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%I M4046 N1679 #522 May 05 2024 19:45:36
%S 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,
%T 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,
%U 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
%N Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - _Clark Kimberling_, Oct 19 2011
%C Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - _Michel Lagneau_, Dec 02 2011
%C Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - _Michel Lagneau_, Apr 17 2012
%C 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
%C 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
%C 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
%C The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - _Giovanni Zedda_, Jun 22 2019
%C Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - _Diego Rattaggi_, Apr 17 2021
%C 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
%C 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
%C 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
%C 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
%C 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
%C The positive solution to x^2 - x - 1 = 0. - _Michal Paulovic_, Jan 16 2023
%D 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.
%D Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge, NJ, 1997.
%D Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, Vol. 94, Cambridge University Press, 2003, Section 1.2.
%D Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.
%D 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.
%D H. E. Huntley, The Divine Proportion, Dover, NY, 1970.
%D Mario Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field]
%D 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.
%D Scott Olsen, The Golden Section, Walker & Co., NY, 2006.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D Hans Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001.
%D Claude-Jacques Willard, Le nombre d'or, Magnard, Paris, 1987.
%H Robert G. Wilson v, <a href="/A001622/b001622.txt">Table of n, a(n) for n = 1..100000</a>
%H Mohammad K. Azarian, <a href="https://doi.org/10.35834/1998/1003176">Problem 123</a>, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; <a href="https://doi.org/10.35834/2000/1201050">Solution</a>, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.
%H John Baez, <a href="http://math.ucr.edu/home/baez/week203.html">This week's finds in mathematical physics, Week 203</a>.
%H John Baez, <a href="http://math.ucr.edu/home/baez/numbers/">The Rankin Lectures 2008, My Favorite Numbers: 5</a>. [<a href="http://www.youtube.com/watch?v=2oPGmxDua2U">video</a>]
%H Murray Berg, <a href="http://www.fq.math.ca/Scanned/4-2/berg.pdf">Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers</a>, Fib. Quart., Vol. 4, No. 2 (1961), pp. 157-162.
%H Ömür Deveci, Zafer Adıgüzel and Taha Doğan, <a href="https://doi.org/10.7546/nntdm.2020.26.1.179-190">On the Generalized Fibonacci-circulant-Hurwitz numbers</a>, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 1, 179-190.
%H T. Eveilleau, <a href="http://perso.orange.fr/therese.eveilleau/pages/truc_mat/textes/rectangle_dor.htm">Le nombre d'or</a> (in French).
%H Abdul Gaffar, Anand B. Joshi, Sonali Singh, and Keerti Srivastava, <a href="https://doi.org/10.1007/s11042-022-12246-y">A high capacity multi-image steganography technique based on golden ratio and non-subsampled contourlet transform</a>, Multimedia Tools and Applications (2022).
%H Gutenberg Project, <a href="http://www.gutenberg.org/etext/633">The golden ratio to 20000 places</a>.
%H ICON Project, <a href="http://www.cs.arizona.edu/icon/oddsends/phi.htm">The golden ratio to 50000 places</a>.
%H The IMO Compendium, <a href="https://imomath.com/othercomp/Can/CanMO75.pdf">Problem 4</a>, 7th Canadian Mathematical Olympiad 1975.
%H L. B. W. Jolley, <a href="https://archive.org/details/summationofserie00joll">Summation of Series</a>, Dover, 1961
%H Franklin H. J. Kenter, <a href="https://arxiv.org/abs/1712.04856">It's good to be phi: a solution to a problem of Gosper and Knuth</a>, arXiv:1712.04856 [math.HO], 2017.
%H Clark Kimberling, <a href="https://www.heldermann.de/JGG/JGG11/JGG112/jgg11014.htm">Two kinds of golden triangles, generalized to match continued fractions</a>, Journal for Geometry and Graphics, Vol. 11 (2007), pp. 165-171.
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Kimberling/kimber12.html">Lucas Representations of Positive Integers</a>, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/">Fibonacci numbers and the golden section</a>.
%H Wolfdieter Lang, <a href="https://www.itp.kit.edu/~wl/EISpub/A333852.pdf">A list of representative simple difference sets of the Singer type for small orders m</a>, Karlsruher Institut für Technologie (Karlsruhe, Germany 2020).
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/2307.10645">Cantor's List of Real Algebraic Numbers of Heights 1 to 7</a>, arXiv:2307.10645 [math.NT], 2023.
%H Simon Litsyn and Vladimir Shevelev, <a href="http://dx.doi.org/10.1142/S1793042105000339">Irrational Factors Satisfying the Little Fermat Theorem</a>, International Journal of Number Theory, Vol. 1, No. 4 (2005), pp. 499-512.
%H Gary B. Meisner, <a href="https://www.goldennumber.net/">Phi, The Golden Number</a>.
%H George Markowsky, <a href="http://www.umcs.maine.edu/~markov/GoldenRatio.pdf">Misconceptions About the Golden Ratio</a>, College Mathematics Journal, 23:1 (January 1992), 2-19.
%H George Markowsky, <a href="http://www.ams.org/notices/200503/rev-markowsky.pdf">Book review: The Golden Ratio</a>, Notices of the AMS, 52:3 (March 2005), 344-347.
%H R. S. Melham and A. G. Shannon, <a href="https://www.fq.math.ca/Scanned/33-1/melham2.pdf">Inverse Trigonometric Hyperbolic Summation Formulas Involving Generalized Fibonacci Numbers</a>, The Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
%H Jean-Christophe Michel, <a href="http://jc.michel.free.fr/nombre_d_or.php">Le nombre d'or</a>.
%H J. J. O'Connor and E. F. Robertson, <a href="https://mathshistory.st-andrews.ac.uk/HistTopics/Golden_ratio/">The Golden ratio</a>.
%H Hugo Pfoertner, <a href="/A001622/a001622.txt">1 million digits of phi</a>, Computed using A. J. Yee's y-cruncher.
%H Simon Plouffe, Plouffe's Inverter, <a href="http://www.plouffe.fr/simon/constants/golden.txt">The golden ratio to 10 million digits</a>. [Only announcement, file truncated]
%H Simon Plouffe, <a href="http://web.archive.org/web/20150911212308/http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap46.html">The golden ratio:(1+sqrt(5))/2 to 20000 places</a>.
%H Fred Richman, Fibonacci sequence with multiprecision Java, <a href="http://math.fau.edu/Richman/fibjava.htm">Successive approximations to phi from ratios of consecutive Fibonacci numbers</a>.
%H Herman P. Robinson, <a href="/A257574/a257574.pdf">The CSR Function</a>, Popular Computing (Calabasas, CA), Vol. 4, No. 35 (Feb 1976), pages PC35-3 to PC35-4. Annotated and scanned copy.
%H E. F. Schubert, <a href="https://www.ecse.rpi.edu/~schubert/Educational-resources/Fibonacci%20series.pdf">The Fibonacci series</a>.
%H Vladimir Shevelev, <a href="http://list.seqfan.eu/oldermail/seqfan/2014-March/012737.html">A property of n-bonacci constant</a>, Seqfan (Mar 23 2014).
%H Jonathan Sondow, <a href="http://arxiv.org/abs/1106.4246">Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers</a>, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, pp. 97-100; arXiv:1106.4246 [math.NT], 2011.
%H Matthew R. Watkins, <a href="http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/goldenmean.htm">The "Golden Mean" in number theory</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GoldenRatio.html">Golden Ratio</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SilverRatio.html">Silver Ratio</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Mark_Barr">Mark Barr</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Golden_ratio">Golden ratio</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Kronecker-Weber_theorem">Kronecker-Weber theorem</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Metallic_mean">Metallic mean</a>.
%H Alexander J. Yee, <a href="http://www.numberworld.org/y-cruncher">y-cruncher - A Multi-Threaded Pi-Program</a>.
%H <a href="/index/Al#algebraic_02">Index entries for algebraic numbers, degree 2</a>
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F 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
%F Equals Hypergeometric2F1([1/5, 4/5], [1/2], 3/4) = 2*cos((3/5)*arcsin(sqrt(3/4))). - _Artur Jasinski_, Oct 26 2008
%F From _Hieronymus Fischer_, Jan 02 2009: (Start)
%F 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).
%F General formula: Provided x>1 satisfies x-x^(-1)=floor(x), where x=phi for this sequence, then:
%F for odd n: x^n - x^(-n) = floor(x^n), hence fract(x^n) = x^(-n),
%F for even n: x^n + x^(-n) = ceiling(x^n), hence fract(x^n) = 1 - x^(-n),
%F for all n>0: x^n + (-x)^(-n) = round(x^n).
%F x=phi is the minimal solution to x - x^(-1) = floor(x) (where floor(x)=1 in this case).
%F 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)
%F Equals 2*cos(Pi/5) = e^(i*Pi/5) + e^(-i*Pi/5). - _Eric Desbiaux_, Mar 19 2010
%F 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
%F 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]
%F 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
%F phi^n = (A000032(n) + A000045(n)*sqrt(5)) / 2. - _Thomas Ordowski_, Jun 09 2013
%F 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
%F 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
%F 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
%F exp(arcsinh(cos(Pi/2-log(phi)*i))) = exp(arcsinh(sin(log(phi)*i))) = (sqrt(3) + i) / 2. - _Geoffrey Caveney_, Apr 23 2014
%F exp(arcsinh(cos(Pi/3))) = phi. - _Geoffrey Caveney_, Apr 23 2014
%F cos(Pi/3) + sqrt(1 + cos(Pi/3)^2). - _Geoffrey Caveney_, Apr 23 2014
%F 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
%F phi = 1/2 + sqrt(1 + (1/2)^2). - _Geoffrey Caveney_, Apr 25 2014
%F Phi is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. - _Chayim Lowen_, Aug 30 2015
%F From _Isaac Saffold_, Feb 28 2018: (Start)
%F 1 = Sum_{k=0..n} binomial(n, k) / phi^(n+k) for all nonnegative integers n.
%F 1 = Sum_{n>=1} 1 / phi^(2n-1).
%F 1 = Sum_{n>=2} 1 / phi^n.
%F phi = Sum_{n>=1} 1/phi^n. (End)
%F From _Christian Katzmann_, Mar 19 2018: (Start)
%F phi = Sum_{n>=0} (15*(2*n)! + 8*n!^2)/(2*n!^2*3^(2*n+2)).
%F phi = 1/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
%F 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
%F 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
%F From _Michal Paulovic_, Jan 16 2023: (Start)
%F Equals the real part of 2 * e^(i * Pi / 5).
%F Equals 2 * sin(3 * Pi / 10) = 2*A019863.
%F Equals -2 * sin(37 * Pi / 10).
%F Equals 1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / ...)))).
%F Equals (2 + 3 * (2 + 3 * (2 + 3 * ...)^(1/4))^(1/4))^(1/4).
%F Equals (1 + 2 * (1 + 2 * (1 + 2 * ...)^(1/3))^(1/3))^(1/3).
%F Equals (1 + phi + (1 + phi + (1 + phi + ...)^(1/3))^(1/3))^(1/3).
%F Equals 13/8 + Sum_{k=0..oo} (-1)^(k+1)*(2*k+1)!/((k+2)!*k!*4^(2*k+3)).
%F (End)
%F phi^n = phi * A000045(n) + A000045(n-1). - _Gary W. Adamson_, Sep 09 2023
%F 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
%F Equals Product_{k>=0} ((5*k + 2)*(5*k + 3))/((5*k + 1)*(5*k + 4)). - _Antonio Graciá Llorente_, Feb 24 2024
%F From _Antonio Graciá Llorente_, Apr 21 2024: (Start)
%F Equals Product_{k>=1} phi^(-2^k) + 1, with phi = A001622.
%F Equals Product_{k>=0} ((5^(k+1) + 1)*(5^(k-1/2) + 1))/((5^k + 1)*(5^(k+1/2) + 1)).
%F Equals Product_{k>=1} 1 - (4*(-1)^k)/(10*k - 5 + (-1)^k) = Product_{k>=1} A047221(k)/A047209(k).
%F Equals Product_{k>=0} ((5*k + 7)*(5*k + 1 + (-1)^k))/((5*k + 1)*(5*k + 7 + (-1)^k)).
%F Equals Product_{k>=0} ((10*k + 3)*(10*k + 5)*(10*k + 8)^2)/((10*k + 2)*(10*k + 4)*(10*k + 9)^2).
%F Equals Product_{k>=5} 1 + 1/(Fibonacci(k) - (-1)^k).
%F Equals Product_{k>=2} 1 + 1/Fibonacci(2*k).
%F Equals Product_{k>=2} (Lucas(k)^2 + (-1)^k)/(Lucas(k)^2 - 4*(-1)^k). (End)
%e 1.6180339887498948482045868343656381177203091798057628621...
%p Digits:=1000; evalf((1+sqrt(5))/2); # _Wesley Ivan Hurt_, Nov 01 2013
%t RealDigits[(1 + Sqrt[5])/2, 10, 130] (* _Stefan Steinerberger_, Apr 02 2006 *)
%t RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][[1]] (* _Robert G. Wilson v_, Mar 01 2008 *)
%t RealDigits[GoldenRatio,10,120][[1]] (* _Harvey P. Dale_, Oct 28 2015 *)
%o (PARI) 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
%o (PARI)
%o /* Digit-by-digit method: write it as 0.5+sqrt(1.25) and start at hundredths digit */
%o r=11; x=400; print(1); print(6);
%o for(dig=1, 110, {d=0; while((20*r+d)*d <= x, d++);
%o d--; /* while loop overshoots correct digit */
%o print(d); x=100*(x-(20*r+d)*d); r=10*r+d})
%o \\ _Michael B. Porter_, Oct 24 2009
%o (PARI)
%o a(n) = floor(10^(n-1)*(quadgen(5))%10);
%o alist(len) = digits(floor(quadgen(5)*10^(len-1))); \\ _Chittaranjan Pardeshi_, Jun 22 2022
%o (Python)
%o from sympy import S
%o def alst(n): # truncate extra last digit to avoid rounding
%o return list(map(int, str(S.GoldenRatio.n(n+1)).replace(".", "")))[:-1]
%o print(alst(105)) # _Michael S. Branicky_, Jan 06 2021
%Y Cf. A000012 (continued fraction coefficients), A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973.
%Y Cf. A102208, A102769, A131595.
%Y Cf. A302973, A303069, A304022.
%K nonn,cons,nice,easy,changed
%O 1,2
%A _N. J. A. Sloane_
%E Additional links contributed by _Lekraj Beedassy_, Dec 23 2003
%E More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 24 2004
%E More terms from _Stefan Steinerberger_, Apr 02 2006
%E Broken URL to Project Gutenberg replaced by _Georg Fischer_, Jan 03 2009
%E Edited by _M. F. Hasler_, Feb 24 2014
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