cp's OEIS Frontend

Midsphere radius of regular icosahedron with unit edges.

Also half of the golden ratio (A001622). - Stanislav Sykora, Jan 30 2014

Andris Ambainis (see Aaronson link) observes that combining the results of Barak-Hardt-Haviv-Rao with Dinur-Steurer yields the maximal probability of winning n parallel repetitions of a classical CHSH game (see A201488) asymptotic to this constant to the power of n, an improvement on the naive probability of (3/4)^n. (All the random bits are received upfront but the players cannot communicate or share an entangled state.) - Charles R Greathouse IV, May 15 2014

This is the height h of the isosceles triangle in a regular pentagon, in length units of the circumscribing radius, formed by a side as base and two adjacent radii. h = sin(3*Pi/10) = cos(Pi/5) (radius 1 unit). - Wolfdieter Lang, Jan 08 2018

Also the limiting value(L) of "r" which is abscissa of the vertex of the parabola F(n)*x^2 - F(n+1)*x + F(n + 2)(where F(n)=A000045(n) are the Fibonacci numbers and n>0). - Burak Muslu, Feb 24 2021

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

This is the height h of the isosceles triangle in a regular pentagon inscribed in the unit circle formed from a diagonal as base and two adjacent pentagon sides. h = sqrt(sqrt(3-phi)^2 - (sqrt(2 + phi)/2)^2) = sqrt(10 - 5*phi)/2 = (3 - phi)/2. - Wolfdieter Lang, Jan 07 2018

(13 - 5*sqrt(5))/2 = lim_{n->oo} -F(n)+(2+F(n+1))*(1+F(n+2))/(3+F(n+3)), where F = A000045 (Fibonacci numbers).

Apart from leading digits the same as A187798 and A187426. - R. J. Mathar, Sep 21 2013

Apart from the first digit the same as A187799.

Essentially the same as A225667 and A132338. - R. J. Mathar, Sep 30 2013

A word constructed by replacing each 00 factor of the Fibonacci word (A003849) with 020. The obtained ternary sequence is a word with Sturmian erasures (by removing each word,the obtained binary sequence is Sturmian)[1]. By removing each of 0's or 2's, the set of replacements on the Fibonacci word, is equal to the morphisms of deriving the Fibonacci word [2]. So the obtained binary word by removing each of 0's,1's or 2's is the Fibonacci word. Since the slope of the sequential projection (sending for example one letter to 1 and all the others to 0) is 1, the factor complexity of this ternary word for each integer n>0, is n+2.[3]

The binary sequence obtained by removing all 0's from the 3-Fibonacci word: 1,2,1,1,2,1,2,1,1,2,1,2,1,1,2,1,...

From Michel Dekking, Oct 19 2016: (Start)

The sequence (a(n)) is fixed point of the morphism zeta given by zeta: 0->01, 1->02, 2->epsilon.

Here epsilon is the empty word. To see this, code the 0’s in the Fibonacci sequence followed by 0 by 5, and the 0’s followed by 1 by 6. Then add 2 after 5. This gives the morphism 1->52, 5->61, 6->61, 2->epsilon. Then injectivize, i.e., map 5 and 6 to 0.

The sequence (a(n)) is related to A108103. Let theta be the standard form of zeta: theta(1)=12, theta(2)=13, theta(3)=epsilon. Let psi be the morphism generating the version of A108103 with 2 and 3 interchanged, psi: 1->2, 2->131, 3->1. Then the unique fixed point of theta is different from the fixed points of psi, but theta and psi generate the same language, i.e., arbitrarily long words occurring in the fixed point of theta occur in the fixed points of psi. This is a nontrivial exercise (prove that 2 theta^{2n}(1) = psi^{2n}(2) 13 for all n>0).

The sequence (a(n)) is not related to A270788, which might be called the ternary Fibonacci sequence. The dynamical system generated by (a(n)) has an eigenvalue -1, whereas the system generated by A270788 is isomorphic to the Fibonacci dynamical system. (End)

The asymptotic density of the occurrences of 0, 1, and 2 is 1/2, 1/(2*phi) = A019827, and 1/(2*phi^2) = A187426 / 10, respectively, where phi is the golden ratio (A001622). The asymptotic mean of this sequence is (3-phi)/2 (A187798). - Amiram Eldar, May 28 2024