

A081090


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.


4




OFFSET

1,2


COMMENTS

log(a(n+1))/log(a(n)) > 1+sqrt(2). The 8th term has 59 digits, while the 9th term has 141 digits.
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(n1), for n>1.  Paul D. Hanna, Mar 05 2003


LINKS



FORMULA

a(n) = a(n2)*(a(n1)^2 + 1) for n>3, with a(1)=0, a(2)=2, a(3)=5. Also, a(n)*a(n1) = A081088(n) for n>2, a(n) = a(n2)*A081089(n1) for n>2.


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



