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

Author

Philippe Deléham, Feb 09 2014

Keywords

• nonn
• tabl

Comments

Riordan array (f(x)/x, f(x)) where f(x) is the g.f. of A000958.

Reversal of A236918.

Row sums are A109262(n+1).

Diagonal sums are A033297(n+2).

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

Columns give A000958, A114495.

Cf. A033297 (diagonal sums), A109262 (row sums), A236918 (row reversal).

Programs

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

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 *)

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)