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%I #31 May 29 2022 07:13:54
%S 1,1,4,18,126,1110,12120,156660,2341500,39701340,752839920,
%T 15785181720,362606123880,9055825538760,244296192460320,
%U 7079382509799600,219321853964413200,7233629128601475600,253054306933115688000,9358989706213886138400,364860828050107348159200
%N Number of partitions of 112233...nn into n pairs.
%D Laszlo Lovasz, Combinatorial Problems and Solutions, AMS Chelsea Publishing, American Mathematical Society.
%H G. C. Greubel, <a href="/A108704/b108704.txt">Table of n, a(n) for n = 0..400</a>
%F E.g.f.: exp(x*x/2)/sqrt(1-2*x).
%F a(n) ~ 2^(n+1/2)*n^n/exp(n-1/8). - _Vaclav Kotesovec_, Sep 26 2013
%F a(n) = 2^n*(n-1/2)!*2F2((1-n)/2,-n/2;1/4 -n/2,3/4 - n/2; 1/8)/sqrt(Pi). - _Benedict W. J. Irwin_, May 25 2016
%F Conjecture: a(n)-(2*n-1)*a(n-1)-(n-1)*a(n-2)+2*(n-1)*(n-2)*a(n-3)=0. - _R. J. Mathar_, Jun 08 2016
%F From _Emanuele Munarini_, May 25 2022: (Start)
%F The exponential generating series A(t) satisfies the differential equation (1-2*t)*A'(t) = (1+t-2*t^2)*A(t), which is equivalent to the conjectured recurrence.
%F a(n) = Sum_{k=0..n/2} binomial(n,k)*binomial(2*n-4*k,n-2*k)*(n-k)!/2^(n-k).
%F (End)
%e Partitions of 1122 into 2 pairs: 11 22, 12 12, 12 21, 21 21 = 4 partitions so a(2)=4.
%p a:= n-> n! *coeff(series(exp(x*x/2)/sqrt(1-2*x), x, n+1), x, n):
%p seq (a(n), n=0..20);
%t CoefficientList[Series[E^(x*x/2)/Sqrt[1-2*x], {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Sep 26 2013 *)
%o (PARI) my(x='x+O('x^50)); Vec(serlaplace(exp(x*x/2)/sqrt(1-2*x))) \\ _G. C. Greubel_, May 24 2017
%Y Cf. A002135.
%K nonn
%O 0,3
%A _Miklos Kristof_, Jun 20 2005
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