%I #19 May 11 2024 16:34:24
%S 1,2,2,2,4,6,12,20,40,70,140,252,504,924,1848,3432,6864,12870,25740,
%T 48620,97240,184756,369512,705432,1410864,2704156,5408312,10400600,
%U 20801200,40116600,80233200,155117520,310235040,601080390,1202160780,2333606220,4667212440,9075135300,18150270600,35345263800
%N a(n) = number of n-lettered words in the alphabet {1, 2} with as many occurrences of the substring (consecutive subword) [1, 1] as of [2, 2].
%H Shalosh B. Ekhad and Doron Zeilberger, <a href="http://arxiv.org/abs/1112.6207">Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type</a>, arXiv preprint arXiv:1112.6207, 2011. See subpages for rigorous derivations of g.f., recurrence, asymptotics for this sequence. [From _N. J. A. Sloane_, Apr 07 2012]
%F G.f.: 1 + x + x*sqrt((1+2*x)/(1-2*x))= 1 + x + x/G(0), where G(k)= 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jul 26 2013
%F E.g.f.: 1 + x - (x*BesselI(1, 2*x)*(2 + Pi*(1 + 2*x)*StruveL(0, 2*x)) - x*(1 + 2*x)*BesselI(0, 2*x)*(2 + Pi*StruveL(1, 2*x)))/2. - _Stefano Spezia_, May 11 2024
%p a:= proc(n) option remember; `if`(n<3, [1,2$3][n+1],
%p (2*a(n-1)+4*(n-3)*a(n-2))/(n-1))
%p end:
%p seq(a(n), n=0..39); # _Alois P. Heinz_, May 11 2024
%Y Apart from initial terms, same as A063886.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Apr 07 2012