A192222 a(n) = Fibonacci(2^n + 1).

1, 2, 5, 34, 1597, 3524578, 17167680177565, 407305795904080553832073954, 229265413057075367692743352179590077832064383222590237

Offset: 0

Author

Jonathan Sondow, Jun 26 2011

Keywords

• nonn
• easy

Comments

a(n) is the numerator of the n-th iterate when Newton's method is applied to the function x^2 - x - 1 with initial guess x = 1. The n-th iterate is a(n)/A058635(n). - Jason Zimba, Jan 20 2023

Links

• John Gill and Matthew Miller, Newton's Method and Ratios of Fibonacci Numbers, Fibonacci Quarterly, 19(1):1-3, February 1981.
• Jonathan Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, No. 1 (2011), pp. 97-100, arXiv preprint, arXiv:1106.4246 [math.NT], 2011.
• Yohei Tachiya, Transcendence of certain infinite products, J. Number Theory, Vol. 125, No. 1 (2007), pp. 182-200.

Crossrefs

Cf. A000045 (Fibonacci numbers F(n)), A001622, A134973 (decimal expansion of 3/phi), A192223 (Lucas(2^n + 1)), A338305.

Programs

Mathematica

Table[Fibonacci[2^n + 1], {n, 0, 10}] (* T. D. Noe, Jan 11 2012 *)

Formula

a(n) = A000045(2^n + 1).

Product_{n>0} (1 + 1/a(n)) = 3/phi = A134973, where phi = (1+sqrt(5))/2 is the golden mean.

Sum_{n>=0} 1/a(n) = A338305. - Amiram Eldar, Oct 22 2020