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A173998 For n>=1, a(n) = n + 2 + Sum_{i=1..n-1} a(i)*a(n-i).

%I A173998 #53 Mar 09 2025 13:19:05
%S A173998 3,13,83,673,6203,61613,642683,6940673,76930803,870136013,10002590883,
%T A173998 116521027873,1372486213803,16318813519213,195599588228683,
%U A173998 2360929398934273,28671940652447203,350089944825571213,4295280755452388083,52926654021145267873
%N A173998 For n>=1, a(n) = n + 2 + Sum_{i=1..n-1} a(i)*a(n-i).
%C A173998 Using induction, it is easy to prove that a(n)==3 (mod 10).
%C A173998 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 A173998 Vincenzo Librandi, <a href="/A173998/b173998.txt">Table of n, a(n) for n = 1..200</a>
%F A173998 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 A173998 a(n) ~ sqrt(13*sqrt(10)-40)*(7+2*sqrt(10))^n/(4*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012
%F A173998 From _Ehren Metcalfe_, Mar 02 2025: (Start)
%F A173998 G.f.: 1/2 + sqrt(9*x^2 - 14*x + 1)/(2*(x - 1)).
%F A173998 a(n) = (1/2)*Sum_{j=0..n} (-1)^(j+1)*Sum_{k=0..j} binomial(1/2,k)*binomial(1/2, j-k)*(7 + 2*sqrt(10))^k*(7 - 2*sqrt(10))^(j-k) = 1 + 2*Sum_{j=0..n} A379103(j), for n>=1. (End)
%t A173998 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 *)
%o A173998 (Python)
%o A173998 from sympy import series, sqrt, Symbol, Poly
%o A173998 x = Symbol("x")
%o A173998 p = Poly(series((1 + sqrt(9*x**2 - 14*x + 1)/(x - 1))/2, n=20).removeO(), x)
%o A173998 print([p.coeff_monomial(x**n) for n in range(1, p.degree())]) # _Ehren Metcalfe_, Mar 03 2025
%o A173998 (PARI) my(x='x+O('x^30)); Vec(1/2 + sqrt(9*x^2 - 14*x + 1)/(2*(x - 1))) \\ _Michel Marcus_, Mar 05 2025
%Y A173998 Cf. A030431, A379103.
%K A173998 nonn
%O A173998 1,1
%A A173998 _Vladimir Shevelev_, Mar 05 2010