A089463 Triangle, read by rows, of coefficients for the third iteration of the hyperbinomial transform.

1, 3, 1, 15, 6, 1, 108, 45, 9, 1, 1029, 432, 90, 12, 1, 12288, 5145, 1080, 150, 15, 1, 177147, 73728, 15435, 2160, 225, 18, 1, 3000000, 1240029, 258048, 36015, 3780, 315, 21, 1, 58461513, 24000000, 4960116, 688128, 72030, 6048, 420, 24, 1, 1289945088

Offset

0,2

Comments

Equals the matrix cube of A088956 when treated as a lower triangular matrix. The 3rd hyperbinomial transform of a sequence {b} is defined to be the sequence {d} given by d(n) = sum(k=0..n, T(n,k)*b(k)), where T(n,k) = 3*(n-k+3)^(n-k-1)*C(n,k). Given a table in which the n-th row is the n-th binomial transform of the first row, then the 3rd hyperbinomial transform of any diagonal results in the 3rd diagonal lower in the table.

Links

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Formula

T(n, k) = 3*(n-k+3)^(n-k-1)*C(n, k).

E.g.f.: exp(x*y)*(-LambertW(-y)/y)^3.

Note: (-LambertW(-y)/y)^3 = sum(n>=0, 3*(n+3)^(n-1)*y^n/n!).

Example

Rows begin:
{1},
{3,1},
{15,6,1},
{108,45,9,1},
{1029,432,90,12,1},
{12288,5145,1080,150,15,1},
{177147,73728,15435,2160,225,18,1},
{3000000,1240029,258048,36015,3780,315,21,1},..

Mathematica

Flatten[Table[3(n-k+3)^(n-k-1) Binomial[n, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jun 26 2013 *)

Prog

(PARI) for(n=0, 10, for(k=0, n, print1(3*(n-k+3)^(n-k-1)*binomial(n, k), ", "))) \\ G. C. Greubel, Nov 17 2017

Crossrefs

Cf. A089464(row sums), A089465(diagonal), A089460, A088956.

Sequence in context: A048966 A297704 A104990 * A136231 A113389 A038553

Adjacent sequences: A089460 A089461 A089462 * A089464 A089465 A089466

Keyword

nonn,tabl

Author

Paul D. Hanna, Nov 05 2003

Status

approved