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 A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2. (Formerly M4046 N1679) 1174
 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 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 REFERENCES Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3, Fall 1998, p. 176. Solution published in Vol. 12, No. 1, Winter 2000, pp. 61-62. R. A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge NJ 1997. S. R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.2. M. Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi:The Golden Ratio", Chapter 8, Simon & Schuster NY 1961. M. 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. H. E. Huntley, The Divine Proportion, Dover NY 1970. L. B. W. Jolley, The summation of series, Dover (1961). M. Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field] S. Olsen, The Golden Section, Walker & Co. NY 2006. 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). H. Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001. C. J. Willard, Le nombre d'or, Magnard Paris 1987. LINKS Robert G. Wilson v, Table of n, a(n) for n = 1..100000 John Baez, The Rankin Lectures 2008, My Favorite Numbers: 5. [video] M. Berg, Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers, Fib. Quart., 4 (1961), 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) Gutenberg Project, The golden ratio to 20000 places ICON Project, The golden ratio to 50000 places 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, 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). S. Litsyn and V. Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, vol.1, no.4 (2005), 499-512. G. Markowsky, Misconceptions About the Golden Ratio, College Mathematics Journal, 23:1 (January 1992), 2-19. G. Markowsky, Book review: The Golden Ratio, Notices of the AMS, 52:3 (March 2005), 344-347. J. C. Michel, Le nombre d'or J. J. O'Connor & 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. F. 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 V. Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014) 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. M. R. Watkins, The "Golden Mean" in number theory Eric Weisstein's World of Mathematics, Golden Ratio Eric Weisstein's World of Mathematics, Silver Ratio Eric Weisstein's World of Mathematics, Fibonacci n-Step Number Wikipedia, Golden ratio Wikipedia, Kronecker Weber theorem Wikipedia, Metallic mean Alexander J. Yee, y-cruncher - A Multi-Threaded Pi-Program FORMULA 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 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*1/5) = e^(i*Pi*1/5)+e^(-i*Pi*1/5). - 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))). 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 a(n) = -10*floor((sqrt(5) + 1)/2*10^(-2 + n)) + floor((sqrt(5) + 1)/2*10^(-1 + n)) for n > 0. - Mariusz Iwaniuk, Apr 28 2017 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_{n>0} (1+2/(-1+2^n*(sqrt(4+(1-2/2^n)^2)+sqrt(4+(1-1/2^n)^2)))). - Gleb Koloskov, Jul 14 2021 EXAMPLE 1.6180339887498948482045868343656381177203091798057628621... MAPLE Digits:=1000; evalf((1+sqrt(5))/2); # Wesley Ivan Hurt, Nov 01 2013 MATHEMATICA RealDigits[(1 + Sqrt)/2, 10, 130] (* Stefan Steinerberger, Apr 02 2006 *) RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][] (* Robert G. Wilson v, Mar 01 2008 *) RealDigits[GoldenRatio, 10, 120][] (* Harvey P. Dale, Oct 28 2015 *) PROG (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 (PARI) /* 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 (Python) 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 CROSSREFS Cf. A000012, A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973. Cf. A102208, A102769, A131595. Cf. A302973, A303069, A304022. Sequence in context: A337369 A156921 A094214 * A186099 A021622 A073228 Adjacent sequences:  A001619 A001620 A001621 * A001623 A001624 A001625 KEYWORD nonn,cons,nice,easy AUTHOR EXTENSIONS 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 STATUS approved

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Last modified August 3 20:43 EDT 2021. Contains 346441 sequences. (Running on oeis4.)