A152149 -id:A152149 - cp's OEIS frontend

A376961 Length of the shortest side of the doubly golden triangle (A152149) that has area 1.

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

Offset: 1

Clark Kimberling, Nov 13 2024

The unique (shape of) triangle ABC that is both angle-golden and side-golden is discussed in A152149. The vertex angles, A,B,C are unique with A = B*tau and C = Pi - C*tau^2, where tau = golden ratio (A001622), but the lengths a,b,c of sides opposite A,B,C are not unique.  Instead, they are proportional to sin A, sin B, sin C. Consequently, if ABC is scaled so that its area is 1, then the sidelengths are unique. In that case,
b = length of shortest side = 1.1891197542670080135...
c = length of longest side = 1.70109767501680105234...
a = length of other side = 1.9240361790979417706848...
area = 1
perimeter = a+b+c = 4.8142536083817508366273974008...
circumradius = 0.972989352363244654532817794159950...
inradius = 0.4154330375362743229952970705418968790...
The following list gives approximate coordinates for five well-known triangle centers in the plane of the doubly golden triangle that has area 1. Approximate distances from each of these points X to the sidelines BC, CA, AB appear under the heading NTC (normalized trilinear coordinates), and approximate areas of the triangles BXC, CXA, AXB appear under NBC (normalized barycentric coordinates).
Triangle center     NTC                          NBC
incenter            (0.6366, 0.6366, 0.6366)   (0.4521, 0.2092, 0.3385)
centroid            (0.4693, 1.0140, 0.6267)   (1/3, 1/3, 1/3)
circumcenter        (0.2396, 1.2668, 0.7772)   (0.1702, 0.4164, 0.4133)
orthocenter         (1.0678, 0.2020, 0.3292)   (0.7196, 0.0841, 0.1961)
nine-point center   (0.0696, 0.0710, 1.7432)   (0.4299, 0.2710, 0.2990)

			1.189119754267008013592915705505374859646258022049...

Cf. A001622, A152149.

r = (1 + 5^(1/2))/2;
b = FindRoot[Sin[r*t + t] == r*Sin[t], {t, 1}, WorkingPrecision -> 120][[1, 2]];
c = r*b ; (* angle C, where b = angle B *)
a = Pi - r^2 b; (* angle A *)
{a1, b1, c1} = {Sin[a], Sin[b], Sin[c]}
k = 2/((a1 + b1 + c1) (-a1 + b1 + c1) (-b1 + c1 + a1) (-c1 + a1 + b1))^(1/4)
{k a1, k b1, k c1} (* sidelengths *)
k*b1 (* length of shortest side *)
RealDigits[k b1][[1]] (* this sequence *)

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

Original entry on oeis.org

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

A188615 Decimal expansion of Brocard angle of side-silver right triangle.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 05 2011

The Brocard angle is invariant of the size of the side-silver right triangle ABC.  The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(silver ratio)=(1+sqrt(2)).  This is the unique right triangle matching the continued fraction [2,2,2,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there are exactly 2 removable subtriangles at each stage.  (This is analogous to the removal of 2 squares at each stage of the partitioning of the silver rectangle as a nest of squares.)
Archimedes's-like scheme: set p(0) = 1/(2*sqrt(2)), q(0) = 1/3; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - A.H.M. Smeets, Jul 12 2018
This angle is also the half-angle at the summit of the Kelvin wake pattern traced by a boat. - Robert FERREOL, Sep 27 2019

			Brocard angle: 0.3398369094541219370963925133917640663882 approx.
Brocard angle: 19.471220634490691369245999 degrees, approx.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.
Wikipedia, Kelvin wake pattern
Index entries for transcendental numbers.

Cf. A188614, A188543, A152149.

[Arccos(Sqrt(8/9))]; // G. C. Greubel, Nov 18 2017

r=1+2^(1/2);
b=1; a=r*b; c=(a^2+b^2)^(1/2);
area=(1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);
brocard=ArcCot[(a^2+b^2+c^2)/(4area)];
N[brocard, 130]
RealDigits[N[brocard,130]][[1]]
N[180 brocard/Pi,130] (* degrees *)
RealDigits[ArcCos[Sqrt[8/9]], 10, 50][[1]] (* G. C. Greubel, Nov 18 2017 *)

acos(sqrt(8/9)) \\ Charles R Greathouse IV, May 02 2013

(Brocard angle) = arccot((a^2+b^2+c^2)/(4*area(ABC))) = arccot(sqrt(8)).
Also equals arcsin(1/3) or arccsc(3). - Jean-François Alcover, May 29 2013
Equals Integral_{x=sqrt(2)/2..sqrt(2)} dx/(x^2 + 1). - Kritsada Moomuang, May 29 2025

A188543 Decimal expansion of the angle B in the doubly silver triangle ABC.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 03 2011

There is a unique (shape of) triangle ABC that is both side-silver and angle-silver. Its angles are B, t*B and pi-B-t*B, where t is the silver ratio, 1+sqrt(2), at A014176. "Side-silver" and "angle-silver" refer to partitionings of ABC, each in a manner that matches the continued fraction [2,2,2,...] of t. For doubly golden and doubly e-ratio triangles, see A152149 and A188544. For the side partitioning and angle partitioning (i,e, constructions in which 2 triangles are removed at each stage, analogous to the removal of 1 square at each stage of the partitioning of the golden rectangle into squares) which match arbitrary continued fractions (of sidelength ratios and angle ratios), see the 2007 reference.

			B=0.4235466615478147887414222095779154 approximately.
B=24.2674 degrees approximately.

Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.

Cf. A152149, A014176, A188544.

r = 1+2^(1/2); Clear[t]; RealDigits[FindRoot[Sin[r*t + t] == r*Sin[t], {t, 1}, WorkingPrecision -> 120][[1, 2]]][[1]]

B is the number in [0,Pi] such that sin(B*t^2)=t*sin(B), where t=1+sqrt(2), the silver ratio.

A188595 Decimal expansion of Brocard angle of side-golden right triangle.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 05 2011

The Brocard angle is invariant of the size of the side-golden right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(golden ratio)=(1+sqrt(5))/2. This is the unique right triangle matching the continued fraction [1,1,1,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there is exactly 1 removable subtriangle at each stage. (This is analogous to the removal of 1 square at each stage of the partitioning of the golden rectangle as a nest of squares.)

Also <3_5> in Conway et al. (1999). - Eric W. Weisstein, Nov 06 2024

			Brocard angle: 0.420534335283965127888262515913215373 approx.

G. C. Greubel, Table of n, a(n) for n = 0..5000
John H. Conway, Charles Radin, and Lorenzo Sadun, On Angles Whose Squared Trigonometric Functions are Rational, arXiv:math-ph/9812019, 1998. See also Discr. Computat. Geom. (1999) Vol. 22, 321-332.
Eric Weisstein's World of Mathematics, Dehn Invariant.
Index entries for transcendental numbers.

Cf. A188594, A152149, A188615, A228496.

[Arctan(Sqrt(1/5))]; // G. C. Greubel, Nov 21 2017

r=(1+5^(1/2))/2; b=1; a=r*b; c=(a^2+b^2)^(1/2); area=(1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2); brocard = ArcCot[(a^2+b^2+c^2)/(4 area)]; RealDigits[N[brocard,130]][[1]]
RealDigits[ArcTan[Sqrt[1/5]], 10, 50][[1]] (* G. C. Greubel, Nov 21 2017 *)

atan(sqrt(1/5)) \\ G. C. Greubel, Nov 21 2017

Brocard angle: arccot((a^2+b^2+c^2)/(4*area(ABC))) = arccot(sqrt(5)).

Equals A228496/2. - Hugo Pfoertner, Nov 06 2024

A188594 Decimal expansion of (circumradius)/(inradius) of side-golden right triangle.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Apr 05 2011

This ratio is invariant of the size of the side-golden right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(golden ratio)=(1+sqrt(5))/2. This is the unique right triangle matching the continued fraction [1,1,1,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there is exactly 1 removable subtriangle at each stage. (This is analogous to the removal of 1 square at each stage of the partitioning of the golden rectangle as a collection of squares.)

Largest root of 4*x^4 - 20*x^2 - 20*x - 5. - Charles R Greathouse IV, May 07 2011

			2.656875757337521549489732...

G. C. Greubel, Table of n, a(n) for n = 1..5000
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.

Cf. A001622, A152149, A188595, A188614.

phi := (1+Sqrt(5))/2; [(Sqrt(5) + phi*Sqrt(2 + phi))/2]; // G. C. Greubel, Nov 23 2017

r=(1+5^(1/2))/2; b=1; a=r*b; c=(a^2+b^2)^(1/2);
area = (1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);
RealDigits[N[a*b*c*(a+b+c)/(8*area^2), 130]][[1]]
RealDigits[(Sqrt[5] + GoldenRatio*Sqrt[2 + GoldenRatio])/(2),10,50][[1]] (* G. C. Greubel, Nov 23 2017 *)

{phi = (1 + sqrt(5))/2}; (sqrt(5) + phi*sqrt(2 + phi))/2 \\ G. C. Greubel, Nov 23 2017

(circumradius)/(inradius)=abc(a+b+c)/(8*area^2), where area=area(ABC).

Equals (sqrt(5) + phi*sqrt(2 + phi))/2, where phi = A001622 is the golden ratio. - G. C. Greubel, Nov 23 2017

A188614 Decimal expansion of (circumradius)/(inradius) of side-silver right triangle.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Apr 05 2011

This ratio is invariant of the size of the side-silver right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(silver ratio)=(1+sqrt(2)). This is the unique right triangle matching the continued fraction [2,2,2,...] of r; i.e., under the side-partitioning procedure described in the 2007 reference, there are exactly 2 removable subtriangles at each stage. (This is analogous to the removal of 2 squares at each stage of the partitioning of the silver rectangle as a nest of squares.)

			ratio=3.26197262739566856105805510300327465221450 approx.

Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.

Cf. A188543, A152149, A188614, A188618.

a179260 := sqrt(2+sqrt(2)) ; a014176 := 1+sqrt(2) ; 1/(a014176/a179260-1) ; evalf(%) ; # R. J. Mathar, Apr 05 2011

r= 1+2^(1/2); b=1; a=r*b; c=(a^2+b^2)^(1/2); area=(1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2); RealDigits[N[a*b*c*(a+b+c)/(8*area^2),130]][[1]]

(circumradius)/(inradius) = abc(a+b+c)/(8*area^2), where area=area(ABC).

A188544 Decimal expansion of the angle B in the doubly e-ratio triangle ABC.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 03 2011

There is a unique (shape of) triangle ABC that is both side-e-ratio and angle-e-ratio. Its angles are B, t*B and pi-B-t*B, where t=e. "Side-e-ratio" and "angle-e-ratio" refer to partitionings of ABC, each in a manner that matches the continued fraction [2,1,2,1,1,4,1,1,6,...] of t. For doubly golden and doubly silver triangles, see A152149 and A188543. For the side partitioning and angle partitioning (i,e, constructions) which match arbitrary continued fractions (of sidelength ratios and angle ratios), see the 2007 reference.

			B=0.36893127494780584265191127268864 approximately.
B=21.1382 degrees approximately.

Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.

Cf. A152149, A188543.

r = E; Clear[t]; RealDigits[FindRoot[Sin[r*t + t] == r*Sin[t], {t, 1}, WorkingPrecision -> 120][[1, 2]]][[1]]

B is the number in [0,Pi] such that sin(B*e^2)=e*sin(B).

a(127) corrected by Sean A. Irvine, Sep 08 2021

A188616 Decimal expansion of angle B of unique side-golden and angle-silver triangle.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 05 2011

Let r=(golden ratio)=(1+sqrt(5))/2 and u=(silver ratio)=1+sqrt(2). A triangle ABC with sidelengths a,b,c is side-golden if a/b=r and angle-silver if C/B=u. There is a unique triangle that has both properties. The quickest way to understand the geometric reasons for the names is by analogy to the golden and silver rectangles. For the former, exactly 1 square is available at each stage of the partitioning of the rectangle into a nest of squares, and for the former, exactly 2 squares are available. Analogously, for ABC, exactly one 1 triangle of a certain kind is available at each stage of a side-partitioning procedure, and exactly 2 triangles of another kind are available for angle-partitioning. For details, see the 2007 reference.

			B=0.59106779970516487976323237419662 approximately

Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.

Cf. A152149, A188543.

Remove["Global`*"]; r=1+2^(1/2); u=(1+5^(1/2))/2; RealDigits[FindRoot[Sin[r*t+t]==u*Sin[t],{t,1}, WorkingPrecision->120][[1,2]]][[1]]

A178988 Decimal expansion of volume of golden tetrahedron.

Original entry on oeis.org

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

Offset: 2

Jonathan Vos Post, Jan 03 2011

Volume of tetrahedron with edges 1, phi, phi^2, phi^3, phi^4, phi^5 where phi is the golden ratio (1+sqrt(5))/2.

A152149 records more recent developments about side-golden and angle-golden triangles, both of which, like the golden rectangle, have generalizations that match continued fractions. There is a unique triangle which is both side-golden and angle-golden. Is there a comparable tetrahedron? - Clark Kimberling, Mar 31 2011

			75.7552212810...

Clark Kimberling, "A New Kind of Golden Triangle." In Applications of Fibonacci Numbers: Proceedings of the Fourth International Conference on Fibonacci Numbers and Their Applications,' Wake Forest University (Ed. G. E. Bergum, A. N. Philippou, and A. F. Horadam). Dordrecht, Netherlands: Kluwer, pp. 171-176, 1991.
Theoni Pappas, "The Pentagon, the Pentagram & the Golden Triangle." The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, pp. 188-189, 1989.

Marjorie Bicknell and Verner E. Hoggatt Jr., Golden Triangles, Rectangles, and Cuboids, Fib. Quart. 7, 73-91, 1969.
Frank M. Jackson and Eric W. Weisstein, Tetrahedron.
Clark Kimberling, A New Kind of Golden Triangle, In: Bergum G.E., Philippou A.N., Horadam A.F. (eds), Applications of Fibonacci Numbers. Springer, Dordrecht, pp. 171-176, 1991.
Robert Schoen, The Fibonacci Sequence in Successive Partitions of a Golden Triangle, Fib. Quart. 20, 159-163, 1982.
Eric W. Weisstein, Golden Triangle.

Cf. A001622, A071399, A171973, A070169, A126766, A010524, A093524, A093525, A093591, A152149.

RealDigits[Sqrt[275465/96 + 369575*Sqrt[5]/288], 10, 120][[1]] (* Amiram Eldar, Jun 12 2023 *)

sqrt(275465/96 + (369575*sqrt(5))/288) \\ Charles R Greathouse IV, May 27 2016

Equals sqrt(275465/96 + (369575*sqrt(5))/288).

The minimal polynomial is 20736*x^4 - 119000880*x^2 + 73225. - Joerg Arndt, Jul 25 2021

a(101) corrected by Georg Fischer, Jul 25 2021

cp's OEIS Frontend

A376961 Length of the shortest side of the doubly golden triangle (A152149) that has area 1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Python

Formula

Extensions

A188615 Decimal expansion of Brocard angle of side-silver right triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A188543 Decimal expansion of the angle B in the doubly silver triangle ABC.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A188595 Decimal expansion of Brocard angle of side-golden right triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A188594 Decimal expansion of (circumradius)/(inradius) of side-golden right triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A188614 Decimal expansion of (circumradius)/(inradius) of side-silver right triangle.

Views