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From Michel Dekking, Sep 16 2019: (Start)

Let sigma be the defining morphism in A287104: 0->10, 1->12, 2->0.

Let u := 10, v := 12, w: = 120 be the return words of the word 1. [See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]

Then

sigma(u) = vu, sigma(v) = w, sigma(w) = wu.

If we code w<->0, u<->1, v<->2, then this morphism turns into the morphism

0 -> 01, 1 -> 21, 2 -> 0.

This is exactly the morphism which has A287072 as unique fixed point.

Since u and v have length 2 and w has length 3, this implies that the sequence d of first differences of (a(n)) equals A287072 with the projection 0 -> 3, 1 -> 2, 2 -> 2. This gives the formula below.

(End)

From Michel Dekking, Sep 16 2019: (Start)

Let sigma be the defining morphism of A287104: 0->10, 1->12, 2->0.

Let u=201, v=2101, w=20101 be the return words of the word 2.

Under sigma u, v, and w are mapped to sigma(201) = 01012, sigma(2101) = 0121012, sigma(20101) = 010121012.

All three images have suffix 2. We can therefore move this suffix to the front of all three images, obtaining the fixed point (a(n+1)) = 20101... when iterating. This induces the morphism 3 -> 5, 4 -> 34, 5 -> 54 on the return words, coded by their lengths.

Coding the symbols according to 3<->2, 4<->1, 5<->0, this leads to the morphism 2->0, 1->21, 0->01 on the alphabet {0,1,2}. This is exactly the morphism which has A287072 as unique fixed point. So the sequence d of first differences of (a(n)) equals A287072 with the coding above. This gives the formula below.

(End)