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

Author

Khalil Ayadi, Jan 09 2025

Keywords

• nonn
• cofr

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).

Examples

			0 + 1/(1 + 1/(1 + 1/(1 + ... ))) = 0.6087912199223083952132365...
		

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.

Crossrefs

Cf. A006280, A019426.

Programs

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