A354668 Triangle read by rows: T(n,k) is the number of tilings of an (n+2*k) X 1 board using k (1,2;3)-combs and n-k squares.
1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 2, 0, 1, 1, 3, 4, 0, 0, 1, 2, 5, 8, 0, 0, 1, 1, 3, 8, 12, 0, 3, 3, 0, 1, 4, 12, 18, 9, 12, 9, 0, 0, 1, 5, 16, 27, 25, 29, 27, 0, 0, 1, 1, 6, 21, 42, 51, 66, 54, 0, 6, 4, 0, 1, 7, 27, 62, 95, 135, 108, 36
Offset: 0
Examples
Triangle begins: 1; 1, 0; 1, 0, 0; 1, 0, 0, 1; 1, 0, 1, 2, 0; 1, 1, 3, 4, 0, 0; 1, 2, 5, 8, 0, 0, 1; 1, 3, 8, 12, 0, 3, 3, 0; 1, 4, 12, 18, 9, 12, 9, 0, 0; 1, 5, 16, 27, 25, 29, 27, 0, 0, 1; 1, 6, 21, 42, 51, 66, 54, 0, 6, 4, 0; 1, 7, 27, 62, 95, 135, 108, 36, 30, 16, 0, 0; ...
Links
- Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
- Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.
Crossrefs
Programs
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Mathematica
f[n_]:=If[n<0, 0, f[n-1]+x*f[n-3]+KroneckerDelta[n,0]]; T[n_, k_]:=Module[{j=Floor[(n+2*k)/3], r=Mod[n+2*k,3]}, Coefficient[f[j]^(3-r)*f[j+1]^r, x, k]]; Flatten@Table[T[n,k], {n, 0, 11}, {k, 0, n}]
Formula
T(n,0) = 1.
T(n,n) = delta(n mod 3,0).
T(n,1) = n-4 for n>3.
T(3*j-r,3*j-p) = 0 for j>0, p=1,2, and r=1-p,...,p.
T(n,2*j) = C(n/2,j)^2 for j>0 and n even and 2*j <= n <= 2*j+8.
T(n,2*j) = C((n-1)/2,j)*C((n+1)/2,j) for j>0 and n odd and 2*j < n < 2*j+8.
T(2*j+3*p,2*j-p) = C(j+3,4)^p for j>0 and p=0,1,2.
G.f. of sums of T(n-2*k,k) over k: (1+x^3-x^4-x^5+x^6-2*x^7-x^8-x^9-2*x^10-x^12-x^13-x^15)/((1-x)*(1+x+x^2)*(1-x-x^3)*(1+3*x^3+7*x^6+9*x^9+7*x^12+3*x^15+x^18)).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=4*k+1 if k>=0.
Comments