A339633 - OEIS

%I #56 Dec 24 2020 04:35:17

%S 1,18,28,58,140,334,824,2038,5084,12730,32004,80694,204004,516902,

%T 1312336,3337682,8502132,21688182,55395140,141651742,362601356,

%U 929084578,2382677360,6115461118,15707982020,40375223374,103846409504,267258086338,688201711116,1773088924494

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

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

%D Doron Zeilberger, Math 454, Section 02 (Combinatorics) Fall 2020 (Rutgers University).

%H Doron Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/Combo20/ClassProjects.html">Class Projects for Combinatorics Fall 2020 (Rutgers University)</a>.

%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>.

%H Doron Zeilberger, <a href="https://sites.math.rutgers.edu/~zeilberg/math454_20.html">Math 454, Section 02 (Combinatorics) Fall 2020 (Rutgers University)</a>.

%F a(n) = [x^(2*n)*t^(3*n)] (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 = 1 it is a 3 X 2 matrix, and so there are 18 ways to have three 1's and three 0's such that there are no 3-streaks of 1's nor 0's in the matrix.

%p # Maple program adapted from EvenTTT3(N) in Project 5 of Doron Zeilberger's Combinatorics Class Fall 2020 (Rutgers University).

%p A339633List:=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)*t^(3*i)

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

%p end:

%Y Bisection of A339631 (even part).

%Y Cf. A339634 (odd part).

%K nonn

%O 0,2

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