cp's OEIS Frontend

This is a transcendental number.

The n-th convergent of a(0..n) has q(n) as denominator.

Thus a(3*n+2) = a(3*n+3)=1 and a(3*n+4) = q(3*n+2)^2 for n>=1 are the results of repeatedly appending a triple of terms 1,1,Q^2 where Q is the convergent denominator after the first new 1.

The recurrence for q follows from this construction, and the alternating series is the continued fraction value for any sequence of convergent denominators.

This structure leads to the series and the recurrence for q.

Sum_{i>=0} (-1)^i/x(i) is another way to write the series, where x(i) = q(i)*q(i+1). When x(0)=1 , x(3n+2) divides x(3n+3), x(3n+2)-x(3n+1)=((x(3n+1))/x(3n))*(x(3n-1)/x(3n-2))*(x(3n-3)/x(3n-4))...(x(2)/x(1)))^2,x(3n+4)-x(3n+3)=(x(3n+3)/x(3n+2))^2*(x(3n+2)-x(3n+1)).

a(19) has 85 decimal digits and a(21) has 170 decimal digits.

This number is transcendental.

q(n) is the denominator of the convergent resulting from terms a(0..n).

The continued fraction is constructed by successively appending a pair of terms 1 and its own q(n) so far, so a(2*n) = 1 and a(2*n+1) = q(2*n-1) for n>=1

The series and the recurrence for q follows from that construction.

The series can also be written Sum_{i>=0} (-1)^i/x(i) where x(i) = q(i)*q(i+1) and in that case x(0)=1, x(2n+1) divides x(2n+2), and x(2n+3) = ((x(2n+2)/x(2n+1))*(x(2n)/x(2n-1))*...*(x(2)/x(1)))^2 + x(2n+2).