cp's OEIS Frontend

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

Also the limiting ratio of Lucas(n)/Fibonacci(n+1), or Fibonacci(n-1)/Fibonacci(n+1) + 1. - Alexander Adamchuk, Oct 10 2007

a(n) is the numerator of the n-th iterate when Newton's method is applied to the function x^2 - x - 1 with initial guess x = 1. The n-th iterate is a(n)/A058635(n). - Jason Zimba, Jan 20 2023

Compare with A192223(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 is taken equal to 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*(1 + sqrt(5)) denote the golden ratio A001622. This sequence, apart from the initial term, provides a Pierce expansion of 1 to the base phi. That is, we have the identity 1 = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ....

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 1 to the base phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] provides a Pierce expansion of the quadratic irrational 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi. Some examples are given below.

Let x and b be positive real numbers. We define an Engel expansion of x to the base b to be a (possibly infinite) nondecreasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the series representation x = b/a(1) + b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) + .... Depending on the values of x and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 we recover the ordinary Engel expansion of x.

Let Phi := 1/2*(sqrt(5) - 1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, gives an Engel expansion of x = Phi^6 to the base b = Phi^2. The associated Engel series expansion of Phi^6 to the base Phi^2 begins Phi^6 = Phi^2/7 + Phi^4/(7*18) + Phi^6/(7*18*123) + Phi^8/(7*18*123*5778) + ....

This result can be extended in two ways. Firstly, for k = 1,2,3,... , the sequence {Lucas(k*(2^n + 2))}n>=1 is an Engel expansion of Phi^(6*k) to the base Phi^(2*k). Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] is an Engel expansion of Phi^(2^n + 4) to the base Phi^2. See below for some examples.

Let phi := 1/2*(1 + sqrt(5)) denote the golden ratio. This sequence, apart from the initial term, gives an Engel expansion of 1 to the base phi^2 (see A230601 for a definition of this term). The associated Engel series expansion of 1 to the base phi^2 begins 1 = phi^2/3 + phi^4/(3*18) + phi^6/(3*18*843) + phi^8/(3*18*843*1860498) + .... This result can be extended in two ways. Firstly, the sequence Lucas(2*k*(2^n - 1)) for k = 1,2,3,... is an Engel expansion of 1 to the base phi^(2*k). Secondly, for n = 1,2,3,... the sequence [a(n),a(n+1),a(n+2),...] is an Engel expansion of phi^(4 - 2^n) to the base phi^2. Some example are given below.

Let r and b be positive real numbers. We define a Pierce expansion of r to the base b to be a (possibly infinite) increasing sequence of positive integers [a(0), a(1), a(2), ...] such that we have the alternating series representation r = b/a(0) - b^2/(a(0)*a(1)) + b^3/(a(0)*a(1)*a(2)) - .... Depending on the values of r and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 and 0 < r < 1 we recover the ordinary Pierce expansion of r.

See A058635, A192223 and A230600 for some predictable Pierce expansions to a base b other than 1.

In the particular case that the base b >= 1 and 0 < r < b then we can find a Pierce expansion of r to the base b as follows:

Define the map f(x) (which depends on the base b) by f(x) = x/b*ceiling(b/x) - 1 and let f^(n)(x) denote the n-th iterate of the map f(x), with the convention that f^(0)(x) = x.

For n = 0,1,2,... define a(n) = ceiling(b/f^(n)(-r)) until f^n(-r) = 0.

Then it can be shown that the sequence of positive integers |a(n)| is a Pierce expansion of r to the base b.

For the present sequence we apply this algorithm with r := 1 and with base b := Pi. See A232325 for an Engel expansion of 1 to the base Pi.