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
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
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
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 is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. -
Chayim Lowen, Aug 30 2015
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)
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
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)
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>=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)
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