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%I #92 Jun 08 2024 00:01:06
%S 1,1,3,21,987,2178309,10610209857723,251728825683549488150424261,
%T 141693817714056513234709965875411919657707794958199867
%N a(n) = Fibonacci(2^n).
%C The next term has 107 digits.
%C From _Peter Bala_, Oct 30 2013: (Start)
%C 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)) - ....
%C 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.
%C 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)
%C 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
%D Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, p. 446.
%H Vincenzo Librandi, <a href="/A058635/b058635.txt">Table of n, a(n) for n = 0..12</a>
%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/37-40-2012/azarianIJCMS37-40-2012.pdf">Fibonacci Identities as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/41-44-2012/azarianIJCMS41-44-2012.pdf">Fibonacci Identities as Binomial Sums II</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
%H S. B. Ekhad and D. Zeilberger, <a href="http://arxiv.org/abs/1303.5306">How To Generate As Many Somos-Like Miracles as You Wish</a>, arXiv preprint arXiv:1303.5306 [math.CO], 2013.
%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 H. Hu, Z.-W. Sun and J.-X. Liu, <a href="http://pweb.nju.edu.cn/zwsun/39f.pdf">Reciprocal sums of second order recurrent sequences</a>, Fib. Quart. 39(2001), no. 3, 214-220.
%H Hideyuki Ohtsuka, <a href="https://www.fq.math.ca/Problems/AdvFeb2015.pdf">Problem H-767</a>, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 1 (2015), p. 88; <a href="https://www.fq.math.ca/Problems/February2017advanced.pdf">Nested Radicals and Fibonacci Numbers</a>, Solution to Problem H-767 by the proposer, ibid., Vol. 55, No. 1 (2017), pp. 90-91.
%F a(n) = a(n-1)*A001566(n-2). - Joe Keane (jgk(AT)jgk.org), May 31 2002
%F Sum_{n>=0} 1/a(n) = (1/2)*(7-sqrt(5)). - _Benoit Cloitre_, Jan 26 2003
%F 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
%F From _Artur Jasinski_, Oct 05 2008: (Start)
%F a(n) = (G^(2^n) - (1 - G)^(2^n))/sqrt(5) where G = GoldenRatio = (1 + sqrt(5))/2.
%F a(n) = sqrt(4/5)*cosh((2^n)*arccosh(sqrt(5/4))). (End)
%F 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
%F Recurrence equations from _Peter Bala_, Oct 30 2013: (Start)
%F a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 3.
%F a(n)/a(n-1) = 5*a(n-2)^2 + 2 for n >= 3.
%F a(n) = a(n-1)*sqrt(5*a(n-1)^2 + 4) for n >= 2. (End)
%F 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
%F From _Amiram Eldar_, Dec 02 2021: (Start)
%F a(n) = A000045(A000079(n)).
%F Limit_{n->oo} sqrt(a(1)^2 + sqrt(a(2)^2 + sqrt(a(3)^2 + ... + sqrt(a(n))))) = 3 (Ohtsuka, 2015). (End)
%F 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
%e Let b = 1/sqrt(5) and x = (1/2)*(3 - sqrt(5)). We have the following Pierce expansions to base b:
%e x = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + ....
%e x^2 = b/3 - b^2/(3*21) + b^3/(3*21*987) - b^4/(3*21*987*2178309) + ....
%e x^4 = b/21 - b^2/(21*987) + b^3/(21*987*2178309) - ....
%e x^8 = b/987 - b^2/(987*2178309) + .... - _Peter Bala_, Oct 30 2013
%p a:= n-> (<<0|1>, <1|1>>^(2^n))[1,2]:
%p seq(a(n), n=0..10); # _Alois P. Heinz_, Nov 21 2014
%t Table[ Fibonacci[ 2^n ], {n, 0, 9} ]
%t G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^n) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 7}] (* _Artur Jasinski_, Oct 05 2008 *)
%t Table[Round[(4/5)^(1/2)*Cosh[2^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 10}] (* _Artur Jasinski_, Oct 05 2008 *)
%o (Magma) [Fibonacci(2^n): n in [0..10]]; // _Vincenzo Librandi_, Mar 25 2014
%o (PARI) a(n)=fibonacci(2^n) \\ _Charles R Greathouse IV_, Oct 03 2016
%Y Cf. A000032, A000045, A000079, A054783, A001566, A071579, A230338.
%K nonn
%O 0,3
%A _Robert G. Wilson v_, Jan 16 2001
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