OFFSET
0,17
COMMENTS
This is the m=4 member in the sequence of triangles A007318, A059259, A350110, A350111, A350112 which give the number of tilings of an (n+k) X 1 board using k (1,m-1)-fences and n-k unit square tiles. A (1,g)-fence is composed of two unit square tiles separated by a gap of width g.
It is also the m=4, t=2 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(4*j+r-k,k) is the coefficient of x^k in (f(j,x))^(4-r)*(f(j+1,x))^r for r=0,1,2,3 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x)=f(n,x)+x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0.
T(n+4-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 4.
Sum of (n+3)-th antidiagonal (counting initial 1 as the 0th) is A031923(n).
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.
Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.
FORMULA
T(n,k) = T(n-1,k) + T(n-2,k-1) - T(n-3,k-1) + T(n-3,k-2) + T(n-4,k-1) + T(n-4,k-3) + 2*T(n-4,k-4) + T(n-5,k-2) + 2*T(n-5,k-3) - T(n-5,k-4) - T(n-6,k-3)-T(n-6,k-5) - T(n-7,k-4)-T(n-7,k-5) - T(n-7,k-6) - T(n-8,k-7)-T(n-8,k-8) + delta(n,0)*delta(k,0) - delta(n,2)*delta(k,1) - delta(n,3)*delta(k,2) - delta(n,4)*delta(k,4) with T(n<k,k)=T(n,k<0)=0.
T(n,0) = 1.
T(n,n) = delta(n mod 4,0).
T(n,1) = n-3 for n>2.
T(4*j-r,4*j-p) = 0 for j>0, p=1,2,3, and r=1,...,p.
T(4*(j-1)+p,4*(j-1)) = T(4*j,4*j-p) = j^p for j>0 and p=0,1,2,3,4.
T(4*j+1,4*j-1) = 4*j(j+1)/2 for j>0.
T(4*j+2,4*j-2) = 4*C(j+2,4) + 6*C(j+1,2)^2 for j>1.
G.f. of row sums: (1-x-x^3)/((1-2*x)*(1-x^2)*(1+2*x^2+x^3+x^4)).
G.f. of antidiagonal sums: (1-x^2-x^3+x^4-x^6)/((1-x-x^2)*(1-x^4)*(1+3*x^4+x^8)).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=3*k+1 if k>=0.
EXAMPLE
Triangle begins:
1;
1, 0;
1, 0, 0;
1, 0, 0, 0;
1, 1, 1, 1, 1;
1, 2, 3, 4, 2, 0;
1, 3, 6, 7, 4, 0, 0;
1, 4, 9, 12, 8, 0, 0, 0;
1, 5, 13, 20, 16, 8, 4, 2, 1;
1, 6, 18, 32, 36, 28, 19, 12, 3, 0;
1, 7, 24, 50, 69, 69, 58, 31, 9, 0, 0;
1, 8, 31, 74, 120, 144, 127, 78, 27, 0, 0, 0;
1, 9, 39, 105, 195, 264, 265, 189, 81, 27, 9, 3, 1;
1, 10, 48, 144, 300, 458, 522, 432, 270, 132, 58, 24, 4, 0;
MATHEMATICA
f[n_]:=If[n<0, 0, f[n-1]+x*f[n-2]+KroneckerDelta[n, 0]];
T[n_, k_]:=Module[{j=Floor[(n+k)/4], r=Mod[n+k, 4]},
Coefficient[f[j]^(4-r)*f[j+1]^r, x, k]];
Flatten@Table[T[n, k], {n, 0, 13}, {k, 0, n}]
(* or *)
T[n_, k_]:=If[k<0 || n<k, 0, T[n-1, k] + T[n-2, k-1] - T[n-3, k-1] + T[n-3, k-2] + T[n-4, k-1] + T[n-4, k-3] + 2*T[n-4, k-4] + T[n-5, k-2] + 2*T[n-5, k-3] - T[n-5, k-4] - T[n-6, k-3] - T[n-6, k-5] - T[n-7, k-4] - T[n-7, k-5] - T[n-7, k-6] - T[n-8, k-7] - T[n-8, k-8] + KroneckerDelta[n, k, 0] - KroneckerDelta[n, 2]*KroneckerDelta[k, 1] - KroneckerDelta[n, 3]*KroneckerDelta[k, 2] - KroneckerDelta[n, k, 4]]; Flatten@Table[T[n, k], {n, 0, 9}, {k, 0, n}]
CROSSREFS
KEYWORD
AUTHOR
Michael A. Allen, Dec 22 2021
STATUS
approved