A236918 Triangle read by rows: Catalan triangle of the k-Fibonacci sequence.

1, 1, 1, 1, 2, 3, 1, 3, 7, 8, 1, 4, 12, 22, 24, 1, 5, 18, 43, 73, 75, 1, 6, 25, 72, 156, 246, 243, 1, 7, 33, 110, 283, 564, 844, 808, 1, 8, 42, 158, 465, 1092, 2046, 2936, 2742, 1, 9, 52, 217, 714, 1906, 4178, 7449, 10334, 9458, 1, 10, 63, 288, 1043, 3096, 7670, 15904, 27231, 36736, 33062

Offset: 1

Author

N. J. A. Sloane, Feb 09 2014

Keywords

• nonn
• tabl

Comments

Reversal of the Riordan array (f(x), x*f(x)) where f(x) is the g.f. of A000958. - Philippe Deléham, Feb 10 2014

Row sums are in A109262. - Philippe Deléham, Feb 10 2014

Examples

			Triangle begins:
  1;
  1,  1;
  1,  2,  3;
  1,  3,  7,   8;
  1,  4, 12,  22,   24;
  1,  5, 18,  43,   73,   75;
  1,  6, 25,  72,  156,  246,  243;
  1,  7, 33, 110,  283,  564,  844,   808;
  1,  8, 42, 158,  465, 1092, 2046,  2936,  2742;
  1,  9, 52, 217,  714, 1906, 4178,  7449, 10334,  9458;
  1, 10, 63, 288, 1043, 3096, 7670, 15904, 27231, 36736, 33062;
  ... - Extended by _Philippe Deléham_, Feb 10 2014
		

Links

• G. C. Greubel, Rows n = 1..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

Diagonals give A000958, A114495.

Cf. A109262 (row sums).

Programs

Mathematica

P[n_, x_]:= P[n,x]= If[n==0, 1, Sum[(j/(2*n-j))*Binomial[2*n-j, n-j]*Fibonacci[j, 1/x] *x^(n-1), {j,0,n}]];
T[n_, k_]:= Coefficient[P[n, x], x, k];
Table[T[n, k], {n,10}, {k, 0, n-1}]//Flatten (* G. C. Greubel, Jun 14 2022 *)

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)*x^(n-1)*f(j, 1/x) for j in (0..n) )
def A236918(n,k): return ( p(n,x) ).series(x, n+1).list()[k]
flatten([[A236918(n,k) for k in (0..n-1)] for n in (1..12)]) # G. C. Greubel, Jun 14 2022

Formula

T(n, k) = coefficient of [x^k]( p(n, x) ), where p(n, x) = Sum_{j=0..n} (j/(2*n-j))*binomial(2*n-j, n-j)*x^(n-1)*Fibonacci(j, 1/x) with p(0, x) = 1 and Fibonacci(n, x) are the Fibonacci polynomials. - G. C. Greubel, Jun 14 2022

Extensions

More terms from Philippe Deléham, Feb 09 2014