A011965 Second differences of Bell numbers.
1, 2, 7, 27, 114, 523, 2589, 13744, 77821, 467767, 2972432, 19895813, 139824045, 1028804338, 7905124379, 63287544055, 526827208698, 4551453462543, 40740750631417, 377254241891064, 3608700264369193, 35613444194346451, 362161573323083920, 3790824599495473121
Offset: 0
References
- Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.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..570(terms 0..250 from Alois P. Heinz).
- Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841.
- Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, 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]
- Adam M. Goyt and Lara K. Pudwell, Avoiding colored partitions of two elements in the pattern sense, arXiv preprint arXiv:1203.3786 [math.CO], 2012.
- Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
Crossrefs
Programs
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Magma
[Bell(n+2) -2*Bell(n+1) + Bell(n): n in [0..40]]; // G. C. Greubel, Jan 07 2025
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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
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Mathematica
Differences[BellB[Range[0, 30]], 2] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
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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
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Python
# or Sagemath b=bell_number print([b(n+2) -2*b(n+1) +b(n) for n in range(41)]) # G. C. Greubel, Jan 07 2025
Formula
E.g.f.: exp(exp(x)-1)*(exp(2*x)-exp(x)+1). - Vladeta Jovovic, Feb 11 2003
a(n) = A000110(n) - 2*A000110(n-1) + A000110(n-2). - 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*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1 - G(0) where G(k) = 1 - 1/(1-k*x-2*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1 - 1/x + (1-x)^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)*(1-1/x) where G(k) = 1 - 1/(1-x*(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 07 2013
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - 2*LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
Conjecture: a(n) = Sum_{k=0..2^n - 1} b(k) for n >= 0 where b(2n+1) = b(n) + b(A025480(n-1)), b(2n) = b(n - 2^f(n)) + b(2n - 2^f(n)) + b(A025480(n-1)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). Also b((4^n - 1)/3) = A141154(n+1). - Mikhail Kurkov, Jan 27 2022
Comments