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-1 < n*r - a(n) < 1 for n >= 1, where r = (5 + sqrt(5))/4.

From Michel Dekking, Dec 28 2017: (Start)

Let (d(n)) be the sequence of first differences: d(n)=a(n+1)-a(n).

CLAIM: d(n) = A108103(n+1) for n=1,2,….

Proof: As a word A287772 = 100110010011001100100110010011... obtained by substituting 0->1, 1->00 in the Fibonacci word F=0100101001001010010100...

This implies that A287772 is a concatenation of 00’s separated by 1’s and 11’s. Moreover, a 0110 occurs iff 1001 occurs in F, and a 010 occurs iff 101 occurs in F. Note also that occurrence of a 00 in A287772 yields a d(n)=1 (and so every other letter in d is a 1), occurrence of a 010 yields a d(n)=2, and occurrence of a 0110 yields a d(n)=3. Since the 1001’s and 101’s occur in 1F according to F itself with 1 prepended (see A001468 and A282162), we must have d(n)=A108103(n+1). (End)

This is the first sequence reached in the infinite process described in the A066983 comment line.

(a(n)) is a morphic sequence, i.e., a letter to letter projection of a fixed point of a morphism. The morphism is 1->121,2->3,1,3->313. The fixed point is the fixed point 121312131312... starting with 1. The letter-to-letter map is 1->1, 2->1, 3->3. See also the comments in A108103. - Michel Dekking, Jan 06 2018

This is the second sequence reached in the infinite process described in A066983 comment line.

(a(n)) is a morphic sequence, i.e., a letter to letter projection of a fixed point of a morphism. The morphism is 1->121,2->3,1,3->313. The fixed point is the fixed point 3131213131213... starting with 3. The letter-to-letter map is 1->1, 2->1, 3->3. See also COMMENTS of A108103. - Michel Dekking, Jan 06 2018

Convolution of the sequences A003263 and A033999.

Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).

From Michel Dekking, Dec 30 2017: (Start)

Proof of the 'positions of 0' conjecture: let (z(n))=1,2,5,6,8,9,12,... be the positions of 0. The crucial observation is that if a number n is the sum of distinct Lucas parts greater than 1, then n+1 is a sum of Lucas parts. This implies that (z(2n))=2,6,9,13,... is the sequence of numbers A054770 that are not a sum of Lucas numbers. We see there that Ian Agol proved that b(n):=A054770(n)=floor(phi*n)+2n-1. But then the sequence of first differences (b(n+1)-b(n)) equals the Fibonacci word on the alphabet {4,3}, yielding that (z(2n)-z(2n-1)) equals the Fibonacci word on {3,2}, and we already know that z(2n+1)-z(2n)=1 for all n. On the other hand, A287775 has the same first difference sequence given by A108103. Since A287775(1)=2, the conjecture follows. (End)

Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).

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

a(n) = Pi(A108103(n+2)), where Pi is the permutation Pi(1)=3, Pi(2)=2, Pi(3)=1. For a proof, see my link. - Michel Dekking, Dec 28 2019