A237596 Convolution triangle of A000958(n+1).
1, 1, 1, 3, 2, 1, 8, 7, 3, 1, 24, 22, 12, 4, 1, 75, 73, 43, 18, 5, 1, 243, 246, 156, 72, 25, 6, 1, 808, 844, 564, 283, 110, 33, 7, 1, 2742, 2936, 2046, 1092, 465, 158, 42, 8, 1, 9458, 10334, 7449, 4178, 1906, 714, 217, 52, 9, 1, 33062, 36736, 27231, 15904, 7670, 3096, 1043, 288, 63, 10, 1
Offset: 0
Examples
Triangle begins: 1; 1, 1; 3, 2, 1; 8, 7, 3, 1; 24, 22, 12, 4, 1; 75, 73, 43, 18, 5, 1; 243, 246, 156, 72, 25, 6, 1; 808, 844, 564, 283, 110, 33, 7, 1; ...
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
- Sergio Falcon, Catalan transform of the K-Fibonacci sequence, Commun. Korean Math. Soc. 28 (2013), No. 4, pp. 827-832.
Crossrefs
Programs
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Maple
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left. PMatrix(10, n -> A000958(n)); # Peter Luschny, Oct 19 2022
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Mathematica
P[n_, x_]:= P[n, x]= If[n==0, 1, Sum[(j/(2*n-j))*Binomial[2*n-j, n-j]*Fibonacci[j, x], {j,0,n}]]; T[n_, k_] := Coefficient[P[n+1, x], x, k]; Table[T[n, k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 14 2022 *)
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SageMath
def f(n,x): return sum( binomial(n-j-1, j)*x^(n-2*j-1) for j in (0..(n-1)//2) ) def p(n,x): if (n==0): return 1 else: return sum( (j/(2*n-j))*binomial(2*n-j, n-j)*f(j, x) for j in (0..n) ) def A237596(n,k): return ( p(n+1,x) ).series(x, n+1).list()[k] flatten([[A237596(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 14 2022
Formula
G.f. for the column k-1: ((1-sqrt(1-4*x))^k/(1+sqrt(1-4*x) + 2*x)^k)/x.
Sum_{k=0..n} T(n,k) = A109262(n+1).
From G. C. Greubel, Jun 14 2022: (Start)
T(n, k) = coefficient of [x^k]( p(n+1, x) ), where p(n, x) = Sum_{j=0..n} (j/(2*n-j))*binomial(2*n-j, n-j)*Fibonacci(j, x) with p(0, x) = 1 and Fibonacci(n, x) are the Fibonacci polynomials.
T(n, k) = A236918(n, n-k). (End)
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