

A053871


a(0)=1; a(1)=0; a(n) = 2*(n1)*(a(n1) + a(n2)).


24



1, 0, 2, 8, 60, 544, 6040, 79008, 1190672, 20314880, 387099936, 8148296320, 187778717632, 4702248334848, 127140703364480, 3691602647581184, 114562300528369920, 3784124901630435328, 132555364873399378432, 4908221631901073295360, 191549525877429961604096
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OFFSET

0,3


COMMENTS

Number of deranged matchings of 2n people with partners (of either sex) other than their spouse. 2n objects are initially paired in some way and then are repaired so that no object is with its original partner (the dancing problem in the article).
Of interest in the "collision problem", where, given a 2to1 function f, we are asked for x, y such that f(x)=f(y).
2^n*n!*a(n) = (2n)! b(n) where b(n) are the probabilities that appear in Margolis (2001). One interpretation is in terms of matchings or 1factors of the complete graph on 2n vertices. The number of these is (2n)!/2^{n}n!. The number of 1factors being disjoint from (that is, having no edges in common with) a given 1factor is then a(n); and then b(n) is the probability of picking such a disjoint one factor at random.
Also n!*a(n) = 2^n * c(n) where c(n) = A001499(n). If we define d(n,k) = k(n1)(d(n1,k) + d(n2,k)), with d(0,k) = 1 and d(1,k) = 0, so d(n,1) are the derangement numbers A000166, then a(n) = d(n,2) (cf. A033030, A088991). On the other hand, taking d*(n,k) = d(n,k)/k^{n}, we have d*(n,k) = (n1)(d*(n1,k) + d*(n2,k)/k), with d*(0,k) = 1 and d*(1,k) = 0 and it is easy to see from Bricard's recurrence for c(n) that c(n)/n! satisfies this for k = 2.
A proof that the description in the first comment as "number of deranged matchings" implies the defining recursion relation: let (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) be the given pairs. In a deranged matching x_1 will be paired with any of the 2(n1) objects x_2, y_2, x_3, y_3, ..., x_n, y_n. It is sufficient to count only those deranged matchings in which x_1, is matched with x_2. They are of two kinds: (i) y_1 is not matched with y_2; their number is a(n1); (ii) y_1 is matched with y_2; their number is a(n2).  Emeric Deutsch, Jan 23 2009
From Lewis Mammel (l_mammel(AT)att.net), Oct 07 2009: (Start)
The formula is given directly by the Principle of Inclusion and Exclusion.
The first term includes all pairings, the second term excludes all pairings containing each pair, the third term includes all pairings containing each pair of pairs, and so on.
Based on na > n for large n, the ratio a(n)/(2n1)!! > exp(1/2) ~= 0.60653.
We find a(n)/(2n1)!! for n= 100, 200, 300, 400 ~= 0.6050124904, 0.6057720290, 0.6060250088, 0.6061514604. (End)
This is a subset of the set of pairings of the first 2n integers (A001147) in another way: forbidding pairs of the form (2k,2k+1) for all k as well as the pair (1,2n) (seeing the constraint as circular by opposition to the linear A165968).  Olivier Gérard, Feb 08 2011
a(n) is the nth central moment of a central chisquared distribution (with 1 degree of freedom), i.e., a(n) = E[ (Y E[Y] )^n ] = E[ (X^2  1 )^n ] where Y is chisquared, X is std normal, X~N(0,1), and the expectation operator is E[].  David Fioramonti, May 11 2016
For n > 1, also the number of maximum matchings in the ncocktail party graph.  Eric W. Weisstein, Jun 15 2017
Number of polygonal tiles with n sides with two exits per side and n edges connecting pairs of exits, with no edges between exits on the same side (cf. A289191, A289269 for nonisomorphic tiles under rotational and rotational and reflectional symmetries).  Marko Riedel, Jun 28 2017
Number of ways n people can hold hands (every hand is holding another hand) where no one is holding their own hand.  Harry Richman, Aug 29 2023


REFERENCES

R. Bricard, L'Intermédiaire des Mathématiciens, 8 (1901), 312313.


LINKS

Eric Weisstein's World of Mathematics, Matching


FORMULA

E.g.f.: 1/(exp(x)*sqrt(12x)).
a(n) = Integral_{x>=0} (x1)^n * (exp(x/2)/sqrt(2*Pi*x)) dt.  Paul Barry, Apr 12 2010
Conjectured g.f.: T(0)/(1+x), where T(k) = 1  x*(k+1)/(x*(k+1)  (1+x)/T(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Nov 13 2013
G.f.: Sum_{k>=0} (2*k  1)!!*x^k/(1 + x)^(k+1).  Ilya Gutkovskiy, Apr 12 2019
Conjecture: if m == n (mod q) for q odd, then (1)^m*a(m) == (1)^n*a(n) (mod q).  Harry Richman, Aug 29 2023


MAPLE

f:= gfun:rectoproc({a(0) = 1, a(1) = 0, a(n) = 2*(n  1)*(a(n  1) + a(n  2))}, a(n), remember):


MATHEMATICA

RecurrenceTable[{a[0]==1, a[1]==0, a[n]==2(n1)(a[n1]+a[n2])}, a[n], {n, 20}] (* Harvey P. Dale, Sep 15 2011 *)
CoefficientList[Assuming[{Element[x, Reals], x>0}, Series[Sqrt[Pi/2] * (I + Erfi[Sqrt[(1+1/x)/2]]) / (E^((1+x)/(2*x)) * Sqrt[x*(x+1)]), {x, 0, 20}]], x] (* Vaclav Kotesovec, Feb 15 2015 *)
Range[0, 20]! CoefficientList[Series[1/(Exp[x] Sqrt[1  2 x]), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 15 2017 *)
Table[(1)^n HypergeometricPFQ[{1/2, n}, {}, 2], {n, 20}] (* Eric W. Weisstein, Jun 15 2017 *)
Table[I (1)^n HypergeometricU[1/2, 3/2 + n, 1/2]/Sqrt[2], {n, 20}] (* Eric W. Weisstein, Dec 31 2017 *)


PROG

(PARI) a(n)=(1)^(n+1)*sum(k=0, n, (1)^k*binomial(n, k)*prod(i=0, k, 2*i1))
(Haskell)
a053871 n = a053871_list !! n
a053871_list = 1 : 0 : zipWith (*)
[2, 4..] (zipWith (+) a053871_list $ tail a053871_list)


CROSSREFS

See A289191 for when rotational symmetries of the tiles are taken into account.  Marko Riedel, Jun 28 2017
Cf. A165968, number of pairings of 2n things disjoint to a given pairing, and containing a given pair not in the given pairing. It is given by a(n)/(2n2).  Lewis Mammel (l_mammel(AT)att.net), Oct 07 2009


KEYWORD

nonn,nice,easy


AUTHOR



EXTENSIONS



STATUS

approved



