cp's OEIS Frontend

Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).

Characteristic function of A022342. - Philippe Deléham, May 03 2004

a(n) = number of 0's between successive 1's (see also A003589 and A007538). - Eric Angelini, Jul 06 2005

With offset 1 this is the characteristic sequence for Wythoff A-numbers A000201=[1,3,4,6,...].

Eric Angelini's comment made me think that if 1 is defined to be the number of 0's between successive 1's in a string of 0's and 1's, then this string is 101. Applying the same operation to the digits of 101 leads to 101101, the iteration leads to successive palindromes of lengths given by A001911, up to a(n). - Rémi Schulz, Jul 06 2010

For generalized Fibonacci words see A221150, A221151, A221152, ... - Peter Bala, Nov 11 2013

The limiting mean of the first n terms is phi - 1; the limiting variance is phi (A001622). - Clark Kimberling, Mar 12 2014

Apply the difference operator to every column of the Wythoff difference array, A080164, to get an array of Fibonacci numbers, F(h). Replace each F(h) with h, and apply the difference operator to every column. In the resulting array, every column is A005614. - Clark Kimberling, Mar 02 2015

Binary expansion of the rabbit constant A014565. - M. F. Hasler, Nov 10 2018

A word that is morphic; but neither pure morphic, uniform morphic, primitive morphic, nor recurrent.

A word that is pure morphic, but neither uniform morphic, primitive morphic, nor recurrent.

The n-th runs of ones is given by: 3 - A003849(n) (infinite Fibonacci word) = A076662(n+1). Runs of zeros are given by: A085358 and are also directly related to the Fibonacci sequence. Coefficients of A(x)^3 are found in A085359.

a(n) = 0 iff some binary digit of n is 1 while the corresponding binary digit of 3*n is 0. - Robert Israel, Jul 12 2016

The Run Length Transform of [0,1,0,0,0,...], A063524, the characteristic function of 1. (See A227349 for the definition). - Antti Karttunen, Oct 15 2016

a(n) mod 2 = A003849(n);

a(n) = A000225(n+1) - A044432(n).

From Daniel Rust, Aug 18 2018: (Start)

This word is the fixed point of the morphism 0->2, 1->01, 2->2201 with the finite string 0, 1, 2, 0, 1 appended to the beginning. This morphism comes from taking the 'proper' version of the Fibonacci morphism 0->01, 1->1, given by 0->001, 1->01 (A189661 but with the rightmost 0 moved to the left of each image word), then replacing 001 with 2 and noting that the new symbol 2 should map to 00100101 = 2201 in order to be consistent.

The finite string appended to the beginning comes from the process of finding a proper version of the Fibonacci morphism using a return word encoding and taking conjugates which causes a shift of the respective fixed points.

(End)

This sequence is the unique fixed point of the morphism 0->01, 1->2, 2->0122. See the paragraph following Lemma 23 in the paper by Allouche and me. - Michel Dekking, Oct 05 2018

The positive integers are partitioned by this sequence together with A383424, A383425, and A383426.

Conjecture: {a(n) - a(n-1), n>=2} = {1, 3, 4, 5, 7}.

The positive integers are partitioned by this sequence together with A383423, A383424, and A383425.

Conjecture: {a(n) - a(n-1), n>=2} = {3, 7, 8, 10}.

From Michel Dekking, Mar 17 2020: (Start)

This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu. In fact, one can take the alphabet {A,B,C,D} with the morphism

mu: A->AB, B->CD, C->ABCD, D->CD,

and the letter-to-letter map lambda defined by

lambda: A->0, B->1, C->2, D->1.

Then (a(n)) = lambda(x), where x = ABCDABCDCD... is the unique fixed point of the morphism mu.

How does one see this? The infinite Fibonacci word

x_F = A003849 = 0100101001001....

can be written as a concatenation of the two words 01 and 001.

In fact, if beta is the Fibonacci morphism 0->01, 1->0, then beta(01)=010, beta(001)=01010, from which this can be read off.

This can also be found in Lemma 23 in Allouche and Dekking, which gives, moreover, that if we define the morphism g on the alphabet {a,b} by

g(a) = ab, g(b) =abb

then a(n) = delta(x_G(n)), where

x_G = ababbababb...

is the unique fixed point of g, and delta is the map

delta(a) = 01, delta(b) = 21.

In my paper "Morphic words,..." such a map delta is called a decoration.

It is well-known that decorated fixed points are morphic sequences, and the 'natural' algorithm to achieve this splits a in AB, and b in CD. This gives the morphism mu on the alphabet {A,B,C,D}, and the letter-to-letter map lambda.

We next prove the first conjecture by Kimberling. We easily see from the form of mu that the letters B and D occur, and only occur, at even positions in the fixed point x of mu. Since lambda(B)=lambda(D)=1, and A and C are not mapped to 1, it follows immediately that the positions of 1 in (a(n)) are given by A005843 = (2n).

For a proof of Kimberling's second conjecture, note that a(n)=2 iff x(n)=C, where x is the fixed point of mu. The return words of C in x are CD and CDAB. Coding these two return words by their lengths, mu induces a descendant morphism tau given by

tau(2) = 24, tau(4) = 244.

We see that tau is just an alphabet change of the morphism g. Let t = 2424424244... be the unique fixed point of tau. It is well-known (see, e.g., Lemma 12 in "Morphic words,..."), that t = 2 x_F, where x_F = x_F(4,2) now is the Fibonacci word on the alphabet {4,2}. The partial sums of x_F(4,2) are equal to the generalized Beatty sequence V given by

V(n) = 2*floor(n*phi) + 1,

according to Lemma 8 in the Allouche and Dekking paper.

This proves Kimberling's conjecture, provided we give the sequence A130568 offset 1, as we should.

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