A339631 -id:A339631 - cp's OEIS frontend

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

3, 16, 30, 72, 178, 444, 1114, 2808, 7098, 17984, 45656, 116106, 295718, 754226, 1926060, 4924188, 12602416, 32284214, 82777240, 212415744, 545495716, 1401849594, 3604921774, 9275890122, 23881602058, 61518226734, 158548607640, 408814563524, 1054590179342

Offset: 0

Doron Zeilberger, Taerim Kim, Karnaa Mistry, Weiji Zheng, Dec 10 2020

nonn

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.

			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.
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.

Doron Zeilberger, Class Projects for Combinatorics Fall 2020 (Rutgers University); Local copy [Pdf file only, no active links]
Doron Zeilberger, Initial Maple Package for Project 5 Combinatorics Fall 2020 (Rutgers University); Local copy
Doron Zeilberger, Math 454, Section 02 (Combinatorics) Fall 2020 (Rutgers University); Local copy [Pdf file only, no active links]

Bisection of A339631 (odd part).

Cf. A339633 (even part).

# Maple program adapted from OddTTT3(N) in Project 5 of Doron Zeilberger's Combinatorics Class Fall 2020 (Rutgers University).
A339634List:=proc(n) local f,i,t,x,N:
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):
N:=n-1:
#Take the Taylor expansion up to x^(2*N+2)
f:=taylor(f,x=0,2*N+3):
#Extract the coefficients of x^(2*i+1)*t^(3*i+2)
[3,seq(coeff(coeff(f,x,2*i+1),t,3*i+2),i=1..N)]:
end:

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.

A339633 Number of final Tic-Tac-Toe positions on a (2*n) X 3 board that resulted in a tie.

Original entry on oeis.org

1, 18, 28, 58, 140, 334, 824, 2038, 5084, 12730, 32004, 80694, 204004, 516902, 1312336, 3337682, 8502132, 21688182, 55395140, 141651742, 362601356, 929084578, 2382677360, 6115461118, 15707982020, 40375223374, 103846409504, 267258086338, 688201711116, 1773088924494

Offset: 0

Doron Zeilberger, Taerim Kim, Karnaa Mistry, Weiji Zheng, Dec 10 2020

nonn

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.

			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.

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

Doron Zeilberger, Class Projects for Combinatorics Fall 2020 (Rutgers University).
Doron Zeilberger, Initial Maple Package for Project 5 Combinatorics Fall 2020 (Rutgers University).
Doron Zeilberger, Math 454, Section 02 (Combinatorics) Fall 2020 (Rutgers University).

Bisection of A339631 (even part).

Cf. A339634 (odd part).

# Maple program adapted from EvenTTT3(N) in Project 5 of Doron Zeilberger's Combinatorics Class Fall 2020 (Rutgers University).
A339633List:=proc(n) local f,i,t,x,N:
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):
N:=n-1:
#Take the Taylor expansion up to x^(2*N+2)
f:=taylor(f,x=0,2*N+3):
#Extract the coefficients of x^(2*i)*t^(3*i)
[1,seq(coeff(coeff(f,x,2*i),t,3*i),i=1..N)]:
end:

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.

cp's OEIS Frontend

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

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