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.

A129247 Invert transform of the Bell numbers.

Original entry on oeis.org

1, 1, 3, 10, 36, 138, 560, 2402, 10898, 52392, 267394, 1450790, 8371220, 51327178, 333759746, 2295276480, 16639104002, 126718172670, 1010487248556, 8411744415418, 72899055533482, 656136245454232, 6120474697035762
Offset: 0

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Author

Thomas Wieder, May 10 2008

Keywords

Comments

The following definition of the invert transform appears in [M. Bernstein & N. J. A. Sloane, Some canonical sequences of integers, Linear Algebra and its Applications, 226-228 (1995), 57-72]: "b_n is the number of ordered arrangements of postage stamps of total value n that can be formed if we have a_i types of stamps of value i, i >= 1."
Hankel transform is A000178. - Paul Barry, Jan 08 2009
Equals INVERT transform of the Bell sequence starting with offset 1: (1, 2, 5, ...), while A137551 = INVERT transform of the Bell sequence starting with offset 0: (1, 1, 2, 5, 15, 52, ...). - Gary W. Adamson, May 24 2009

Examples

			We have Bell(i) types of an integer i with i=1,2,...,n, where Bell(i) is the i-th Bell number.
We write i_j for integer i of type j.
a(2)=3 because of the 3 ordered arrangements
  {1_1,1_1}
  {2_1}, {2_2}.
a(3)=10 because of the 10 ordered arrangements
  {1_1,1_1,1_1},
  {1_1,2_1}, {2_1,1_1},
  {1_1,2_2}, {2_2,1_1}
  {3_1}, {3_2}, {3_3}, {3_4}, {3_5}.
		

Crossrefs

Programs

  • Maple
    A129247 := proc(n) option remember ; local i ; if n <= 1 then 1 ; else add(combinat[bell](i)*procname(n-i),i=1..n) ; fi ; end: for n from 0 to 40 do printf("%d,",A129247(n)) ; od: # R. J. Mathar, Aug 25 2008
  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[BellB[i]*a[n - i], {i, 1, n}];
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 09 2017 *)

Formula

a(n) = Sum_{i=1..n} Bell(i)*a(n-i).
G.f.: 1/(U(0) - 2*x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Nov 12 2012
G.f.: 1/( Q(0) - 2*x ) where Q(k) = 1 + x/(x*k - 1 )/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: 1/(Q(0) - x), where Q(k) = 1 - x - x/(1 - x*(2*k+1)/(1 - x - x/(1 - 2*x*(k+1)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, May 12 2013

Extensions

Extended by R. J. Mathar, Aug 25 2008
a(0)=1 prepended by Alois P. Heinz, Sep 22 2017