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A057873 a(1) = 1; a(n+1) = sum of terms in continued fraction for sum{k=1 to n}[a(n+1-k)/a(k)]. 0

%I

%S 1,1,2,5,13,29,149,217,449,855,1578,2834,5445,9425,17054,30095,53610,

%T 94905,170505,300335,532606,942870,1669907,2957734,5236935,9271871,

%U 16416945,29066281,51463071,91587523,161792680,286563514,507342270

%N a(1) = 1; a(n+1) = sum of terms in continued fraction for sum{k=1 to n}[a(n+1-k)/a(k)].

%e Sum{k=1 to 4}[a(5-k)/a(k)] = 5/1 +2/1 +1/2 +1/5 = 77/10 =7 +1/(1 +1/(2 +1/3)). So a(5) = 7 +1 +2 +3 = 13.

%K easy,nonn

%O 1,3

%A _Leroy Quet_, Nov 19 2000

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Last modified October 23 10:17 EDT 2021. Contains 348211 sequences. (Running on oeis4.)