A380013 Continued fraction expansion of Sum_{i>=0} (-1)^i/(q(i)*q(i+1)) where q(0)=q(1)=1, q(2n+2)=q(2n+1)+q(2n), and q(2n+3)=q(2n+1)*(q(2n+2)+1).

0, 1, 1, 1, 1, 3, 1, 18, 1, 432, 1, 196992, 1, 38895676416, 1, 1512881323731695591424, 1, 2288809899755012359448064967916189926490112, 1

Offset

0,6

Comments

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).

Links

Andrew Howroyd, Table of n, a(n) for n = 0..25

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 + ... ))) = 0.6087912199223083952132365...

Prog

(PARI)
Q(n) = {my(v=vector(n+1)); v[1]=v[2]=1; for(i=2, n, v[i+1] = if(i%2==0, v[i]+v[i-1], v[i-1]*(v[i]+1))); v}
seq(n)=my(q=Q(max(2, n-2))); vector(n+1, n, if(n%2 || n<4, n>1, q[n-2])) \\ Andrew Howroyd, Jan 13 2025

Crossrefs

Cf. A006280, A019426.

Sequence in context: A089974 A346039 A143849 * A370233 A335689 A105626

Adjacent sequences: A380010 A380011 A380012 * A380014 A380015 A380016

Keyword

nonn,cofr

Author

Khalil Ayadi, Jan 09 2025

Status

approved