A073387 Convolution triangle of A002605(n) (generalized (2,2)-Fibonacci), n>=0.
1, 2, 1, 6, 4, 1, 16, 16, 6, 1, 44, 56, 30, 8, 1, 120, 188, 128, 48, 10, 1, 328, 608, 504, 240, 70, 12, 1, 896, 1920, 1872, 1080, 400, 96, 14, 1, 2448, 5952, 6672, 4512, 2020, 616, 126, 16, 1, 6688, 18192, 23040, 17856, 9352, 3444, 896, 160, 18, 1
Offset: 0
Examples
Lower triangular matrix, T(n,k), n >= k >= 0, else 0: 1; 2, 1; 6, 4, 1; 16, 16, 6, 1; 44, 56, 30, 8, 1; 120, 188, 128, 48, 10, 1; 328, 608, 504, 240, 70, 12, 1; 896, 1920, 1872, 1080, 400, 96, 14, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
- Wolfdieter Lang, First 10 rows.
Crossrefs
Programs
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Magma
A073387:= func< n,k | (&+[2^(n-k-j)*Binomial(n-j,k)*Binomial(n-k-j,j): j in [0..Floor((n-k)/2)]]) >; [A073387(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 03 2022
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Maple
T := (n,k) -> `if`(n=0,1,2^(n-k)*binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -2)): seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Apr 25 2016
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Mathematica
T[n_, k_]:=T[n,k]=Sum[2^(n-k-j)*Binomial[n-j,k]*Binomial[n-k-j,j], {j,0,(n-k)/2}]; Table[T[n,k], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 04 2019 *)
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SageMath
def A073387(n,k): return sum(2^(n-k-j)*binomial(n-j,k)*binomial(n-k-j,j) for j in range(((n-k+2)//2))) flatten([[A073387(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Oct 03 2022
Formula
T(n, k) = 2*(p(k-1, n-k)*(n-k+1)*T(n-k+1) + q(k-1, n-k)*(n-k+2)*T(n-k))/(k!*12^k), n >= k >= 1, with T(n) = T(n, k=0) = A002605(n), else 0; p(m, n) = Sum_{j=0..m} A(m, j)*n^(m-j) and q(m, n) = Sum_{j=0..m} B(m, j)*n^(m-j) with the number triangles A(k, m) = A073403(k, m) and B(k, m) = A073404(k, m).
T(n, k) = Sum_{j=0..floor((n-k)/2)} 2^(n-k-j)*binomial(n-j, k)*binomial(n-k-j, j) if n > k, else 0.
T(n, k) = ((n-k+1)*T(n, k-1) + 2*(n+k)*T(n-1, k-1))/(6*k), n >= k >= 1, T(n, 0) = A002605(n+1), else 0.
Sum_{k=0..n} T(n, k) = A007482(n).
G.f. for column m (without leading zeros): 1/(1-2*x*(1+x))^(m+1), m>=0.
T(n,k) = 2^(n-k)*binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -2) for n>=1. - Peter Luschny, Apr 25 2016
From G. C. Greubel, Oct 03 2022: (Start)
T(n, n-1) = A005843(n), n >= 1.
T(n, n-2) = 2*A005563(n-1), n >= 2.
T(n, n-3) = 4*A159920(n-1), n >= 2.
Sum_{k=0..n} (-1)^k*T(n, k) = A001045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A015518(n+1). (End)
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