A089465 -id:A089465 - cp's OEIS frontend

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

Paul D. Hanna, Nov 05 2003

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.

			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},..

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

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

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

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

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!.

A144304 Square array A(n,m), n>=0, m>=0, read by antidiagonals: A(n,m) = n-th number of the m-th iteration of the hyperbinomial transform on sequence A001858.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 7, 7, 1, 4, 14, 38, 38, 1, 5, 23, 93, 291, 291, 1, 6, 34, 178, 822, 2932, 2932, 1, 7, 47, 299, 1763, 9193, 36961, 36961, 1, 8, 62, 462, 3270, 21504, 125292, 561948, 561948, 1, 9, 79, 673, 5523, 43135, 313585, 2022555, 10026505, 10026505, 1

Offset: 0

Alois P. Heinz, Sep 17 2008

			Square array begins:
   1,   1,   1,    1,    1, ...
   1,   2,   3,    4,    5, ...
   2,   7,  14,   23,   34, ...
   7,  38,  93,  178,  299, ...
  38, 291, 822, 1763, 3270, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
N. J. A. Sloane, Transforms

Columns m=0-3 give: A001858, A001858(n+1), A089462, A089465.
Rows n=0-2 give: A000012, A000027, A008865(m+2).
Main diagonal gives A252727.

hymtr:= proc(p) proc(n,m) `if`(m=0, p(n), m*add(p(k) *binomial(n, k) *(n-k+m)^(n-k-1), k=0..n)) end end: f:= proc(n) option remember; add(add(binomial(m, j) *binomial(n-1, n-m-j) *n^(n-m-j) *(m+j)!/ (-2)^j/ m!, j=0..m), m=0..n) end: A:= hymtr(f): seq(seq(A(n, d-n), n=0..d), d=0..12);

hymtr[p_] := Function[{n, m}, If[m == 0, p[n], m*Sum[p[k]*Binomial[n, k]*(n-k+m)^(n-k-1), {k, 0, n}]]]; f[0] = 1; f[n_] := f[n] = Sum[Sum[Binomial[m, j]*Binomial[n-1, n-m-j]*n^(n-m-j)*(m+j)!/(-2)^j/m!, {j, 0, m}], {m, 0, n}]; A[0, ] = 1; A[1, k] := k+1; A[n_, m_] := hymtr[f][n, m]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

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

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