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%I #12 Dec 27 2022 03:26:39
%S 1,1,0,1,0,1,1,0,2,0,1,0,4,0,1,1,1,6,0,3,0,1,2,9,0,9,0,1,1,3,12,5,18,
%T 0,4,0,1,4,16,12,36,0,16,0,1,1,5,20,25,60,15,40,0,5,0,1,6,25,42,100,
%U 42,100,0,25,0,1,1,7,31,66,150,112,200
%N Triangle read by rows: T(n,k) is the number of tilings of an (n+4*k) X 1 board using k (1,1;5)-combs and n-k squares.
%C This is the m=2, t=5 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(2*j+r-4*k,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for r=0,1, where f(n,x) is a (1,5)-bonacci polynomial defined by f(n,x)=f(n-1,x)+x*f(n-5,x)+delta(n,0) where f(n<0,x)=0.
%C T(n+8-4*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 2, 4, 6, or 8.
%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.
%F T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + 2*T(n-2,k-2) + T(n-3,k-1) - T(n-3,k-2) - 2*T(n-3,k-3) - T(n-4,k-1) + T(n-4,k-2) + T(n-4,k-3) - T(n-4,k-4) + T(n-5,k-1) - 2*T(n-5,k-3) + T(n-5,k-5) + delta(n,0)*delta(k,0) - delta(n,1)*delta(k,1) - delta(n,2)*delta(k,2) - delta(n,3)*(delta(k,1) - delta(k,3)) with T(n,k<0) = T(n<k,k) = 0.
%F T(n,0) = 1.
%F T(n,n) = delta(n mod 2,0).
%F T(n,1) = n-4 for n>3.
%F T(2*j+r,2*j-1) = 0 for j>0, r=-1,0,1,2.
%F T(n,2*j) = C(n/2,j)^2 for j>0 and n even and 2*j <= n <= 2*j+8.
%F 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.
%F T(2*j+3*p,2*j-p) = C(j+3,4)^p for j>0 and p=0,1,2.
%F G.f. of row sums: (1-x-x^2)/(1-2*x-x^2+2*x^3).
%F G.f. of sums of T(n-4*k,k) over k: (1-x^5-x^7-x^10+x^15)/(1-x-x^5+x^6-x^7+x^8-x^9-2*x^10+x^11-x^12+2*x^15-x^16+2*x^17+x^20-x^25).
%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, 1;
%e 1, 0, 2, 0;
%e 1, 0, 4, 0, 1;
%e 1, 1, 6, 0, 3, 0;
%e 1, 2, 9, 0, 9, 0, 1;
%e 1, 3, 12, 5, 18, 0, 4, 0;
%e 1, 4, 16, 12, 36, 0, 16, 0, 1;
%e 1, 5, 20, 25, 60, 15, 40, 0, 5, 0;
%e 1, 6, 25, 42, 100, 42, 100, 0, 25, 0, 1;
%e 1, 7, 31, 66, 150, 112, 200, 35, 75, 0, 6, 0;
%e ...
%t T[n_,k_]:=If[k<0 || n<k, 0, T[n-1,k] + T[n-1,k-1] - T[n-2,k-1] + 2*T[n-2,k-2] + T[n-3,k-1] - T[n-3,k-2] -2*T[n-3,k-3] - T[n-4,k-1] + T[n-4,k-2] + T[n-4,k-3] - T[n-4,k-4] + T[n-5,k-1] - 2*T[n-5,k-3] + T[n-5,k-5] + KroneckerDelta[n,k,0] - KroneckerDelta[n,k,1] - KroneckerDelta[n,k,2] - KroneckerDelta[n,3]*KroneckerDelta[k,1] + KroneckerDelta[n,k,3]]; Flatten@Table[T[n,k], {n, 0, 11}, {k, 0, n}]
%t f[n_]:=If[n<0, 0, f[n-1]+x*f[n-5]+KroneckerDelta[n,0]]; T[n_, k_]:=Module[{j=Floor[(n+4*k)/2], r=Mod[n+4*k,2]}, Coefficient[f[j]^(2-r)*f[j+1]^r, x, k]]; Flatten@Table[T[n,k], {n, 0, 11}, {k, 0, n}]
%Y Row sums are A005578.
%Y Sums over k of T(n-4*k,k) are A224811.
%Y Other members of the family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354668 (m=3,t=3).
%Y Other triangles related to tiling using combs: A059259, A123521, A157897, A335964.
%K easy,nonn,tabl
%O 0,9
%A _Michael A. Allen_, Jun 05 2022