A081090 - OEIS

%I #11 Mar 12 2015 19:01:06

%S 0,2,5,52,13525,9512132552,1223751003414213892335125,

%T 14245051228051734585272181044575005954679284643762013257552

%N a(n)^2 + 1 = A081089(n), where A081089(n) = A081088(n+1)/A081088(n); involves the partial quotients of a series of continued fractions that sum to unity.

%C log(a(n+1))/log(a(n)) -> 1+sqrt(2). The 8th term has 59 digits, while the 9th term has 141 digits.

%C sum(n>0, a(n)/a(n+1) ) = 1/2; the ratio of the terms a(n+1)/a(n), for n>1, form the convergents of the continued fraction series described by A081088; thus a(n+1) = A081088(n)*a(n) + a(n-1), for n>1. - _Paul D. Hanna_, Mar 05 2003

%F a(n) = a(n-2)*(a(n-1)^2 + 1) for n>3, with a(1)=0, a(2)=2, a(3)=5. Also, a(n)*a(n-1) = A081088(n) for n>2, a(n) = a(n-2)*A081089(n-1) for n>2.

%Y Cf. A081086, A081088, A081089.

%K nonn,easy

%O 1,2

%A _Hans Havermann_ and _Paul D. Hanna_, Mar 05 2003