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A350111
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Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,3)-fences and n-k squares.
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7
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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
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OFFSET
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0,17
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COMMENTS
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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).
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LINKS
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FORMULA
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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.
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EXAMPLE
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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;
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MATHEMATICA
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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}]
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CROSSREFS
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Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).
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KEYWORD
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AUTHOR
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STATUS
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approved
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