A089464 -id:A089464 - cp's OEIS frontend

A144303 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 the sequence of 1's.

1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 29, 1, 1, 5, 22, 81, 212, 1, 1, 6, 33, 163, 689, 2117, 1, 1, 7, 46, 281, 1564, 7553, 26830, 1, 1, 8, 61, 441, 2993, 18679, 101961, 412015, 1, 1, 9, 78, 649, 5156, 38705, 268714, 1639529, 7433032, 1, 1, 10, 97, 911, 8257, 71801, 592489, 4538209, 30640257, 154076201, 1

Offset: 0

Alois P. Heinz, Sep 17 2008, revised Oct 30 2012

See A088956 for the definition of the hyperbinomial transform.

A(n,m), n>=0, m>=0, is the number of subtrees of the complete graph K_{n+m} on n+m vertices containing a given, fixed subtree on m vertices. - Alex Chin, Jul 25 2013

			Square array begins:
  1,     1,      1,      1,      1,       1,       1, ...
  1,     2,      3,      4,      5,       6,       7, ...
  1,     6,     13,     22,     33,      46,      61, ...
  1,    29,     81,    163,    281,     441,     649, ...
  1,   212,    689,   1564,   2993,    5156,    8257, ...
  1,  2117,   7553,  18679,  38705,   71801,  123217, ...
  1, 26830, 101961, 268714, 592489, 1166886, 2120545, ...

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

Columns m=0-10 give: A000012, A088957, A089461, A089464, A218496, A218497, A218498, A218499, A218500, A218501, A218502.
Rows n=0-2 give: A000012, A000027, A028872.
Main diagonal gives A252766.
Cf. A007318, A088956.

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:
A:= hymtr(1):
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[, 0] = 1; a[n, k_] := Sum[k*(n - j + k)^(n - j - 1)*Binomial[n, j], {j, 0, n}]; Table[a[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jun 24 2013 *)

E.g.f. of column k: exp(x) * (-LambertW(-x)/x)^k.

A(n,k) = Sum_{j=0..n} k * (n-j+k)^(n-j-1) * C(n,j).

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

Original entry on oeis.org

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!.

A362523 a(n) = n! * Sum_{k=0..floor(n/3)} (k+1)^(k-1) / (k! * (n-3*k)!).

Original entry on oeis.org

1, 1, 1, 7, 25, 61, 1201, 7771, 30577, 1058905, 9904321, 53722351, 2708688841, 33126146197, 228967340785, 15262865820931, 230517745701601, 1936173471789361, 161021598306402817, 2894434429492525015, 28614958982310290041

Offset: 0

Seiichi Manyama, Apr 23 2023

Seiichi Manyama, Table of n, a(n) for n = 0..427
Eric Weisstein's World of Mathematics, Lambert W-Function.

Cf. A088957, A362522.

Cf. A089464, A362348.

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x-lambertw(-x^3))))

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

a(n) ~ sqrt(3) * (exp(3*exp(-1/3)/2) + 2*cos(sqrt(3)*exp(-1/3)/2 - 2*Pi*n/3)) * n^(n-1) / exp(2*n/3 + exp(-1/3)/2 - 1). - Vaclav Kotesovec, Aug 05 2025

A218496 4th iteration of the hyperbinomial transform on the sequence of 1's.

Original entry on oeis.org

1, 5, 33, 281, 2993, 38705, 592489, 10516441, 212841889, 4845154913, 122664558905, 3421333467689, 104297273041969, 3451364116327249, 123251578626936841, 4725537745859375705, 193647372258547916609, 8447809104669814884545, 390938955429073736493145

Offset: 0

Alois P. Heinz, Oct 30 2012

nonn

See A088956 for the definition of the hyperbinomial transform.

Alois P. Heinz, Table of n, a(n) for n = 0..150

Column k=4 of A144303.

a:= n-> add(4*(n-j+4)^(n-j-1)*binomial(n,j), j=0..n):
seq (a(n), n=0..20);

E.g.f.: exp(x) * (-LambertW(-x)/x)^4.
a(n) = Sum_{j=0..n} 4 * (n-j+4)^(n-j-1) * C(n,j).
Hyperbinomial transform of A089464.
a(n) ~ 4*exp(4+exp(-1))*n^(n-1). - Vaclav Kotesovec, Aug 16 2013

cp's OEIS Frontend

A144303 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 the sequence of 1's.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A362523 a(n) = n! * Sum_{k=0..floor(n/3)} (k+1)^(k-1) / (k! * (n-3*k)!).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A218496 4th iteration of the hyperbinomial transform on the sequence of 1's.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula