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Let r_1 = 1; r_{n+1} = [r_1; r_2, r_3,..., r_n]; n-th term is numerator of r_n.
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%I #6 Apr 09 2014 10:16:29

%S 1,1,2,5,31,1231,1917121,4667676800641,27640053867460068128570881,

%T 969475805641849161579046591814284573846079748476161

%N Let r_1 = 1; r_{n+1} = [r_1; r_2, r_3,..., r_n]; n-th term is numerator of r_n.

%C [r_1; r_2, r_3,..., r_n] is a continued fraction, where the r's are rationals. limit{n -> infinity} r_n = 1.7118691868...

%Y Cf. A053978, A064846.

%K easy,frac,nonn

%O 1,3

%A _Leroy Quet_, Oct 31 2001