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

0,2

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Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.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

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)

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 *)

Prog

(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

Crossrefs

Cf. A000110, A005493. A106436, A011965.

Sequence in context: A026661 A099014 A219672 * A271096 A269716 A192249

Adjacent sequences: A011963 A011964 A011965 * A011967 A011968 A011969

Keyword

nonn

Author

N. J. A. Sloane

Status

approved