A350112 Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,4)-fences and n-k squares.

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 2, 0, 1, 3, 6, 10, 9, 4, 0, 0, 1, 4, 10, 16, 16, 8, 0, 0, 0, 1, 5, 14, 25, 28, 16, 0, 0, 0, 0, 1, 6, 19, 38, 48, 32, 16, 8, 4, 2, 1, 1, 7, 25, 56, 80, 80, 60, 40, 25, 15, 3, 0, 1, 8, 32, 80, 136, 166, 157, 128, 95, 40, 9, 0, 0

Offset: 0

Author

Michael A. Allen, Dec 22 2021

Keywords

• easy
• nonn
• tabl

Comments

This is the m=5 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=5, 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(5*j+r-k,k) is the coefficient of x^k in (f(j,x))^(5-r)*(f(j+1,x))^r for r=0,1,2,3,4 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+5-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 5.

Sum of (5j+r)-th antidiagonal (counting initial 1 as the 0th) is f(j)^(5-r)*f(j+1)^r where j=0,1,..., r=0,1,2,3,4, and f(n) is the Fibonacci number A000045(n+1).

Examples

			Triangle begins:
  1;
  1,   0;
  1,   0,   0;
  1,   0,   0,   0;
  1,   0,   0,   0,   0;
  1,   1,   1,   1,   1,   1;
  1,   2,   3,   4,   5,   2,   0;
  1,   3,   6,  10,   9,   4,   0,   0;
  1,   4,  10,  16,  16,   8,   0,   0,   0;
  1,   5,  14,  25,  28,  16,   0,   0,   0,   0;
  1,   6,  19,  38,  48,  32,  16,   8,   4,   2,   1;
  1,   7,  25,  56,  80,  80,  60,  40,  25,  15,   3,   0;
  1,   8,  32,  80, 136, 166, 157, 128,  95,  40,   9,   0,   0;
  1,   9,  40, 112, 217, 309, 346, 330, 223, 105,  27,   0,   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.
• Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Crossrefs

Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).

Other triangles related to tiling using fences: A123521, A157897, A335964.

Programs

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)/5], r=Mod[n+k,5]},
  Coefficient[f[j]^(5-r)*f[j+1]^r,x,k]];
Flatten@Table[T[n,k], {n, 0, 13}, {k, 0, n}]

Formula

T(n,0) = 1.

T(n,n) = delta(n mod 5,0).

T(n,1) = n-4 for n>3.

T(5*j-r,5*j-p) = 0 for j>0, p=1,2,3,4, and r=1,...,p.

T(5*(j-1)+p,5*(j-1)) = T(5*j,5*j-p) = j^p for j>0 and p=0,1,...,5.

T(5*j+1,5*j-1) = 5*j(j+1)/2 for j>0.

T(5*j+2,5*j-2) = 5*C(j+2,4) + 10*C(j+1,2)^2 for j>1.

T(n,k) = T(n-1,k) + T(n-1,k-1) for n >= 4*k+1 if k >= 0.