A136216 Triangle T, read by rows, where T(n,k) = A008544(n-k)*C(n,k) where A008544 equals the triple factorials in column 0.

1, 2, 1, 10, 4, 1, 80, 30, 6, 1, 880, 320, 60, 8, 1, 12320, 4400, 800, 100, 10, 1, 209440, 73920, 13200, 1600, 150, 12, 1, 4188800, 1466080, 258720, 30800, 2800, 210, 14, 1, 96342400, 33510400, 5864320, 689920, 61600, 4480, 280, 16, 1

Offset: 0

Author

Paul D. Hanna, Feb 07 2008

Keywords

• nonn
• tabl

Comments

This array is the particular case P(2,3) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown in the comments to A094587. - Peter Bala, Jul 10 2008

The row polynomials form an Appell sequence. - Tom Copeland, Dec 03 2013

Examples

			Triangle begins:
1;
2, 1;
10, 4, 1;
80, 30, 6, 1;
880, 320, 60, 8, 1;
12320, 4400, 800, 100, 10, 1;
209440, 73920, 13200, 1600, 150, 12, 1;
4188800, 1466080, 258720, 30800, 2800, 210, 14, 1; ...
		

Links

• Wikipedia, Appell sequence
• Wikipedia, Sheffer sequence

Crossrefs

Cf. A136215 (square-root), A112333, A008544, A136212, A136213.

Cf. A094587.

Programs

Mathematica

(* The function RiordanArray is defined in A256893. *)
RiordanArray[1/(1 - 3 #)^(2/3)&, #&, 9, True] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

PARI

{T(n,k) = binomial(n,k)*if(n-k==0,1, prod(j=0,n-k-1,3*j+2))}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

Formula

Column k of T = column 0 of V^(k+1) for k>=0 where V = A112333.

Equals the matrix square of triangle A136215.

T(n,k) = (3*n-3*k-1)*T(n-1,k) + T(n-1,k-1). - Peter Bala, Jul 10 2008

Using the formalism of A132382 modified for the triple rather than the double factorial (replace 2 by 3 in basic formulas), the e.g.f. for the row polynomials is exp(x*t)*(1-3x)^(-2/3). - Tom Copeland, Aug 18 2008

From Peter Bala, Aug 28 2013: (Start)

Exponential Riordan array [1/(1 - 3*y)^(2/3), y].

The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = sum {k = 0..n} binomial(n,k)*y^(n-k)*R(k,x).

Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = product {k = 0..n-1} (2*x + 3*k) with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (80, 30, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 5)*(2*x + 8) = 80 + 20*(2*x) + 6*(2*x*(2*x + 3)) + (2*x)*(2*x + 3)*(2*x + 6). (End)