

A011965


Second differences of Bell numbers.


15



1, 2, 7, 27, 114, 523, 2589, 13744, 77821, 467767, 2972432, 19895813, 139824045, 1028804338, 7905124379, 63287544055, 526827208698, 4551453462543, 40740750631417, 377254241891064, 3608700264369193, 35613444194346451, 362161573323083920, 3790824599495473121
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OFFSET

0,2


COMMENTS

Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 3. Alternatively, number of partitions of n+3 with at least one singleton and with the largest element in a singleton equal to n+1.  Olivier GERARD, Oct 29 2007
Out of the A005493(n) set partitions with a specific two elements clustered separately, number that have a different set of two elements clustered separately.  Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007


REFERENCES

Olivier GĂ©rard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.


LINKS

Chai Wah Wu, Table of n, a(n) for n = 0..1000 n = 0..250 from Alois P. Heinz.
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, 782785. 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, 782785. MR1531841. [Annotated scanned copy]
Adam M. Goyt and Lara K. Pudwell, Avoiding colored partitions of two elements in the pattern sense, arXiv preprint arXiv:1203.3786, 2012.
Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.


FORMULA

a(n) = A005493(n)A005493(n1).
E.g.f.: exp(exp(x)1)*(exp(2*x)exp(x)+1).  Vladeta Jovovic, Feb 11 2003
a(n) = A000110(n)  2*A000110(n1) + A000110(n2).  Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
G.f.: G(0) where G(k) = 1  2*x*(k+1)/((2*k+1)*(2*x*k+2*x1)  x*(2*k+1)*(2*k+3)*(2*x*k+2*x1)/(x*(2*k+3)  2*(k+1)*(2*x*k+3*x1)/G(k+1) )); (recursively defined continued fraction).  Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1  G(0) where G(k) = 1  1/(1k*x2*x)/(1x/(x1/G(k+1) )); (recursively defined continued fraction).  Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1  1/x + (1x)^2/x/(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
G.f.: G(0)*(11/x) where G(k) = 1  1/(1x*(k+1))/(1x/(x1/G(k+1) )); (recursively defined continued fraction).  Sergei N. Gladkovskii, Feb 07 2013


MAPLE

a:= n> add((1)^k*binomial(2, k)*combinat['bell'](n+k), k=0..2): seq(a(n), n=0..20); # Alois P. Heinz, Sep 05 2008


MATHEMATICA

Differences[BellB[Range[0, 30]], 2] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)


PROG

(Python)
# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A011965_list, blist, b = [1], [1, 2], 2
for _ in range(1000):
....blist = list(accumulate([b]+blist))
....b = blist[1]
....A011965_list.append(blist[3])
# Chai Wah Wu, Sep 02 2014


CROSSREFS

Cf. A000110, A005493, A106436.
Sequence in context: A127897 A180473 A154108 * A150629 A150630 A150631
Adjacent sequences: A011962 A011963 A011964 * A011966 A011967 A011968


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane.


STATUS

approved



