

A058635


(2^n)th Fibonacci number.


19




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)


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. 18711876. 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. 20532059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
S. B. Ekhad and D. Zeilberger, How To Generate As Many SomosLike Miracles as You Wish, arXiv preprint arXiv:1303.5306 [math.CO], 2013.
H. Hu, Z.W. Sun and J.X. Liu, Reciprocal sums of second order recurrent sequences, Fib. Quart. 39(2001), no. 3, 214220.


FORMULA

a(n) = a(n1)*A001566(n2).  Joe Keane (jgk(AT)jgk.org), May 31 2002
Sum(n>=0, 1/a(n)) = (1/2)*(7sqrt(5)).  Benoit Cloitre, Jan 26 2003
1/phi^2 = (.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
a(n) = (G^(2^n)  (1  G)^(2^n))/sqrt(5) where G = GoldenRatio = (1 + sqrt(5))/2.  Artur Jasinski, Oct 05 2008
a(n) = sqrt(4/5)*cosh((2^n)*arccosh(sqrt(5/4))).  Artur Jasinski, Oct 05 2008
a(n) = (a(n1)^3 / a(n2)^2 + 5 * a(n1) * a(n2)^2) / 2, for n > 1.  Lee A. Newberg, Jul 20 2010
Recurrence equations:
a(n)/a(n1) = (a(n1)/a(n2))^2  2 for n >= 3.
a(n)/a(n1) = 5*a(n2)^2 + 2 for n >= 3.
a(n) = a(n1)*sqrt(5*a(n1)^2 + 4) for n >= 2.  Peter Bala, Oct 30 2013
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


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> (<<01>, <11>>^(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

Cf. A000045, A054783, A001566, A071579, A230338.
Sequence in context: A111433 A111435 A111438 * A077260 A290766 A290872
Adjacent sequences: A058632 A058633 A058634 * A058636 A058637 A058638


KEYWORD

nonn


AUTHOR

Robert G. Wilson v, Jan 16 2001


STATUS

approved



