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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)] and [0; a(1), a(2), a(3), ..., a(n)].
1

%I #14 Sep 04 2018 00:34:47

%S 1,2,8,18,55,84,127,146,179,224,297,548,684,960,1627,1669,1144,9307,

%T 20650,2234,3977,2311,2393,2523,4949,2856,3118,11000,3324,3230,3084,

%U 3555,4173,3936,7738,3786,4472,4491,5188,4100,4849,4674,6089,4955,5862

%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)] and [0; a(1), a(2), a(3), ..., a(n)].

%H Robert Israel, <a href="/A058082/b058082.txt">Table of n, a(n) for n = 1..1000</a>

%e [a(1); a(2)] + [0; a(1), a(2)] = 13/6 = [2; 6]. So a(3) = 2 + 6 = 8.

%p A[1]:= 1: v:= 1: P:= <<1,0>|<0,1>>:

%p for n from 2 to 100 do

%p P:= P . <<v,1>|<1,0>>;

%p t:= P[1,1]/P[2,1];

%p v:=convert(numtheory:-cfrac(t+1/t,`quotients`),`+`);

%p A[n]:= v;

%p od:

%p seq(A[i],i=1..100); # _Robert Israel_, Sep 03 2018

%K nonn

%O 1,2

%A _Leroy Quet_, Nov 24 2000