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%I #13 May 30 2013 06:32:44
%S 3,13,83,673,6203,61613,642683,6940673,76930803,870136013,10002590883,
%T 116521027873,1372486213803,16318813519213,195599588228683,
%U 2360929398934273,28671940652447203,350089944825571213,4295280755452388083,52926654021145267873
%N For n>=1, a(n) = n + 2 + sum(i=1..n-1, a(i)*a(n-i) ).
%C Using induction, it is easy to prove that a(n)==3 (mod 10).
%C The largest prime factors of these terms are large (they start 3, 13, 83, 673, 6203, 61613, 642683, 161411, 9221, 870136013, 751453, 4016443, 6267060337, 16318813519213,..)
%H Vincenzo Librandi, <a href="/A173998/b173998.txt">Table of n, a(n) for n = 1..200</a>
%F Recurrence: n*a(n) = 3*(5*n-7)*a(n-1) - (23*n-48)*a(n-2) + 9*(n-3)*a(n-3). - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) ~ sqrt(13*sqrt(10)-40)*(7+2*sqrt(10))^n/(4*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012
%t aa=ConstantArray[0,20];aa[[1]]=3;Do[aa[[n]]=n+2+Sum[aa[[i]]*aa[[n-i]],{i,1,n-1}],{n,2,20}];aa (* _Vaclav Kotesovec_, Oct 20 2012 *)
%Y Cf. A030431.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Mar 05 2010
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