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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.
7
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
0,23
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).
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,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.
EXAMPLE
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;
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}]
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.
Sequence in context: A030105 A255591 A324395 * A037893 A037901 A037839
KEYWORD
easy,nonn,tabl
AUTHOR
Michael A. Allen, Dec 22 2021
STATUS
approved