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a(n) = Fibonacci(2^n + 1).
7

%I #34 Jan 21 2023 09:35:08

%S 1,2,5,34,1597,3524578,17167680177565,407305795904080553832073954,

%T 229265413057075367692743352179590077832064383222590237

%N a(n) = Fibonacci(2^n + 1).

%C 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

%H John Gill and Matthew Miller, <a href="https://www.fq.math.ca/Scanned/19-1/gill.pdf">Newton's Method and Ratios of Fibonacci Numbers</a>, Fibonacci Quarterly, 19(1):1-3, February 1981.

%H Jonathan Sondow, <a href="https://doi.org/10.1063/1.3630044">Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers</a>, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, No. 1 (2011), pp. 97-100, <a href="http://arxiv.org/abs/1106.4246">arXiv preprint</a>, arXiv:1106.4246 [math.NT], 2011.

%H Yohei Tachiya, <a href="http://dx.doi.org/10.1016/j.jnt.2006.11.006">Transcendence of certain infinite products</a>, J. Number Theory, Vol. 125, No. 1 (2007), pp. 182-200.

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

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

%F Sum_{n>=0} 1/a(n) = A338305. - _Amiram Eldar_, Oct 22 2020

%t Table[Fibonacci[2^n + 1], {n, 0, 10}] (* _T. D. Noe_, Jan 11 2012 *)

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

%K nonn,easy

%O 0,2

%A _Jonathan Sondow_, Jun 26 2011