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A005493
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2-Bell numbers: a(n) = number of partitions of [n+1] with a distinguished block.
(Formerly M2851)
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87
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1, 3, 10, 37, 151, 674, 3263, 17007, 94828, 562595, 3535027, 23430840, 163254885, 1192059223, 9097183602, 72384727657, 599211936355, 5150665398898, 45891416030315, 423145657921379, 4031845922290572, 39645290116637023, 401806863439720943, 4192631462935194064
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OFFSET
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0,2
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COMMENTS
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Number of Boolean sublattices of the Boolean lattice of subsets of {1..n}.
a(n) = p(n+1) where p(x) is the unique degree n polynomial such that p(k) = A000110(k+1) for k = 0, 1, ..., n. - Michael Somos, Oct 07 2003
With offset 1, number of permutations beginning with 12 and avoiding 21-3.
Number of blocks in all set partitions of an (n+1)-set. Example: a(2)=10 because the set partitions of {1,2,3} are 1|2|3, 1|23, 12|3, 13|2 and 123, with a total of 10 blocks. - Emeric Deutsch, Nov 13 2006
Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 2. - Olivier Gérard, Oct 29 2007
See page 29, Theorem 5.6 of my paper on the arXiv: These numbers are the dimensions of the homogeneous components of the operad called ComTrip associated with commutative triplicial algebras. (Triplicial algebras are related to even trees and also to L-algebras, see A006013.) - Philippe Leroux, Nov 17 2007
Number of set partitions of (n+2) elements where two specific elements are clustered separately. Example: a(1)=3 because 1/2/3, 1/23, 13/2 are the 3 set partitions with 1, 2 clustered separately. - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
Equals A008277 * [1,2,3,...], i.e., the product of the Stirling number of the second kind triangle and the natural number vector. a(n+1) = row sums of triangle A137650. - Gary W. Adamson, Jan 31 2008
Number of embedded coalitions in an (n+1)-person game. - David Yeung (wkyeung(AT)hkbu.edu.hk), May 08 2008
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REFERENCES
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Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation. Unpublished as of 2017.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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A. M. Odlyzko, Asymptotic enumeration methods, pp. 1063-1229 of R. L. Graham et al., eds., Handbook of Combinatorics, 1995; see Example 12.16 (pdf, ps)
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FORMULA
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a(n-1) = Sum_{k=1..n} k*Stirling2(n, k) for n>=1.
E.g.f.: exp(exp(x) + 2*x - 1). First differences of Bell numbers (if offset 1). - Michael Somos, Oct 09 2002
G.f.: Sum_{k>=0} (x^k/Product_{l=1..k} (1-(l+1)x)). - Ralf Stephan, Apr 18 2004
a(n) = Sum_{i=0..n} 2^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n). - Fred Lunnon, Aug 04 2007 [Written umbrally, a(n) = (B+2)^n. - N. J. A. Sloane, Feb 07 2009]
Representation as an infinite series: a(n-1) = Sum_{k>=2} (k^n*(k-1)/k!)/exp(1), n=1, 2, ... This is a Dobinski-type summation formula. - Karol A. Penson, Mar 14 2002
Recurrence: a(n+1) = 1 + Sum_{j=1..n} (1+binomial(n, j))*a(j). - Jon Perry, Apr 25 2005
Binomial transform of Bell numbers 1, 2, 5, 15, 52, 203, 877, 4140, ... (see A000110).
Define f_1(x), f_2(x), ... such that f_1(x)=x*e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n-1) = e^(-1)*f_n(1). - Milan Janjic, May 30 2008
Representation of numbers a(n), n=0,1..., as special values of hypergeometric function of type (n)F(n), in Maple notation: a(n)=exp(-1)*2^n*hypergeom([3,3...3],[2.2...2],1), n=0,1..., i.e., having n parameters all equal to 3 in the numerator, having n parameters all equal to 2 in the denominator and the value of the argument equal to 1. Examples: a(0)= 2^0*evalf(hypergeom([],[],1)/exp(1))=1 a(1)= 2^1*evalf(hypergeom([3],[2],1)/exp(1))=3 a(2)= 2^2*evalf(hypergeom([3,3],[2,2],1)/exp(1))=10 a(3)= 2^3*evalf(hypergeom([3,3,3],[2,2,2],1)/exp(1))=37 a(4)= 2^4*evalf(hypergeom([3,3,3,3],[2,2,2,2],1)/exp(1))=151 a(5)= 2^5*evalf(hypergeom([3,3,3,3,3],[2,2,2,2,2],1)/exp(1)) = 674. - Karol A. Penson, Sep 28 2007
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i <= j), and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = (-1)^(n)charpoly(A,-2). - Milan Janjic, Jul 08 2010
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x*(k+3) - x^2*(k+1)/U(k+1).
G.f.: 1/(U(0)-x) where U(k) = 1 - x - x*(k+1)/(1 - x/U(k+1)).
G.f.: G(0)/(1-x) 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) )).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*x-k*x)/(1-x/(x-1/G(k+1) )).
G.f.: -G(0)/x where G(k) = 1 - 1/(1-k*x-x)/(1-x/(x-1/G(k+1) )).
G.f.: 1 - 2/x + (1/x-1)*S where S = sum(k>=0, ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k / prod(i=0..k-1, (1-x-x*i)/(1-x) ) ).
G.f.: (1-x)/x/(G(0)-x) - 1/x where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ).
G.f.: (1/G(0) - 1)/x^3 where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )).
G.f.: 1/Q(0), where Q(k)= 1 - 2*x - x/(1 - x*(k+1)/Q(k+1)).
G.f.: G(0)/(1-3*x), where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1 - x*(k+3))*(1 - x*(k+4))/G(k+1) ). (End)
a(n) ~ exp(n/LambertW(n) + 3*LambertW(n)/2 - n - 1) * n^(n + 1/2) / LambertW(n)^(n+1). - Vaclav Kotesovec, Jun 09 2020
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Ilya Gutkovskiy, Jul 02 2020
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
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EXAMPLE
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For example, a(1) counts (12), (1)-2, 1-(2) where dashes separate blocks and the distinguished block is parenthesized.
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MAPLE
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with(combinat): seq(bell(n+2)-bell(n+1), n=0..22); # Emeric Deutsch, Nov 13 2006
seq(add(binomial(n, k)*(bell(n-k)), k=1..n), n=1..23); # Zerinvary Lajos, Dec 01 2006
a := [seq(3, i=1..n)]; b := [seq(2, i=1..n)];
2^n*exp(-x)*hypergeom(a, b, x); round(evalf(subs(x=1, %), 66)) end:
BT := proc(n, k) option remember; if n = 0 and k = 0 then 1
elif k = n then BT(n-1, 0) else BT(n, k+1)+BT(n-1, k) fi end:
A005493 := n -> add(BT(n, k), k=0..n):
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MATHEMATICA
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a=Exp[x]-1; Rest[CoefficientList[Series[a Exp[a], {x, 0, 20}], x] * Table[n!, {n, 0, 20}]]
a[ n_] := If[ n<0, 0, With[ {m = n+1}, m! SeriesCoefficient[ # Exp@# &[ Exp@x - 1], {x, 0, m}]]]; (* Michael Somos, Nov 16 2011 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) + 2*x - 1), n))}; /* Michael Somos, Oct 09 2002 */
(PARI) {a(n) = if( n<0, 0, n+=2; subst( polinterpolate( Vec( serlaplace( exp( exp( x + O(x^n)) - 1) - 1))), x, n))}; /* Michael Somos, Oct 07 2003 */
(Python)
# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A005493_list, blist, b = [], [1], 1
for _ in range(1001):
blist = list(accumulate([b]+blist))
b = blist[-1]
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CROSSREFS
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Cf. A000110, A005494, A008277, A011971, A011968, A049020, A106436, A124323, A137650, A152433, A159573.
A row or column of the array A108087.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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