A058635 a(n) = Fibonacci(2^n).

1, 1, 3, 21, 987, 2178309, 10610209857723, 251728825683549488150424261, 141693817714056513234709965875411919657707794958199867

Offset: 0

Author

Robert G. Wilson v, Jan 16 2001

Keywords

• nonn

Comments

The next term has 107 digits.

From Peter Bala, Oct 30 2013: (Start)

Apart from giving the numerators in the Engel series representation of (1/2)*(7 - sqrt(5)), as noted below by Cloitre, this sequence (excluding the initial term) is also a generalized Pierce expansion defined as follows. Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a nondecreasing sequence [a(1), a(2), a(3), ...] of positive integers such that we have an alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - ....

The present sequence, apart from the initial term, is a Pierce expansion of the real number x := (1/2)*(3 - sqrt(5)) to the base b := 1/sqrt(5). The associated series representation begins (1/2)*(3 - sqrt(5)) = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + .... Cf. A071579 and A230338.

More generally, for n >= 0, the sequence [a(n+1), a(n+2), a(n+3), ...] gives a Pierce expansion of ( (1/2)*(3 - sqrt(5)) )^(2^n) to the base b = 1/sqrt(5). Some examples are given below. (End)

a(n) is the denominator 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 A192222(n)/a(n). - Jason Zimba, Jan 20 2023

Examples

			Let b = 1/sqrt(5) and x = (1/2)*(3 - sqrt(5)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + ....
x^2 = b/3 - b^2/(3*21) + b^3/(3*21*987) - b^4/(3*21*987*2178309) + ....
x^4 = b/21 - b^2/(21*987) + b^3/(21*987*2178309) - ....
x^8 = b/987 - b^2/(987*2178309) + .... - _Peter Bala_, Oct 30 2013
		

References

• Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, p. 446.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..12
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
• S. B. Ekhad and D. Zeilberger, How To Generate As Many Somos-Like Miracles as You Wish, arXiv preprint arXiv:1303.5306 [math.CO], 2013.
• John Gill and Matthew Miller, Newton's Method and Ratios of Fibonacci Numbers, Fibonacci Quarterly, 19(1):1-3, February 1981.
• H. Hu, Z.-W. Sun and J.-X. Liu, Reciprocal sums of second order recurrent sequences, Fib. Quart. 39(2001), no. 3, 214-220.
• Hideyuki Ohtsuka, Problem H-767, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 1 (2015), p. 88; Nested Radicals and Fibonacci Numbers, Solution to Problem H-767 by the proposer, ibid., Vol. 55, No. 1 (2017), pp. 90-91.

Crossrefs

Cf. A000032, A000045, A000079, A054783, A001566, A071579, A230338.

Programs

Magma

[Fibonacci(2^n): n in [0..10]]; // Vincenzo Librandi, Mar 25 2014

Maple

a:= n-> (<<0|1>, <1|1>>^(2^n))[1,2]:
seq(a(n), n=0..10);  # Alois P. Heinz, Nov 21 2014

Mathematica

Table[ Fibonacci[ 2^n ], {n, 0, 9} ]
G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^n) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[2^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 10}] (* Artur Jasinski, Oct 05 2008 *)

PARI

a(n)=fibonacci(2^n) \\ Charles R Greathouse IV, Oct 03 2016

Formula

a(n) = a(n-1)*A001566(n-2). - Joe Keane (jgk(AT)jgk.org), May 31 2002

Sum_{n>=0} 1/a(n) = (1/2)*(7-sqrt(5)). - Benoit Cloitre, Jan 26 2003

1/phi^2 = (0.6180339...)^2 = 2/(3+sqrt(5)) = Sum_{n>=2} 1/a(n) = 1/3 + 1/21 + 1/987 + 1/2178309 + ... - Gary W. Adamson, Jun 12 2003

From Artur Jasinski, Oct 05 2008: (Start)

a(n) = (G^(2^n) - (1 - G)^(2^n))/sqrt(5) where G = GoldenRatio = (1 + sqrt(5))/2.

a(n) = sqrt(4/5)*cosh((2^n)*arccosh(sqrt(5/4))). (End)

a(n) = (a(n-1)^3 / a(n-2)^2 + 5 * a(n-1) * a(n-2)^2) / 2, for n > 1. - Lee A. Newberg, Jul 20 2010

Recurrence equations from Peter Bala, Oct 30 2013: (Start)

a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 3.

a(n)/a(n-1) = 5*a(n-2)^2 + 2 for n >= 3.

a(n) = a(n-1)*sqrt(5*a(n-1)^2 + 4) for n >= 2. (End)

0 = a(n)^2 * ( a(n+3) - 2*a(n+2) ) - a(n+1)*a(n+2) * ( a(n+2) - 2*a(n+1)) if n > 0. - Michael Somos, Mar 24 2014

From Amiram Eldar, Dec 02 2021: (Start)

a(n) = A000045(A000079(n)).

Limit_{n->oo} sqrt(a(1)^2 + sqrt(a(2)^2 + sqrt(a(3)^2 + ... + sqrt(a(n))))) = 3 (Ohtsuka, 2015). (End)

a(n) = Product_{k=0..n-1} L(2^k), for n >= 1, where L(k) is the k-th Lucas number (A000032). - Amiram Eldar, Mar 30 2023