%I #6 Apr 09 2014 10:16:38
%S 1,2,6,19,42,142,223,326,496,715,1000,1394,9958,10626,12033,12597,
%T 13805,14816,20158,21264,37088,41875,45425,47766,49571,327635,330187,
%U 332469,341477,344570,450050,457785,460921,463931,469795,473848,479976
%N a(1) = 1; a(n+1) = sum of terms in continued fraction for the sum of the continued fractions, [a(1); a(2), a(3),...,a(n-1),a(n)] and [a(n); a(n-1), a(n-2),...,a(2), a(1)].
%e [a(1); a(2)] +[a(2); a(1)] = 9/2 = [4; 2]. So a(3) = 4 + 2 = 6.
%K nonn
%O 1,2
%A _Leroy Quet_, Nov 24 2000