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%I #25 Jun 14 2016 12:09:31
%S 1,0,1,3,15,90,645,5355,50505,532980,6219045,79469775,1103335695,
%T 16533226710,265888247625,4566885297975,83422361847825,
%U 1614626682669000,33003508539026025,710350201433547675,16057073233633006575
%N Number of simple labeled graphs on n nodes of degree 1 or 2 without cycles.
%D Herbert S. Wilf, Generatingfunctionology, p. 104.
%H Alois P. Heinz, <a href="/A185369/b185369.txt">Table of n, a(n) for n = 0..400</a>
%F E.g.f.: exp(1/(2*(1-x))-x/2-1/2).
%F a(n) = 1-n if n<2, else a(n) = Sum_{k=2..n} C(n-1,k-1) * k!/2 * a(n-k).
%F a(n) ~ 2^(-3/4)*n^(n-1/4)*exp(-3/4+sqrt(2*n)-n). - _Vaclav Kotesovec_, Sep 25 2013
%F Conjecture: +2*a(n) +4*(-n+1)*a(n-1) +2*(n-1)*(n-3)*a(n-2) +(n-1)*(n-2)*a(n-3)=0. - _R. J. Mathar_, Jun 14 2016
%e a(4) = 15 because there are 15 simple labeled graphs on 4 nodes of degree 1 or 2 without cycles: 1-2 3-4, 1-3 2-4, 1-4 2-3, 1-2-3-4, 1-2-4-3, 1-3-2-4, 1-3-4-2, 1-4-2-3, 1-4-3-2, 2-1-3-4, 2-1-4-3, 3-1-2-4, 3-1-4-2, 4-1-2-3, 4-1-3-2.
%p a:= proc(n) option remember;
%p `if`(n<2, 1-n, add(binomial(n-1, k-1) *k!/2 *a(n-k), k=2..n))
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Feb 24 2011
%t a=1/(2(1-x))-1/2-x/2;
%t Range[0,20]! CoefficientList[Series[Exp[a],{x,0,20}],x]
%K nonn
%O 0,4
%A _Geoffrey Critzer_, Feb 20 2011
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