A094874 -id:A094874 - cp's OEIS frontend

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

1, 6, 1, 8, 0, 3, 3, 9, 8, 8, 7, 4, 9, 8, 9, 4, 8, 4, 8, 2, 0, 4, 5, 8, 6, 8, 3, 4, 3, 6, 5, 6, 3, 8, 1, 1, 7, 7, 2, 0, 3, 0, 9, 1, 7, 9, 8, 0, 5, 7, 6, 2, 8, 6, 2, 1, 3, 5, 4, 4, 8, 6, 2, 2, 7, 0, 5, 2, 6, 0, 4, 6, 2, 8, 1, 8, 9, 0, 2, 4, 4, 9, 7, 0, 7, 2, 0, 7, 2, 0, 4, 1, 8, 9, 3, 9, 1, 1, 3, 7, 4, 8, 4, 7, 5

Offset: 1

N. J. A. Sloane

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004
The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005
Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011
The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011
Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011
If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011
The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011
Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011
Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012
This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012
Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014
For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014
The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019
Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021
Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021
The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021
The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022
In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022
Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022
The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023
The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025
This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

			1.6180339887498948482045868343656381177203091798057628621...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 112, 123, 184, 190, 203.
Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993 - Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1975, pages 76-77, 1993.
Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge, NJ, 1997.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, Vol. 94, Cambridge University Press, 2003, Section 1.2.
Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.
Martin Gardner, Weird Water and Fuzzy Logic: More Notes of a Fringe Watcher, "The Cult of the Golden Ratio", Chapter 9, Prometheus Books, 1996, pages 90-97.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 287.
H. E. Huntley, The Divine Proportion, Dover, NY, 1970.
Mario Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field]
Gary B. Meisner, The Golden Ratio: The Divine Beauty of Mathematics, Race Point Publishing (The Quarto Group), 2018. German translation: Der Goldene Schnitt, Librero, 2023.
Scott Olsen, The Golden Section, Walker & Co., NY, 2006.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 137-139.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Hans Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 36-40.
Claude-Jacques Willard, Le nombre d'or, Magnard, Paris, 1987.

Robert G. Wilson v, Table of n, a(n) for n = 1..100000
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; Solution, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.
John Baez, This week's finds in mathematical physics, Week 203.
John Baez, The Rankin Lectures 2008, My Favorite Numbers: 5. [video]
Murray Berg, Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers, Fib. Quart., Vol. 4, No. 2 (1961), pp. 157-162.
Ömür Deveci, Zafer Adıgüzel and Taha Doğan, On the Generalized Fibonacci-circulant-Hurwitz numbers, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 1, 179-190.
T. Eveilleau, Le nombre d'or (in French).
Abdul Gaffar, Anand B. Joshi, Sonali Singh, and Keerti Srivastava, A high capacity multi-image steganography technique based on golden ratio and non-subsampled contourlet transform, Multimedia Tools and Applications (2022).
Gutenberg Project, The golden ratio to 20000 places.
ICON Project, The golden ratio to 50000 places.
The IMO Compendium, Problem 4, 7th Canadian Mathematical Olympiad 1975.
L. B. W. Jolley, Summation of Series, Dover, 1961
Franklin H. J. Kenter, It's good to be phi: a solution to a problem of Gosper and Knuth, arXiv:1712.04856 [math.HO], 2017.
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, Vol. 11 (2007), pp. 165-171.
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
Ron Knott, Fibonacci numbers and the golden section.
Wolfdieter Lang, A list of representative simple difference sets of the Singer type for small orders m, Karlsruher Institut für Technologie (Karlsruhe, Germany 2020).
Wolfdieter Lang, Cantor's List of Real Algebraic Numbers of Heights 1 to 7, arXiv:2307.10645 [math.NT], 2023.
Simon Litsyn and Vladimir Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, Vol. 1, No. 4 (2005), pp. 499-512.
Gary B. Meisner, Phi, The Golden Number.
George Markowsky, Misconceptions About the Golden Ratio, College Mathematics Journal, 23:1 (January 1992), 2-19.
George Markowsky, Book review: The Golden Ratio, Notices of the AMS, 52:3 (March 2005), 344-347.
R. S. Melham and A. G. Shannon, Inverse Trigonometric Hyperbolic Summation Formulas Involving Generalized Fibonacci Numbers, The Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
Jean-Christophe Michel, Le nombre d'or.
J. J. O'Connor and E. F. Robertson, The Golden ratio.
Hugo Pfoertner, 1 million digits of phi, Computed using A. J. Yee's y-cruncher.
Simon Plouffe, Plouffe's Inverter, The golden ratio to 10 million digits. [Only announcement, file truncated]
Simon Plouffe, The golden ratio:(1+sqrt(5))/2 to 20000 places.
Fred Richman, Fibonacci sequence with multiprecision Java, Successive approximations to phi from ratios of consecutive Fibonacci numbers.
Herman P. Robinson, The CSR Function, Popular Computing (Calabasas, CA), Vol. 4, No. 35 (Feb 1976), pages PC35-3 to PC35-4. Annotated and scanned copy.
E. F. Schubert, The Fibonacci series.
Vladimir Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014).
Jonathan Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, pp. 97-100; arXiv:1106.4246 [math.NT], 2011.
Matthew R. Watkins, The "Golden Mean" in number theory.
Eric Weisstein's World of Mathematics, Golden Ratio.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
Eric Weisstein's World of Mathematics, Pisot Number.
Eric Weisstein's World of Mathematics, Silver Ratio.
Wikipedia, Mark Barr.
Wikipedia, Golden ratio.
Wikipedia, Kronecker-Weber theorem.
Wikipedia, Metallic mean.
Alexander J. Yee, y-cruncher - A Multi-Threaded Pi-Program.
Index entries for algebraic numbers, degree 2.
Index to sequences related to Olympiads.

Cf. A000012 (continued fraction coefficients), A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973.
Cf. A102208, A102769, A131595.
Cf. A302973, A303069, A304022.

Digits:=1000; evalf((1+sqrt(5))/2); # Wesley Ivan Hurt, Nov 01 2013

RealDigits[(1 + Sqrt[5])/2, 10, 130] (* Stefan Steinerberger, Apr 02 2006 *)
RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][[1]] (* Robert G. Wilson v, Mar 01 2008 *)
RealDigits[GoldenRatio,10,120][[1]] (* Harvey P. Dale, Oct 28 2015 *)

default(realprecision, 20080); x=(1+sqrt(5))/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b001622.txt", n, " ", d));  \\ Harry J. Smith, Apr 19 2009

/* Digit-by-digit method: write it as 0.5+sqrt(1.25) and start at hundredths digit */
r=11; x=400; print(1); print(6);
for(dig=1, 110, {d=0; while((20*r+d)*d <= x, d++);
d--; /* while loop overshoots correct digit */
print(d); x=100*(x-(20*r+d)*d); r=10*r+d})
\\ Michael B. Porter, Oct 24 2009

a(n) = floor(10^(n-1)*(quadgen(5))%10);
alist(len) = digits(floor(quadgen(5)*10^(len-1))); \\ Chittaranjan Pardeshi, Jun 22 2022

from sympy import S
def alst(n): # truncate extra last digit to avoid rounding
  return list(map(int, str(S.GoldenRatio.n(n+1)).replace(".", "")))[:-1]
print(alst(105)) # Michael S. Branicky, Jan 06 2021

Equals Sum_{n>=2} 1/A064170(n) = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + ... - Gary W. Adamson, Dec 15 2007
Equals Hypergeometric2F1([1/5, 4/5], [1/2], 3/4) = 2*cos((3/5)*arcsin(sqrt(3/4))). - Artur Jasinski, Oct 26 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of phi^n equals phi^(-n), if n is odd. For even n, the fractional part of phi^n is equal to 1-phi^(-n).
General formula: Provided x>1 satisfies x-x^(-1)=floor(x), where x=phi for this sequence, then:
for odd n: x^n - x^(-n) = floor(x^n), hence fract(x^n) = x^(-n),
for even n: x^n + x^(-n) = ceiling(x^n), hence fract(x^n) = 1 - x^(-n),
for all n>0: x^n + (-x)^(-n) = round(x^n).
x=phi is the minimal solution to x - x^(-1) = floor(x) (where floor(x)=1 in this case).
Other examples of constants x satisfying the relation x - x^(-1) = floor(x) include A014176 (the silver ratio: where floor(x)=2) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
Equals 2*cos(Pi/5) = e^(i*Pi/5) + e^(-i*Pi/5). - Eric Desbiaux, Mar 19 2010
The solutions to x-x^(-1)=floor(x) are determined by x=(1/2)*(m+sqrt(m^2+4)), m>=1; x=phi for m=1. In terms of continued fractions the solutions can be described by x=[m;m,m,m,...], where m=1 for x=phi, and m=2 for the silver ratio A014176, and m=3 for the bronze ratio A098316. - Hieronymus Fischer, Oct 20 2010
Sum_{n>=1} x^n/n^2 = Pi^2/10 - (log(2)*sin(Pi/10))^2 where x = 2*sin(Pi/10) = this constant here. [Jolley, eq 360d]
phi = 1 + Sum_{k>=1} (-1)^(k-1)/(F(k)*F(k+1)), where F(n) is the n-th Fibonacci number (A000045). Proof. By Catalan's identity, F^2(n) - F(n-1)*F(n+1) = (-1)^(n-1). Therefore,(-1)^(n-1)/(F(n)*F(n+1)) = F(n)/F(n+1) - F(n-1)/F(n). Thus Sum_{k=1..n} (-1)^(k-1)/(F(k)*F(k+1)) = F(n)/F(n+1). If n goes to infinity, this tends to 1/phi = phi - 1. - Vladimir Shevelev, Feb 22 2013
phi^n = (A000032(n) + A000045(n)*sqrt(5)) / 2. - Thomas Ordowski, Jun 09 2013
Let P(q) = Product_{k>=1} (1 + q^(2*k-1)) (the g.f. of A000700), then A001622 = exp(Pi/6) * P(exp(-5*Pi)) / P(exp(-Pi)). - Stephen Beathard, Oct 06 2013
phi = i^(2/5) + i^(-2/5) = ((i^(4/5))+1) / (i^(2/5)) = 2*(i^(2/5) - (sin(Pi/5))i) = 2*(i^(-2/5) + (sin(Pi/5))i). - Jaroslav Krizek, Feb 03 2014
phi = sqrt(2/(3 - sqrt(5))) = sqrt(2)/A094883. This follows from the fact that ((1 + sqrt(5))^2)*(3 - sqrt(5)) = 8, so that ((1 + sqrt(5))/2)^2 = 2/(3 - sqrt(5)). - Geoffrey Caveney, Apr 19 2014
exp(arcsinh(cos(Pi/2-log(phi)*i))) = exp(arcsinh(sin(log(phi)*i))) = (sqrt(3) + i) / 2. - Geoffrey Caveney, Apr 23 2014
exp(arcsinh(cos(Pi/3))) = phi. - Geoffrey Caveney, Apr 23 2014
cos(Pi/3) + sqrt(1 + cos(Pi/3)^2). - Geoffrey Caveney, Apr 23 2014
2*phi = z^0 + z^1 - z^2 - z^3 + z^4, where z = exp(2*Pi*i/5). See the Wikipedia Kronecker-Weber theorem link. - Jonathan Sondow, Apr 24 2014
phi = 1/2 + sqrt(1 + (1/2)^2). - Geoffrey Caveney, Apr 25 2014
Phi is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. - Chayim Lowen, Aug 30 2015
From Isaac Saffold, Feb 28 2018: (Start)
1 = Sum_{k=0..n} binomial(n, k) / phi^(n+k) for all nonnegative integers n.
1 = Sum_{n>=1} 1 / phi^(2n-1).
1 = Sum_{n>=2} 1 / phi^n.
phi = Sum_{n>=1} 1/phi^n. (End)
From Christian Katzmann, Mar 19 2018: (Start)
phi = Sum_{n>=0} (15*(2*n)! + 8*n!^2)/(2*n!^2*3^(2*n+2)).
phi = 1/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
phi = Product_{k>=1} (1 + 2/(-1 + 2^k*(sqrt(4+(1-2/2^k)^2) + sqrt(4+(1-1/2^k)^2)))). - Gleb Koloskov, Jul 14 2021
Equals Product_{k>=1} (Fibonacci(3*k)^2 + (-1)^(k+1))/(Fibonacci(3*k)^2 + (-1)^k) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
From Michal Paulovic, Jan 16 2023: (Start)
Equals the real part of 2 * e^(i * Pi / 5).
Equals 2 * sin(3 * Pi / 10) = 2*A019863.
Equals -2 * sin(37 * Pi / 10).
Equals 1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / ...)))).
Equals (2 + 3 * (2 + 3 * (2 + 3 * ...)^(1/4))^(1/4))^(1/4).
Equals (1 + 2 * (1 + 2 * (1 + 2 * ...)^(1/3))^(1/3))^(1/3).
Equals (1 + phi + (1 + phi + (1 + phi + ...)^(1/3))^(1/3))^(1/3).
Equals 13/8 + Sum_{k=0..oo} (-1)^(k+1)*(2*k+1)!/((k+2)!*k!*4^(2*k+3)).
(End)
phi^n = phi * A000045(n) + A000045(n-1). - Gary W. Adamson, Sep 09 2023
The previous formula holds for integer n, with F(-n) = (-1)^(n+1)*F(n), for n >= 0, with F(n) = A000045(n), for n >= 0. phi^n are integers in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Sep 16 2023
Equals Product_{k>=0} ((5*k + 2)*(5*k + 3))/((5*k + 1)*(5*k + 4)). - Antonio Graciá Llorente, Feb 24 2024
From Antonio Graciá Llorente, Apr 21 2024: (Start)
Equals Product_{k>=1} phi^(-2^k) + 1, with phi = A001622.
Equals Product_{k>=0} ((5^(k+1) + 1)*(5^(k-1/2) + 1))/((5^k + 1)*(5^(k+1/2) + 1)).
Equals Product_{k>=1} 1 - (4*(-1)^k)/(10*k - 5 + (-1)^k) = Product_{k>=1} A047221(k)/A047209(k).
Equals Product_{k>=0} ((5*k + 7)*(5*k + 1 + (-1)^k))/((5*k + 1)*(5*k + 7 + (-1)^k)).
Equals Product_{k>=0} ((10*k + 3)*(10*k + 5)*(10*k + 8)^2)/((10*k + 2)*(10*k + 4)*(10*k + 9)^2).
Equals Product_{k>=5} 1 + 1/(Fibonacci(k) - (-1)^k).
Equals Product_{k>=2} 1 + 1/Fibonacci(2*k).
Equals Product_{k>=2} (Lucas(k)^2 + (-1)^k)/(Lucas(k)^2 - 4*(-1)^k). (End)

Additional links contributed by Lekraj Beedassy, Dec 23 2003
More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 24 2004
More terms from Stefan Steinerberger, Apr 02 2006
Broken URL to Project Gutenberg replaced by Georg Fischer, Jan 03 2009
Edited by M. F. Hasler, Feb 24 2014

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

Original entry on oeis.org

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607

Offset: 0

N. J. A. Sloane

Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first, 2nd, 3rd, 6th, 11th, ?, ... terms). - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011
We have Sum_{n=0..N} a(n)^2 = 2*(N+1) + Sum_{n=1..N+1} a(n), Sum_{n=0..N} a(n)^4 = 5*(Sum_{n=1..N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect to the fact that a(n) = Lucas(2^(n+1)); see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = Product_{n>=0} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = Product_{n>=0} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022
The number of digits of a(n) is given by A094057(n+1). - Hans J. H. Tuenter, Jul 29 2025

			From _Cino Hilliard_, Sep 28 2008: (Start)
Init x=1;
x = (5/1 + 1)/2 = 3/1;
x = (5/3 + 3)/2 = 7/3;
x = ((5/7)/3 + 7/3)/2 = 47/21;
x = ((5/47)/21 + 47/21)/2 = 2207/987;
(2207/987)^2 = 5.000004106... (End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 7.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..12
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 2.
Chance Sanford, Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae, arXiv:1603.03765 [math.NT], 2016.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Wikipedia, Engel Expansion.
Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.

a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq(a(n), n=0..8);

NestList[#^2-2&,3,10] (* Harvey P. Dale, Dec 17 2014 *)
Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *)

a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n),n,0,7); /* Martin Ettl, Nov 12 2012 */

{a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
g(5,8) \\ Cino Hilliard, Sep 28 2008

{a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */

{a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */

a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^(n+1)*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1) + Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas). - Philippe Deléham, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - Vladimir Shevelev, Nov 28 2010
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
From Peter Bala, Oct 31 2012: (Start)
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul.
sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
a(n) == 2 (mod 9), for n > 1. - Ivan N. Ianakiev, Dec 25 2013
From Amiram Eldar, Oct 22 2020: (Start)
a(n) = A000032(2^(n+1)).
Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021
a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022
a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022
a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - Jianglin Luo, Sep 21 2023
Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - Amiram Eldar, Mar 01 2024

A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).

Original entry on oeis.org

0, 3, 24, 168, 1155, 7920, 54288, 372099, 2550408, 17480760, 119814915, 821223648, 5628750624, 38580030723, 264431464440, 1812440220360, 12422650078083, 85146110326224, 583600122205488, 4000054745112195, 27416783093579880, 187917426909946968

Offset: 0

N. J. A. Sloane, Jun 09 2002

Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002
From Paul Weisenhorn, May 17 2009: (Start)
a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2
which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.
(End)
Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011

			G.f. = 3*x + 24*x^2 + 168*x^3 + 1155*x^4 + 7920*x^5 + 54288*x^6 + ... - _Michael Somos_, Jan 23 2025

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 29.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A033888, A004187, A094874.
Bisection of A059929, A064831 and A080097.
Related to sum of fibonacci(kn) over n; cf. A000071, A099919, A027941, A138134, A053606.

[Fibonacci(2*n)*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

fs4:=n->sum(fibonacci(4*k),k=0..n):seq(fs4(n),n=0..21); # Gary Detlefs, Dec 07 2010

Table[Fibonacci[2 n]*Fibonacci[2 n + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Accumulate[Fibonacci[4*Range[0,30]]] (* or *) LinearRecurrence[{8,-8,1},{0,3,24},30] (* Harvey P. Dale, Jul 25 2013 *)

a(n)=fibonacci(2*n)*fibonacci(2*n+2) \\ Charles R Greathouse IV, Jul 02 2013

a(n) = -3/5 + (1/5*sqrt(5)+3/5)*(2*1/(7+3*sqrt(5)))^n/(7+3*sqrt(5)) + (1/5*sqrt(5)-3/5)*(-2*1/(-7+3*sqrt(5)))^n/(-7+3*sqrt(5)). Recurrence: a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). G.f.: 3*x/(1-7*x+x^2)/(1-x). - Vladeta Jovovic, Jun 09 2002
a(n) = A081068(n) - 1.
a(n) is the next integer from ((3+sqrt(5))*((7+3*sqrt(5))/2)^(n-1)-6)/10. - Paul Weisenhorn, May 17 2009
a(n) = 7*a(n-1) - a(n-2) + 3, n>1. - Gary Detlefs, Dec 07 2010
a(n) = sum_{k=0..n} Fibonacci(4k). - Gary Detlefs, Dec 07 2010
a(n) = (Lucas(4n+2)-3)/5, where Lucas(n)= A000032(n). - Gary Detlefs, Dec 07 2010
a(n) = (1/5)*(Fibonacci(4n+4) - Fibonacci(4n)-3). - Gary Detlefs, Dec 08 2010
a(n) = 3*A092521(n). - R. J. Mathar, Nov 03 2011
a(0)=0, a(1)=3, a(2)=24, a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Harvey P. Dale, Jul 25 2013
a(n) = A001906(n)*A001906(n+1). - R. J. Mathar, Jul 09 2019
Sum_{n>=1} 1/a(n) = 2/(3 + sqrt(5)) = A094874 - 1. - Amiram Eldar, Oct 05 2020
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025

A132338 Decimal expansion of 1 - 1/phi.

Original entry on oeis.org

3, 8, 1, 9, 6, 6, 0, 1, 1, 2, 5, 0, 1, 0, 5, 1, 5, 1, 7, 9, 5, 4, 1, 3, 1, 6, 5, 6, 3, 4, 3, 6, 1, 8, 8, 2, 2, 7, 9, 6, 9, 0, 8, 2, 0, 1, 9, 4, 2, 3, 7, 1, 3, 7, 8, 6, 4, 5, 5, 1, 3, 7, 7, 2, 9, 4, 7, 3, 9, 5, 3, 7, 1, 8, 1, 0, 9, 7, 5, 5, 0, 2, 9, 2, 7, 9, 2, 7, 9, 5, 8, 1, 0, 6, 0, 8, 8, 6, 2, 5, 1, 5, 2, 4

Offset: 0

N. J. A. Sloane, Nov 07 2007

Density of 1's in Fibonacci word A003849.
Also decimal expansion of Sum_{n>=1} ((-1)^(n+1))*1/phi^n. - Michel Lagneau, Dec 04 2011
The Lambert series evaluated at this point is 0.8828541617125076... [see André-Jeannin]. - R. J. Mathar, Oct 28 2012
Because this equals 2 - phi, this is an integer in the quadratic number field Q(sqrt(5)). (Note that this is also sqrt(5 - 3*phi).) - Wolfdieter Lang, Jan 08 2018
When m >= 1, the equation m*x^m + (m-1)*x^(m-1) + ... + 2*x^2 + x - 1 = 0 has only one positive root, u(m) (say); then lim_{m->oo} u(m) = (3-sqrt(5))/2 (see Aubonnet). - Bernard Schott, May 12 2019
Cosine of the zenith angle at which a string should be cut so that a ball tied to one of its ends, set moving without friction around a vertical circle with the minimum speed in a uniform gravitational field, will then travel through the fixed center of the circle. - Stefano Spezia, Oct 25 2020
Algebraic number of degree 2 with minimal polynomial x^2 - 3*x + 1. The other root is 1 + phi = A104457. - Wolfdieter Lang, Aug 29 2022

			0.38196601125010515179541316563436188...

F. Aubonnet, D. Guinin and A. Ravelli, Oral, Concours d'entrée des Grandes Ecoles Scientifiques, Exercices résolus, "Crus" 1982-83, Bréal, 1983, Exercice 210, 40-42.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
R. André-Jeannin, Lambert series and the summation of reciprocals in certain Fibonacci-Lucas-Type sequences, Fib. Quart. 28 (1990) 223-226.
Yiyan Ni, Myron Hlynka, and Percy H. Brill, Urn Models and Fibonacci Series, arXiv:1806.09150 [math.CO], 2018. See (9) p. 7.
Index entries for algebraic numbers, degree 2

Cf. A001622, A003849, A094874, A079585, A094214, A104457.

RealDigits[N[1/GoldenRatio^2,200]] (* Vladimir Joseph Stephan Orlovsky, May 27 2010 *)
RealDigits[1-1/GoldenRatio,10,120][[1]] (* Harvey P. Dale, Mar 30 2024 *)

(3-sqrt(5))/2 \\ Michel Marcus, Oct 26 2020

Equals 1 - 1/phi = 2 - phi, with phi from A001622.
Equals A094874 - 1, or A079585 - 2, or the square of A094214.
Equals (5-sqrt(5))^2/20 = 1/phi^2 = 1/A104457. - Joost Gielen, Sep 28 2013 [corrected by Joerg Arndt, Sep 29 2013]
Equals (3-sqrt(5))/2. - Bernard Schott, May 12 2019
Equals Sum_{k >= 2} (-1)^k/(Fibonacci(k)*Fibonacci(k+1)). See Ni et al. - Michel Marcus, Jun 26 2018

A187798 Decimal expansion of (3-phi)/2, where phi is the golden ratio.

Original entry on oeis.org

6, 9, 0, 9, 8, 3, 0, 0, 5, 6, 2, 5, 0, 5, 2, 5, 7, 5, 8, 9, 7, 7, 0, 6, 5, 8, 2, 8, 1, 7, 1, 8, 0, 9, 4, 1, 1, 3, 9, 8, 4, 5, 4, 1, 0, 0, 9, 7, 1, 1, 8, 5, 6, 8, 9, 3, 2, 2, 7, 5, 6, 8, 8, 6, 4, 7, 3, 6, 9, 7, 6, 8, 5, 9, 0, 5, 4, 8, 7, 7, 5, 1, 4, 6, 3, 9, 6, 3, 9, 7, 9, 0, 5, 3, 0, 4, 4, 3, 1, 2, 5, 7, 6, 2, 2

Offset: 0

Joost Gielen, Aug 30 2013

This is the height h of the isosceles triangle in a regular pentagon inscribed in the unit circle formed from a diagonal as base and two adjacent pentagon sides. h = sqrt(sqrt(3-phi)^2 - (sqrt(2 + phi)/2)^2) = sqrt(10 - 5*phi)/2 = (3 - phi)/2. - Wolfdieter Lang, Jan 07 2018

			0.6909830056250525758977065828171809411398454100971185689322756886473697685905...

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for algebraic numbers, degree 2.

Cf. A001622, A081011, A187426, A226765, A094874, A344212.

RealDigits[(3 - GoldenRatio)/2, 10, 111][[1]] (* or *)
RealDigits[(5 - Sqrt[5])/4, 10, 111][[1]] (* Robert G. Wilson v, Jan 07 2018 *)

(5-sqrt(5))/4 \\ Charles R Greathouse IV, Aug 31 2013

Equals (3-phi)/2 = A094874/2 with phi from A001622.
From Amiram Eldar, Nov 28 2024: (Start)
Equals 1/A344212.
Equals Product_{k>=0} (1 - 1/A081011(k)). (End)

Extended by Charles R Greathouse IV, Aug 31 2013

A242671 Decimal expansion of k2, a Diophantine approximation constant such that the area of the "critical parallelogram" (in this case a square) is 4*k2.

Original entry on oeis.org

7, 2, 3, 6, 0, 6, 7, 9, 7, 7, 4, 9, 9, 7, 8, 9, 6, 9, 6, 4, 0, 9, 1, 7, 3, 6, 6, 8, 7, 3, 1, 2, 7, 6, 2, 3, 5, 4, 4, 0, 6, 1, 8, 3, 5, 9, 6, 1, 1, 5, 2, 5, 7, 2, 4, 2, 7, 0, 8, 9, 7, 2, 4, 5, 4, 1, 0, 5, 2, 0, 9, 2, 5, 6, 3, 7, 8, 0, 4, 8, 9, 9, 4, 1, 4, 4, 1, 4, 4, 0, 8, 3, 7, 8, 7, 8, 2, 2, 7, 4

Offset: 0

Jean-François Alcover, May 20 2014

Quoting Steven Finch: "The slopes of the 'critical parallelogram' are (1+sqrt(5))/2 [phi] and (1-sqrt(5))/2 [-1/phi]."
Essentially the same as A229780, A134972, A134945, A098317 and A002163. - R. J. Mathar, May 23 2014
Let W_n be the collection of all binary words of length n that do not contain two consecutive 0's. Let r_n be the ratio of the total number of 1's in W_n divided by the total number of letters in W_n. Then lim_{n->oo} r_n = 0.723606... Equivalently, lim_{n->oo} A004798(n)/(n*A000045(n+2)) = 0.723606... - Geoffrey Critzer, Feb 04 2022
The limiting frequency of the digit 0 in the base phi representation of real numbers in the range [0,1], where phi is the golden ratio (A001622) (Rényi, 1957). - Amiram Eldar, Mar 18 2025

			k2 = 0.723606797749978969640917366873127623544...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.23, p. 176.

Alfréd Rényi, Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hungar., Vol.8, No. 3-4 (1957), pp. 477-493.
Index entries for algebraic numbers, degree 2.

Cf. A000032, A000045, A001622, A002163, A004798, A020762, A081007, A094214, A094874, A098317, A134972, A229780, A244847, A296184, A300074, A344212.

RealDigits[(1+1/Sqrt[5])/2, 10, 100] // First

(1 + 1/sqrt(5))/2 \\ Stefano Spezia, Dec 07 2024

Equals (1 + 1/sqrt(5))/2.
Equals 1/A094874. - Michel Marcus, Dec 01 2018
From Amiram Eldar, Feb 11 2022: (Start)
Equals phi/sqrt(5), where phi is the golden ratio (A001622).
Equals lim_{k->oo} Fibonacci(k+1)/Lucas(k). (End)
From Amiram Eldar, Nov 28 2024: (Start)
Equals A344212/2 = A296184/5 = A300074^2 = sqrt(A229780).
Equals Product_{k>=1} (1 - 1/A081007(k)). (End)
Equals 1 - A244847. - Amiram Eldar, Mar 18 2025

A187799 Decimal expansion of 20/phi^2, where phi is the golden ratio. Also (with a different offset), decimal expansion of 3 - sqrt(5).

Original entry on oeis.org

7, 6, 3, 9, 3, 2, 0, 2, 2, 5, 0, 0, 2, 1, 0, 3, 0, 3, 5, 9, 0, 8, 2, 6, 3, 3, 1, 2, 6, 8, 7, 2, 3, 7, 6, 4, 5, 5, 9, 3, 8, 1, 6, 4, 0, 3, 8, 8, 4, 7, 4, 2, 7, 5, 7, 2, 9, 1, 0, 2, 7, 5, 4, 5, 8, 9, 4, 7, 9, 0, 7, 4, 3, 6, 2, 1, 9, 5, 1, 0, 0, 5, 8, 5, 5, 8, 5, 5, 9, 1, 6, 2, 1, 2, 1, 7, 7, 2, 5, 0, 3, 0, 4, 9, 1, 8, 2, 3, 8, 4, 9

Offset: 1

Joost Gielen, Aug 30 2013

			20/phi^2 = 7.6393202250021030359082633...
3 - sqrt(5) = 0.76393202250021030359082633... (with offset 0).

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Mohammad K. Azarian, The Value of a Series of Reciprocal Fibonacci Numbers, Problem B-1133, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275; Solution published in Vol. 52, No. 3, August 2014, pp. 277-278.

Cf. A187426, A187798, A094874.

5*(Sqrt(5)-1)^2; // Vincenzo Librandi, Feb 24 2015

First@ RealDigits[N[5*(Sqrt[5] - 1)^2, 111]] (* Michael De Vlieger, Feb 25 2015 *)

5*(sqrt(5)-1)^2 \\ Charles R Greathouse IV, Aug 31 2013

10*(3 - sqrt(5)) = 30 - 10*sqrt(5) = (5 - sqrt(5))^2 = 20/phi^2.

2 * Sum_{i > 1} (-1)^i/(F(i)F(i + 1)) = 3 - sqrt(5), where F(i) is the i-th Fibonacci number. This formula comes from John D. Watson, Jr.'s solution to Azarian's Problem B-1133 in the Fibonacci Quarterly. Azarian originally posed the problem as an infinite alternating sum explicitly written out for the first dozen terms or so. See the Azarian links above. - Alonso del Arte, Aug 25 2016

Extended by Charles R Greathouse IV, Aug 31 2013

A192223 a(n) = Lucas(2^n + 1).

Original entry on oeis.org

3, 4, 11, 76, 3571, 7881196, 38388099893011, 910763447271179530132922476, 512653048485188394162163283930413917147479973138989971

Offset: 0

Jonathan Sondow, Jun 26 2011

Product_{n>0} (1 + 1/a(n)) = 3 - phi = A094874, where phi = (1+sqrt(5))/2 is the golden mean.
From Peter Bala, Oct 28 2013: (Start)
Compare with A230600(n) = Lucas(2^n - 1).
Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a (possibly infinite) increasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - .... This definition generalizes the ordinary Pierce expansion of a real number 0 < x < 1, where the base b has the value 1. Depending on the values of x and b such a generalized Pierce expansion to the base b may not exist, and if it does exist it may not be unique.
Let Phi := 1/2*(sqrt(5) - 1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, provides a Pierce expansion of Phi^4 to the base Phi. That is we have the identity Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ....
This result can be extended in two ways. Firstly, for k odd, the sequence {Lucas(k*(2^n + 1))}n>=1 gives a Pierce expansion of Phi^(4*k) to the base Phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] gives a Pierce expansion of Phi^(2^n + 2) to the base Phi. See below for some examples. (End)

			Pierce series expansion of Phi^(2^n + 2) to the base Phi for n = 1 to 4:
n = 1:
Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ...
n = 2:
Phi^6 = Phi/11 - Phi^2/(11*76) + Phi^3/(11*76*3571) - ...
n = 3:
Phi^10 = Phi/76 - Phi^2/(76*3571) + Phi^3/(76*3571*7881196) - ...
n = 4:
Phi^18 = Phi/3571 - Phi^2/(3571*7881196) + ...

J. Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, vol. 1385, pp. 97-100.
Y. Tachiya, Transcendence of certain infinite products, J. Number Theory 125 (2007), 182-200.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A000032 (Lucas numbers L(n)), A094874 (decimal expansion of 3 - phi), A192222 (Fibonacci(2^n + 1)). A001622, A058635, A230600, A230601, A230602.

Table[LucasL[2^n + 1], {n, 0, 10}] (* T. D. Noe, Jan 11 2012 *)

a(n) = A000032(2^n + 1).
From Peter Bala, Oct 28 2013: (Start)
a(n) = phi^(2^n + 1) - (1/phi)^(2^n + 1), where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio A001622.
Recurrence equation: a(0) = 3, a(1) = 4 and a(n) = floor(1/phi*a(n-1)^2) + 2 for n >= 2. (End)

A115339 a(2n-1)=F(n+1), a(2n)=L(n), where F(n) and L(n) are the Fibonacci and the Lucas sequences.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 7, 8, 11, 13, 18, 21, 29, 34, 47, 55, 76, 89, 123, 144, 199, 233, 322, 377, 521, 610, 843, 987, 1364, 1597, 2207, 2584, 3571, 4181, 5778, 6765, 9349, 10946, 15127, 17711, 24476, 28657, 39603, 46368, 64079, 75025, 103682, 121393, 167761

Offset: 1

Giuseppe Coppoletta, Mar 06 2006

Alternate Fibonacci and Lucas sequence respecting their natural order.
See A116470 for an essentially identical sequence.
The ratio a(n+1)/a(n) increasingly approximates two constants: (5-sqrt(5))/2 (A094874) and (5+3*sqrt(5))/10 (A176015) according to whether n is odd or even. - Davide Rotondo, Oct 27 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number.
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A000045, A000032.
Cf. A000930.
Cf. A094874, A176015.

a115339 n = a115339_list !! (n-1)
a115339_list = [1, 1, 2, 3] ++
               zipWith (+) a115339_list (drop 2 a115339_list)
-- Reinhard Zumkeller, Aug 03 2013

f[n_] := If[OddQ@n, Fibonacci[(n + 3)/2], Fibonacci[n/2 - 1] + Fibonacci[n/2 + 1]]; Array[f, 50] (* Robert G. Wilson v, Apr 29 2006 *)

x='x+O('x^50); Vec(x*(-1-x-x^2-2*x^3)/(-1+x^2+x^4)) \\ G. C. Greubel, Apr 27 2017

a(n+2) = a(n) + a(n-2).

G.f.: x*( -1-x-x^2-2*x^3 ) / ( -1+x^2+x^4 ). - R. J. Mathar, Mar 08 2011

More terms from Robert G. Wilson v, Apr 29 2006

A225667 Decimal expansion of 13-5*sqrt(5).

Original entry on oeis.org

1, 8, 1, 9, 6, 6, 0, 1, 1, 2, 5, 0, 1, 0, 5, 1, 5, 1, 7, 9, 5, 4, 1, 3, 1, 6, 5, 6, 3, 4, 3, 6, 1, 8, 8, 2, 2, 7, 9, 6, 9, 0, 8, 2, 0, 1, 9, 4, 2, 3, 7, 1, 3, 7, 8, 6, 4, 5, 5, 1, 3, 7, 7, 2, 9, 4, 7, 3, 9, 5, 3, 7, 1, 8, 1, 0, 9, 7, 5, 5, 0, 2, 9, 2, 7, 9

Offset: 1

Clark Kimberling, Jul 21 2013

Let d(n) = - 2*F(n) + h(2 + F(n+1), 1 + F(n+2)), where h = harmonic mean, F = A000045 (Fibonacci numbers). Then floor(d(n)) = 2F(n) + 1 for n>1, and limit(d(n)) = 13 - 5*sqrt(5).

Apart from leading digits the same as A132338, A109866, A094874 and A079585. - R. J. Mathar, Jul 30 2013

			13-5*sqrt(5) = 1.819660112501051517954131656343618822797...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A000045, A226765.

f[n_] := Fibonacci[n]; h[n_] := HarmonicMean[{2 + f[n + 1], 1 + f[n + 2]}]; x = Limit[-2 f[n] + h[n], n -> Infinity] (* "proof" *)
d = RealDigits[x, 10, 120][[1]] (* A225667 *)

cp's OEIS Frontend

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

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A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

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A058038 a(n) = Fibonacci(2*n)*Fibonacci(2*n+2).

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A132338 Decimal expansion of 1 - 1/phi.

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A187798 Decimal expansion of (3-phi)/2, where phi is the golden ratio.

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A242671 Decimal expansion of k2, a Diophantine approximation constant such that the area of the "critical parallelogram" (in this case a square) is 4*k2.

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A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).