OFFSET
1,2
COMMENTS
The hyperbolas x^2 - y^2 = 1 and xy = 1 meet at (c, 1/c) and (-c, -1/c), where c = sqrt(golden ratio); see the Mathematica program for a graph. - Clark Kimberling, Oct 19 2011
An algebraic integer of degree 4. Minimal polynomial: x^4 - x^2 - 1. - Charles R Greathouse IV, Jan 07 2013
Also the limiting value of the ratio of the slopes of the tangents drawn to the function y=sqrt(x) from the abscissa F(n) points (where F(n)=A000045(n) are the Fibonacci numbers and n > 0). - Burak Muslu, Apr 04 2021
The length of the base of the isosceles triangle of smallest perimeter which circumscribes a unit-diameter semicircle (DeTemple, 1992). - Amiram Eldar, Jan 22 2022
The unique real solution to arcsec(x) = arccot(x). - Wolfe Padawer, Apr 14 2023
REFERENCES
B. Muslu, Sayılar ve Bağlantılar 2, Luna, 2021, pages 45-48.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
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.
Duane W. DeTemple, The Triangle of Smallest Perimeter which Circumscribes a Semicircle, The Fibonacci Quarterly, Vol. 30, No. 3 (1992), p. 274.
FORMULA
Equals sqrt((1 + sqrt(5))/2).
Equals 1/sqrt(A094214). - Burak Muslu, Apr 04 2021
From Amiram Eldar, Feb 07 2022: (Start)
Equals 1/A197762.
Equals tan(arccos(1/phi)).
Equals cot(arcsin(1/phi)). (End)
From Gerry Martens, Jul 30 2023: (Start)
Equals 5^(1/4)*cos(arctan(2)/2).
Equals Re(sqrt(1+2*i)) (the imaginary part is A197762). (End)
EXAMPLE
1.2720196495140689642524224617374914917156080418400...
MAPLE
Digits:=100: evalf(sqrt((1+sqrt(5))/2)); # Muniru A Asiru, Sep 11 2018
MATHEMATICA
N[Sqrt[GoldenRatio], 100]
FindRoot[x*Sqrt[-1 + x^2] == 1, {x, 1.2, 1.3}, WorkingPrecision -> 110]
Plot[{Sqrt[-1 + x^2], 1/x}, {x, 0, 3}] (* Clark Kimberling, Oct 19 2011 *)
PROG
(PARI) sqrt((1+sqrt(5))/2) \\ Charles R Greathouse IV, Jan 07 2013
(PARI) a(n) = sqrtint(10^(2*n-2)*quadgen(5))%10; \\ Chittaranjan Pardeshi, Aug 24 2024
CROSSREFS
KEYWORD
AUTHOR
Mohammad K. Azarian, Apr 14 2008
STATUS
approved