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A058635 a(n) = Fibonacci(2^n). 23
1, 1, 3, 21, 987, 2178309, 10610209857723, 251728825683549488150424261, 141693817714056513234709965875411919657707794958199867 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
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
REFERENCES
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, p. 446.
LINKS
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.
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+sqrt5) = 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
EXAMPLE
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
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 *)
PROG
(Magma) [Fibonacci(2^n): n in [0..10]]; // Vincenzo Librandi, Mar 25 2014
(PARI) a(n)=fibonacci(2^n) \\ Charles R Greathouse IV, Oct 03 2016
CROSSREFS
Sequence in context: A111433 A111435 A111438 * A077260 A367999 A290766
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jan 16 2001
STATUS
approved

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Last modified March 18 22:56 EDT 2024. Contains 370952 sequences. (Running on oeis4.)