cp's OEIS Frontend

A Sturmian word.

Define strings S(0)=0, S(1)=01, S(n)=S(n-1)S(n-2); iterate; sequence is S(infinity). If the initial 0 is omitted from S(n) for n>0, we obtain A288582(n+1).

The 0's occur at positions in A022342 (i.e., A000201 - 1), the 1's at positions in A003622.

Replace each run (1;1) with (1;0) in the infinite Fibonacci word A005614 (and add 0 as prefix) A005614 begins: 1,0,1,1,0,1,0,1,1,0,1,1,... changing runs (1,1) with (1,0) produces 1,0,0,1,0,1,0,0,1,0,0,1,... - Benoit Cloitre, Nov 10 2003

Characteristic function of A003622. - Philippe Deléham, May 03 2004

The fraction of 0's in the first n terms approaches 1/phi (see for example Allouche and Shallit). - N. J. A. Sloane, Sep 24 2007

The limiting mean and variance of the first n terms are 2-phi and 2*phi-3, respectively. - Clark Kimberling, Mar 12 2014, Aug 16 2018

Let S(n) be defined as above. Then this sequence is S(1) + Sum_{n=0..} S(n), where the addition of strings represents concatenation. - Isaac Saffold, May 03 2019

The word is a concatenation of three runs: 0, 1, and 00. The limiting proportions of these are respectively 1 - phi/2, 1/2, and (phi - 1)/2. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010

From Amiram Eldar, Mar 10 2021: (Start)

a(n) is the number of the trailing 0's in the dual Zeckendorf representation of (n+1) (A104326).

The asymptotic density of the occurrences of k (0 or 1) is 1/phi^(k+1), where phi is the golden ratio (A001622).

The asymptotic mean of this sequence is 1/phi^2 (A132338). (End)

A "non-commutative Fibonacci" (or "reverse Fibonacci") sequence. Often written as: a, b, ab, bab, abbab, bababbab, abbabbababbab, bababbababbabbababbab, abbabbababbabbababbababbabbababbab, bababbababbabbababbababbabbababbabbababbababbabbababbab, ...

Converges in the appropriate topology. - Dylan Thurston, Jan 28 2005

Do a web search on bababbababbabbababbab to get further links.

From Andrew Hone, Jan 28 2005: (Start)

Write the recurrence symbolically as g_{n+1} = g_{n-1}g_n. Then the determinant d_n = det g_n is given by d_n = d_0^{f_{n-2}} d_1^{f_{n-1}} where f_{n+1} = f_n+f_{n-1}, f_0 = f_1 = 1 are the Fibonacci numbers.

To avoid getting involved with the Baker-Campbell-Hausdorff identity, I now restrict to SL(2), or to make life easier make it SU(2) (which is isomorphic over C). Then we can explicitly write g as an exponential of Lie algebra elements:

g_n = exp (i theta_n v_n cdot sigma ), where theta_n is an angle, v_n is a unit vector and sigma = ( sigma_1, sigma_2, sigma_3)^T is a vector of Pauli spin matrices.

Moreover the adjoint action on su(2) (viewing the coordinates in su(2) as giving points in 3D space) means that g_n gives a rotation through - theta_n /2 about the v_n axis.

So from the double cover of SO(3) by SU(2), we can view the g_n as a sequence of "Fibonacci rotations."

Furthermore, in SU(2) we can write explicitly g_n = cos theta_n + i sin theta_n v_n cdot sigma so the recurrence can be decoupled as

cos theta_{n+1} = cos theta_n + cos theta_{n-1} - sin theta_{n-1} sin theta_n (v_{n-1} cdot v_n),

sin theta_{n+1} v_{n+1} = cos theta_{n-1} sin theta_n v_n + cos theta_n sin theta_{n-1} v_{n-1} - sin theta_{n-1} sin theta_n ( v_{n-1} wedge v_n ). (End)

Changing the offset to 1, the sum of the digits of a(n) is 2*Fib(n-1)+Fib(n-2), where Fib(n) means A000045(n), the n-th Fibonacci number. - Stefan Steinerberger, Feb 05 2006

Let beta be the reversed, mirrored Fibonacci morphism on the alphabet {1,2} given by beta(1)=2, beta(2)=12. Then a(n) = beta^n(1), since from the formula beta^2(1)= 12 = 1 beta(1), one sees directly that the iterates of the letter 1 under beta satisfy the defining recursion a(n) = a(n-2)a(n-1). It follows that the a(2n) converge to A189479 with 1,2 replaced by 0 and 1, and the a(2n+1) converge to A287523 with 1,2 replaced by 0 and 1. - Michel Dekking, Sep 30 2019