A350112 - OEIS

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.

%I #17 Jan 20 2023 16:46:51

%S 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,

%T 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,

%U 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

%N 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.

%C 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.

%C 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.

%C 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.

%C 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.

%C 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).

%H Michael A. Allen, <a href="https://arxiv.org/abs/2209.01377">On a Two-Parameter Family of Generalizations of Pascal's Triangle</a>, arXiv:2209.01377 [math.CO], 2022.

%H Michael A. Allen, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL25/Allen2/allen8.html">On A Two-Parameter Family of Generalizations of Pascal's Triangle</a>, J. Int. Seq. 25 (2022) Article 22.9.8.

%H Michael A. Allen and Kenneth Edwards, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL25/Allen/allen3.html">On Two Families of Generalizations of Pascal's Triangle</a>, J. Int. Seq. 25 (2022) Article 22.7.1.

%F T(n,0) = 1.

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

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

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

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

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

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

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

%e Triangle begins:

%e 1;

%e 1, 0;

%e 1, 0, 0;

%e 1, 0, 0, 0;

%e 1, 0, 0, 0, 0;

%e 1, 1, 1, 1, 1, 1;

%e 1, 2, 3, 4, 5, 2, 0;

%e 1, 3, 6, 10, 9, 4, 0, 0;

%e 1, 4, 10, 16, 16, 8, 0, 0, 0;

%e 1, 5, 14, 25, 28, 16, 0, 0, 0, 0;

%e 1, 6, 19, 38, 48, 32, 16, 8, 4, 2, 1;

%e 1, 7, 25, 56, 80, 80, 60, 40, 25, 15, 3, 0;

%e 1, 8, 32, 80, 136, 166, 157, 128, 95, 40, 9, 0, 0;

%e 1, 9, 40, 112, 217, 309, 346, 330, 223, 105, 27, 0, 0, 0;

%t f[n_]:=If[n<0,0,f[n-1]+x*f[n-2]+KroneckerDelta[n,0]];

%t T[n_, k_]:=Module[{j=Floor[(n+k)/5], r=Mod[n+k,5]},

%t Coefficient[f[j]^(5-r)*f[j+1]^r,x,k]];

%t Flatten@Table[T[n,k], {n, 0, 13}, {k, 0, n}]

%Y 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).

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

%K easy,nonn,tabl

%O 0,23

%A _Michael A. Allen_, Dec 22 2021