*t^5*x^3 + 16*t^4*x^3 + 9*t^4*x^2 + 18*t^3*x^2 + 9*t^2*x^2) / (t^12*x^8 + t^11*x^7 + t^10*x^7 + t^9*x^6 - 2*t^6*x^4 - t^5*x^3 - t^4*x^3 - t^3*x^2 + 1):

%p N:=n-1:

%p #Take the Taylor expansion up to x^(2*N+2)

%p f:=taylor(f,x=0,2*N+3):

%p #Extract the coefficients of x^(2*i+1)*t^(3*i+2)

%p [3,seq(coeff(coeff(f,x,2*i+1),t,3*i+2),i=1..N)]:

%p end:

%Y Bisection of A339631 (odd part).

%Y Cf. A339633 (even part).

%K nonn

%O 0,1

%A _Doron Zeilberger_, Taerim Kim, _Karnaa Mistry_, Weiji Zheng, Dec 10 2020

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