%I #25 May 02 2021 10:55:13

%S 2,1,1,3,2,1,2,1,1,3,1,3,2,1,1,3,2,1,2,1,1,3,2,1,1,3,1,3,2,1,2,1,1,3,

%T 1,3,2,1,1,3,2,1,2,1,1,3,1,3,2,1,2,1,1,3,2,1,1,3,1,3,2,1,1,3,2,1,2,1,

%U 1,3,2,1,1,3,1,3,2,1,2,1,1,3,1,3,2,1,1,3,2,1,2,1,1,3,2,1,1,3,1,3

%N a(1) = 2; for n >= 2, if a(n-1) has not yet been assigned, then a(n-1) = 1 and a(2*n-1) = 2, otherwise a(2*n) = 3.

%C A self-generating sequence.

%C By calculation it looks as though lim_{m->oo} (1/m)*Sum_{n=1..m} a(n) = sqrt(3).

%C The 3's are at positions 4, 10, 12, 16, 22, 26, 28, 34, 36, 40, 46, 48, 54, 58, 60, 64, 70, 74, 76, 82, 84, 88, 94, 98,..., which is 2*A059009(n).

%C From _Michel Dekking_, Sep 11 2020: (Start)

%C The conjectured limit above is incorrect, this limit equals 7/4 = 1.75, instead of sqrt(3) = 1.732... This follows from the CLAIM below.

%C Let nu be the following extension of the 'exchanged' Thue-Morse morphism to a morphism on 3 letters:

%C nu(1) = 21, nu(2) = 13, nu(3) = 13.

%C CLAIM: a(567...) = nu(a(234...)).

%C Here a(567...) denotes the word associated to the sequence a(5),a(6),....

%C Proof: If a(n)=1 then a(2n+1)=2, and it also follows that a(2n+2)=1. If a(n)>1 then a(2n+2)=3, and it also follows that a(2n+1)=1.

%C This can also be expressed as nu(a(n))=a(2n+1)a(2n+2). This gives the claim.

%C (End)

%C From _Michel Dekking_, Sep 12 2020: (Start)

%C The second conjecture above is correct. For a proof, we consider nu^2:

%C nu^2(1) = 1321, nu^2(2) = 2113, nu^2(3) = 1321.

%C Let delta be the decoration 0 -> 21, 1 -> 13.

%C CLAIM: delta(b) = a, where b := A059448.

%C A proof is involved, as (a(n)) is not a fixed point of nu^2.

%C First we note that nu(a(n))=a(2n+1)a(2n+2) implies that for all n

%C nu^2(a(2n-1) )= a(8n-1)a(8n)...a(8n+6).

%C Let mu be the square of the Thue-Morse morphism:

%C mu(0) = 0110, mu(1) = 1001.

%C We use the following property, noted by Shallit in the comments of A059448:

%C b = 010 mu(b).

%C First we prove the following lemma on compositions of morphisms.

%C LEMMA delta mu = nu^2 delta.

%C Proof: delta(mu(0)) = delta(0110) = 21131321 = nu^2(21) = nu^2(delta(0)), and delta(mu(1)) = delta(1001) = 13212113 = nu^2(13) = nu^2(delta(1)).

%C To prove the CLAIM, start with

%C delta(b(123)) = delta(010) = 211321 = a(123456).

%C We continue with induction, using a recursion formula for b from the Comments of A059448, the LEMMA, and the induction hypothesis:

%C delta(b(4n)...b(4n+3)) = delta(mu(b(n)) = nu^2(delta(b(n))) =

%C nu^2(a(2n-1)a(2n)) = a(8n-1)....a(8n+6).

%C Looking at the positions of 3 in (a(n)), and the positions of 1 in (b(n)), one sees that the claim implies the second conjecture.

%C (End)

%H Michel Dekking, <a href="/A323756/b323756.txt">Table of n, a(n) for n = 1..210</a>

%H J.-P. Allouche and M. Mendes France, <a href="https://webusers.imj-prg.fr/~jean-paul.allouche/allmendeshouches.pdf">Automata and Automatic Sequences</a>, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11.

%H J.-P. Allouche and M. Mendes France, <a href="/A003842/a003842.pdf">Automata and Automatic Sequences</a>, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11. [Local copy]

%e a(1) = 2;

%e for n = 2, a(1) has been assigned: a(1) = 2, thus a(4) = 3;

%e for n = 3, a(2) is unassigned, thus a(2) = 1 and a(5) = 2;

%e for n = 4, a(3) is unassigned, thus a(3) = 1 and a(7) = 2; etc.

%Y Cf. A059009. Related to A298952 and A059448.

%K nonn

%O 1,1

%A _Ctibor O. Zizka_, Jan 26 2019

%E New first element: a(1)=2 instead of a(1)=1. _Michel Dekking_, Sep 12 2020