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

0,5

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)).

Links

Table of n, a(n) for n=0..18.

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.

Example

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

Prog

(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

Crossrefs

Cf. A006280, A019426, A003417.

Sequence in context: A102602 A156951 A357052 * A121066 A343635 A287647

Adjacent sequences: A380191 A380192 A380193 * A380196 A380197 A380198

Keyword

nonn,cofr

Author

Khalil Ayadi, Jan 15 2025

Status

approved