A011966 Third differences of Bell numbers.

1, 5, 20, 87, 409, 2066, 11155, 64077, 389946, 2504665, 16923381, 119928232, 888980293, 6876320041, 55382419676, 463539664643, 4024626253845, 36189297168874, 336513491259647, 3231446022478129, 32004743929977258, 326548129128737469, 3428663026172389201

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Number of partitions of n+4 with at least one singleton and with the smallest element in a singleton equal to 4. Alternatively, number of partitions of n+4 with at least one singleton and with the largest element in a singleton equal to n+1. - Olivier GERARD, Oct 29 2007

References

• Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..215
• Cohn, Martin; Even, Shimon; Menger, Karl, Jr.; Hooper, Philip K.; On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841.
• Cohn, Martin; Even, Shimon; Menger, Karl, Jr.; Hooper, Philip K.; On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841. [Annotated scanned copy]
• Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.

Crossrefs

Cf. A000110, A005493. A106436, A011965.

Programs

Maple

a:= n-> add((-1)^(k+1)*binomial(3,k)*combinat['bell'](n+k), k=0..3):
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 05 2008

Mathematica

Differences[BellB[Range[0,30]],3]  (* Harvey P. Dale, Apr 21 2011 *)

Python

# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A011966_list, blist, b = [1], [2, 3, 5], 5
for _ in range(1000):
    blist = list(accumulate([b]+blist))
    b = blist[-1]
    A011966_list.append(blist[-4]) # Chai Wah Wu, Sep 20 2014

Formula

G.f.: -(1-x+x^2)/x^2 + (1-x)^3/x^2/(G(0)-x) where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013

From Vaclav Kotesovec, Jul 28 2021: (Start)

a(n) = Bell(n+3) - 3*Bell(n+2) + 3*Bell(n+1) - Bell(n).

a(n) ~ n^3 * Bell(n) / LambertW(n)^3 * (1 - 3*LambertW(n)/n). (End)