A380194 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(3n+2)=q(3n+1)+q(3n), q(3n+3)=q(3n+2)+q(3n+1), and q(3n+4)=q(3n+2)*(q(3n+2)*q(3n+3)+1).

0, 1, 1, 1, 4, 1, 1, 289, 1, 1, 81126049, 1, 1, 2128359349797626142548649, 1, 1, 38565134716822109850786884343127955049217538196275147632486387905655060249, 1, 1

Offset: 0

Author

Khalil Ayadi, Jan 15 2025

Keywords

• nonn
• cofr

Comments

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)).

Examples

			0 + 1/(1 + 1/(1 + 1/(1 + ... ))) = 0.645164877940276...
		

Links

• Khalil Ayadi, Chiheb Ben Bechir, and Maher Saadaoui, Continued Fractions with Predictable Patterns and Transcendental Numbers, Journal of Integer Sequences, Vol. 28 (2025), Article 25.1.4.

Crossrefs

Cf. A006280, A019426, A003417.

Programs

PARI

q(n) = if (n<=1, 1, if (n%3==1, q(n-2)*(q(n-2)*q(n-1)+1), q(n-1)+q(n-2)));
a(n) = if (n==0, 0, if ((n%3)==1, q(n-2)^2, 1)); \\ Michel Marcus, Jan 17 2025