cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A089465 3rd hyperbinomial transform of A001858; also the hyperbinomial transform of A089462.

Original entry on oeis.org

1, 4, 23, 178, 1763, 21504, 313585, 5342068, 104376201, 2304582544, 56807530871, 1547599725720, 46202052688603, 1500629138909632, 52697989385197137, 1990117967149595824, 80440669725095395025, 3465573101368534916928
Offset: 0

Views

Author

Paul D. Hanna, Nov 05 2003

Keywords

Comments

A001858 enumerates forests of labeled trees with n nodes and shifts 1 place left under the hyperbinomial transform.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[Binomial[m, j]*Binomial[n, n - m - j + 1]*(n + 3)^(n - m - j + 1)*(m + j)!/(-2)^j, {j, 0, m}]/m!, {m, 0, n + 1}], {n, 0, 50}] (* G. C. Greubel, Nov 18 2017 *)
  • PARI
    a(n)=if(n<0,0,sum(m=0,n+1,sum(j=0,m,binomial(m,j)*binomial(n,n-m-j+1)*(n+3)^(n-m-j+1)*(m+j)!/(-2)^j)/m!))

Formula

a(n) = Sum_{k=0..n} 3*(n-k+3)^(n-k-1)*C(n, k)*A001858(k).
a(n) = Sum_{m=0..(n+1)} ( Sum_{j=0..m} C(m, j)*C(n, n-m-j+1)*(n+3)^(n-m-j+1)*(m+j)!/(-2)^j )/m!.
a(n) ~ 3 * exp(7/2) * n^(n-1). - Vaclav Kotesovec, Oct 11 2020

A089460 Triangle, read by rows, of coefficients for the second iteration of the hyperbinomial transform.

Original entry on oeis.org

1, 2, 1, 8, 4, 1, 50, 24, 6, 1, 432, 200, 48, 8, 1, 4802, 2160, 500, 80, 10, 1, 65536, 28812, 6480, 1000, 120, 12, 1, 1062882, 458752, 100842, 15120, 1750, 168, 14, 1, 20000000, 8503056, 1835008, 268912, 30240, 2800, 224, 16, 1, 428717762, 180000000, 38263752, 5505024, 605052, 54432, 4200, 288, 18, 1
Offset: 0

Views

Author

Paul D. Hanna, Nov 05 2003

Keywords

Comments

Equals the matrix square of A088956 when treated as a lower triangular matrix. The 2nd 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) = 2*(n-k+2)^(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 2nd hyperbinomial transform of any diagonal results in the diagonal located 2 diagonals lower in the table.

Examples

			Rows begin:
  {1},
  {2,1},
  {8,4,1},
  {50,24,6,1},
  {432,200,48,8,1},
  {4802,2160,500,80,10,1},
  {65536,28812,6480,1000,120,12,1},
  {1062882,458752,100842,15120,1750,168,14,1},..
		

Crossrefs

Cf. A089461(row sums), A089462(diagonal), A089463, A088956.

Programs

  • Mathematica
    Join[{1}, Table[Binomial[n, k]*2*(n - k + 2)^(n - k - 1), {n, 1, 49}, {k, 0, n}]] // Flatten (* G. C. Greubel, Nov 18 2017 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(2*(n-k+2)^(n-k-1)*binomial(n,k), ", "))) \\ G. C. Greubel, Nov 18 2017

Formula

T(n, k) = 2*(n-k+2)^(n-k-1)*C(n, k).
E.g.f.: exp(x*y)*(-LambertW(-y)/y)^2.
Note: (-LambertW(-y)/y)^2 = Sum_{n>=0} 2*(n+2)^(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

Views

Author

Alois P. Heinz, Sep 17 2008

Keywords

Examples

			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, ...
		

Crossrefs

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.

Programs

  • Maple
    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);
  • Mathematica
    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 *)
Showing 1-3 of 3 results.