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a(1) = 1; a(n+1) = sum of terms in continued fraction for sum of continued fractions, [a(n); a(n-1), a(n-2),...,a(1)] and [0; a(n), a(n-1), a(n-2),...,a(1)].
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%I #6 Apr 09 2014 10:16:38

%S 1,2,6,36,62,130,377,621,1022,2068,3133,4728,7183,11271,17453,25246,

%T 37047,53382,83321,117518,167979,238923,339650,481615,683160,982182,

%U 1389733,1970191,2788008,3946477,5591147,7907444,11188660,15825176

%N a(1) = 1; a(n+1) = sum of terms in continued fraction for sum of continued fractions, [a(n); a(n-1), a(n-2),...,a(1)] and [0; a(n), a(n-1), a(n-2),...,a(1)].

%e [a(2); a(1)] + [0; a(2), a(1)] = 10/3 =[3; 3]. So a(3) = 3 + 3 = 6.

%K nonn

%O 1,2

%A _Leroy Quet_, Nov 24 2000