cp's OEIS Frontend

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 17 2019: (Start)

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

Let u=01, v=012, w=0121 be the return words of the word 0.

[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]

Then under sigma u, v and w are mapped to

sigma(01) = 1012, sigma(012) = 10120, sigma(0121) = 1012012.

Moving the prefix 1 of these three images to the end, the sequence 0 a (i.e., (a(n)) prefixed by the symbol 0), is a fixed point when iterating.

This iteration process induces a morphism 2->4, 3->32, 4->34 on the return words, coded by their lengths.

Coding the symbols according to 2<->2, 4<->0, 3<->1, this leads to the morphism 2->0, 1->12, 0->10 on the alphabet {0,1,2}.

This is simply sigma, which has A287104 as its unique fixed point. So the sequence d of first differences of (a(n)) equals A287104 with the coding above. Noting that the code can be written as x->4-x, 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)

A fixed point of the morphism 0->01, 1->21, 2->0. Let u be the sequence of positions of 0, and likewise, v for 1 and w for 2. Let U, V, W be the limits of u(n)/n, v(n)/n, w(n)/n, respectively. Then 1/U + 1/V + 1/W = 1, where

U = 3.079595623491438786010417...,

V = 2.324717957244746025960908...,

W = U + 1 = 4.079595623491438786010417....

Since the morphism 0->01, 1->21, 2->0 is the time reversal of the morphism 0->10, 1->12 2->0, which has fixed point A287104, in particular the incidence matrices of these two morphisms are equal. Thus the algebraic expressions found for U, V and W in A287104 do also apply to the U, V and W above. - Michel Dekking, Sep 15 2019

If n >=2, then u(n) - u(n-1) is in {2,3,4}, v(n) - v(n-1) is in {2,3}, and w(n) - w(n-1) is in {3,4,5}.