A114552 Odd-indexed bisection of A114551, which is the continued fraction expansion of Sum_{n>=0} 1/A112373(n); also, a(n) = A112373(n+1)/A112373(n) for n>=0.

1, 2, 6, 78, 73086, 4999703411742, 1710009514450915230711940280907486, 1000118217480414644596377710149364954622541926875237885435586055466747491994623223953758

Offset: 0

Author

Paul D. Hanna, Dec 08 2005

Keywords

• nonn

Comments

A112373 is defined by the recurrence: let b(n) = A112373(n), then

b(n) =(b(n-1)^3 + b(n-1)^2)/b(n-2) for n>=2 with b(0)=b(1)=1.

The next term 58499...45086 has 228 digits and is too large to include.

Examples

			a(5) = 4999703411742 = 73086*(1 + 1*2*6*78*73086).
a(5) = a(4)*(1 - a(4) + a(4)^2/a(3)) = 73086*(1 - 73086 + 73086^2/78).
		

Links

• Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015 and J. Int. Seq. 18 (2015) # 15.8.4.

Crossrefs

Cf. A112373, A114550 (constant), A114551 (continued fraction).

Programs

PARI

a(n)=if(n==0,1,if(n==1,2,a(n-1)*(1-a(n-1)+a(n-1)^2/a(n-2))))

Formula

a(n) = a(n-1)*(1 + a(0)*a(1)*a(2)*...*a(n-1)) for n>=1, with a(0)=1.

a(n) = a(n-1)*(1 - a(n-1) + a(n-1)^2/a(n-2)) for n>=2, with a(0)=1, a(1)=2.

Extensions

a(7) corrected by Georg Fischer, Aug 26 2022