

A114938


Number of permutations of the multiset {1,1,2,2,....,n,n} with no two consecutive terms equal.


9



0, 2, 30, 864, 39480, 2631600, 241133760, 29083420800, 4467125013120, 851371260364800, 197158144895712000, 54528028997584665600, 17752366094818747392000, 6720318485119046923315200, 2927066537906697348594432000, 1453437879238150456164433920000
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OFFSET

1,2


COMMENTS

a(n) is also the number of $(0,1)$~ matrices $A=(a_{ij})$ of size $n\times (2n)$ such that each row has exactly two $1$'s and each column has exactly one $1$'s and with the restriction that no $1$ stands on the line from $a_{11}$ to $% a_{22}$ . [From Shanzhen Gao, Feb 24 2010]


REFERENCES

R. P. Stanley, Enumerative Combinatorics Volume I, Cambridge University Press, 1997. Chapter 2, Sieve Methods, Example 2.2.3, page 68.
Shanzhen Gao, Sequences Arising from Integer Matrix Enumeration (in preparation) [From Shanzhen Gao, Feb 24 2010]


LINKS

Andrew Woods, Table of n, a(n) for n = 1..100


FORMULA

a(n) = Sum_{k=0..n} ((C(n, k)*(1)^(nk)*(n+k)!)/2^k).
a(n) = (1)^n * n! * A000806(n), n>0. [Vladeta Jovovic, Nov 19 2009]
a(n) = n*(2*n1)*a(n1) + (n1)*n*a(n2).  Vaclav Kotesovec, Aug 07 2013
a(n) ~ 2^(n+1)*n^(2*n)*sqrt(Pi*n)/exp(2*n+1).  Vaclav Kotesovec, Aug 07 2013


EXAMPLE

a(2) = 2 because there are two permutations of {1,1,2,2} avoiding equal consecutive terms: 1212 and 2121.


MATHEMATICA

Table[Sum[Binomial[n, i](2ni)!/2^(ni) (1)^i, {i, 0, n}], {n, 20}] (* Geoffrey Critzer, Jan 02 2013 *)


CROSSREFS

Cf. A114939 = preferred seating arrangements of n couples.
Cf. A007060 = arrangements of n couples with no adjacent spouses; A007060(n) = 2^n * A114938(n) (this sequence).
Sequence in context: A160694 A013525 A229781 * A082653 A186292 A140174
Adjacent sequences: A114935 A114936 A114937 * A114939 A114940 A114941


KEYWORD

nonn


AUTHOR

Hugo Pfoertner, Jan 08 2006


STATUS

approved



