A171224 Riordan array (f(x),x*f(x)) where f(x) is the g.f. of A117641.

1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 11, 6, 3, 0, 1, 42, 23, 9, 4, 0, 1, 167, 90, 36, 12, 5, 0, 1, 684, 365, 144, 50, 15, 6, 0, 1, 2867, 1518, 595, 204, 65, 18, 7, 0, 1, 12240, 6441, 2511, 858, 270, 81, 21, 8, 0, 1, 53043, 27774, 10782, 3672, 1155, 342, 98, 24, 9, 0, 1

Offset: 0

Author

Philippe Deléham, Dec 05 2009

Keywords

• nonn
• tabl

Examples

			Triangle begins
    1;
    0,  1;
    1,  0,  1;
    3,  2,  0,  1;
   11,  6,  3,  0,  1;
   42, 23,  9,  4,  0,  1;
  167, 90, 36, 12,  5,  0,  1;
  ...
Production array begins
    0,  1;
    1,  0,  1;
    3,  1,  0,  1;
    9,  3,  1,  0,  1;
   27,  9,  3,  1,  0,  1;
   81, 27,  9,  3,  1,  0,  1;
  243, 81, 27,  9,  3,  1,  0,  1;
  ... - _Philippe Deléham_, Mar 04 2013
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened

Crossrefs

Cf. A053121, A097609, A065600.

Programs

Magma

[[((k+1)/(n+1))*(&+[3^(n-k-2*j)*Binomial(n+1,j)*Binomial(n-k-j-1, n-k-2*j): j in [0..Floor((n-k)/2)]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 04 2019

Mathematica

T[n_, k_]:= (k+1)/(n+1)*Sum[3^(n-k-2*j)*Binomial[n+1,j]*Binomial[n-k-j-1, n-k-2*j], {j, 0, Floor[(n-k)/2]}]; Table[T[n, k], {n,0,10}, {k,0,n} ]//Flatten (* G. C. Greubel, Apr 04 2019 *)

Maxima

T(n,k):=(k+1)/(n+1)*sum(3^(n-k-2*j)*binomial(n+1,j)*binomial(n-k-j-1,n-k-2*j),j,0,floor((n-k)/2)); /* Vladimir Kruchinin, Apr 04 2019 */

PARI

{T(n,k) = ((k+1)/(n+1))*sum(j=0, floor((n-k)/2), 3^(n-k-2*j) *binomial(n+1,j)*binomial(n-k-j-1, n-k-2*j))}; \\ G. C. Greubel, Apr 04 2019

Sage

[[((k+1)/(n+1))*sum(3^(n-k-2*j)*binomial(n+1,j)*binomial(n-k-j-1, n-k-2*j) for j in (0..floor((n-k)/2))) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 04 2019

Formula

Sum_{k=0..n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n+1), A002212(n+1), A026378(n+1) for x = 0, 1, 2, 3, 4 respectively.

Triangle equals B*A065600*B^(-1) = B^2*A097609*B^(-2) = B^3*A053121*B^(-3), product considered as infinite lower triangular arrays and B = A007318. - Philippe Deléham, Dec 08 2009

T(n,k) = T(n-1,k-1) + Sum_{i>=0} T(n-1,k+1+i)*3^i, T(0,0) = 1. - Philippe Deléham, Feb 23 2012

T(n,k) = ((k+1)/(n+1))*Sum_{j=0..floor((n-k)/2)} 3^(n-k-2*j)*C(n+1,j)*C(n-k-j-1,n-k-2*j). - Vladimir Kruchinin, Apr 04 2019

Extensions

Terms a(55) onward added by G. C. Greubel, Apr 04 2019