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A011969
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Apply (1+Shift)^2 to Bell numbers.
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5
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1, 3, 5, 10, 27, 87, 322, 1335, 6097, 30304, 162409, 931667, 5686712, 36750201, 250401793, 1792401626, 13436958559, 105208112643, 858286687914, 7279760687179, 64071719451645, 584150874832552, 5508179528996197
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OFFSET
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0,2
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COMMENTS
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Starting with n=2 (a(2)=5), number of set partitions of n+2 with at least one singleton and the smallest element in any singleton is exactly n-1. The maximum number of singletons is therefore 4. Alternatively, starting with n=2, number of set partitions of n+2 with at least one singleton and the largest element in any singleton is exactly 4. E.g. a(3)=10 counts the following set partitions of [5]: {1345, 2}, {13, 2, 45}, {145, 2, 3}, {134, 2, 5}, {15, 2, 34}, {135, 2, 4}, {14, 2, 35}, {13, 2, 4, 5}, {14, 2, 3, 5}, {15, 2, 3, 4}. - Olivier Gérard, Oct 29 2007
Let V(N)={v(1),v(2),...,v(N)} denote an ordered set of increasing positive integers containing 2 pairs of adjacent elements that differ by at least 2, that is, v(i),v(i+1) with v(i+1)-v(i)>1. Then for n>1, a(n) is the number of partitions of V(n+1) into blocks of nonconsecutive integers. - Augustine O. Munagi, Jul 17 2008
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REFERENCES
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Olivier Gérard and Karol Penson, A budget of set partitions statistics, in preparation, unpublished as of Sep 22 2011
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LINKS
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FORMULA
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For n >= 1, a(n+2) = exp(-1)*Sum_{k>=0} (k+1)^2/k!*k^n. - Benoit Cloitre, Mar 09 2008
G.f.: -(1+2*x)*(1+x)^2*Sum_{k>=0} x^(2*k)*(4*x*k^2-2*k-2*x-1)/((2*k+1)*(1*x*k-1))*A(k)/B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1)*(2*x*p-x-1)*(2*x*p-2*x-1)). - Sergei N. Gladkovskii, Jan 03 2013
G.f.: G(0)*(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 03 2013
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EXAMPLE
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a(3)=10 because the set {1,3,5,6} has 10 different partitions into blocks of nonconsecutive integers: 15/36, 16/35, 135/6, 136/5, 1/35/6, 1/36/5, 13/5/6, 15/3/6, 16/3/5, 1/3/5/6.
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MAPLE
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with(combinat): 1, 3, seq(`if`(n>1, bell(n)+2*bell(n-1)+bell(n-2), NULL), n=2..22); # Augustine O. Munagi, Jul 17 2008
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MATHEMATICA
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Join[{1, 3}, #[[1]]+2#[[2]]+#[[3]]&/@Partition[BellB[Range[0, 30]], 3, 1]] (* Harvey P. Dale, May 05 2023 *)
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PROG
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(Python)
# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A011969_list, blist, b, b2 = [1, 3], [1], 1, 1
for _ in range(10**2):
....blist = list(accumulate([b]+blist))
....A011969_list.append(2*b+b2+blist[-1])
....b2, b = b, blist[-1]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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