[seqfan] Re: Rewriting squares

[M. F. Hasler]

On Sun, May 12, 2019 at 12:08 PM jean-paul allouche wrote:

> By the definition itself, the infinite sequence is obtained by iterating a
> morphism
> (in the usual sens in combinatorics on words). For example, starting with 6
> is exactly iterating the morphism
> 6 --> 63
> 3 --> 9
> 9 --> 18
> 8 --> 46
> 1 --> 1
> 4 --> 61
> which gives 6 --> 63 --> 639 --> 63918 --> 63918146...
>

Indeed! In particular,
the digit 7 (surprisingly chosen as initial value, rather than 3 or 9) will
never occur.
Similarly, when starting with 5 (A308171), we have to amend the above with
5 -> 52 ; 2 -> 4
but the digits (5,2) = A308171(1..2) will never occur elsewhere again.

Can it be proved or disproved that we can have A308170(n) = A308171(n+k)
for some k and all n sufficiently large?
What can be said / proved about the respective densities and /or positions
- of the digits that occur infinitely often ?
- where finite subsequence a(m..n), n>m>2, will occur again in the sequence?

The sequence clearly is not squarefree (we have the cube 1, 6, 3, 1, 6, 3,
1, 6, 3 quite early)
but can one make some other statement concerning squares, cubes... of given
/ minimal length ?

On 12/05/2019 à 14:22, Neil Sloane wrote:
> > It is certainly interesting.  We should have the two limiting sequences
> in the OEIS:
> >
> 6,3,9,1,8,1,4,6,1,6,1,6,3,1,6,3,1,6,3,9,1,6,3,9,1,6,3,9,1,8,1,6,3,9,1,8,1,6,
> > 3,9,1,8,1,4,6,1,6,3,9,1,8,1,4,6,1,6,3,9,1,8, ... ===> now  A308170
> > or in the case of 25,
> >
> 5,2,4,6,1,6,3,1,6,3,9,1,6,3,9,1,8,1,6,3,9,1,8,1,4,6,1,6,3,9,1,8,1,4,6,1,6,1,
> > 6,3,1,6,3,9,1,8,1,4,6,1,6,1,6,3,1,6,3,1,6,3, ... ===> now A308171
>


> > Also the number of steps to reach the limit cycle when starting from n,


It's unclear to me what could mean to "reach a limit cycle".
Up to where the sequence(word) must coincide with the limit (which is never
reached, except for the fixed point "1") ?
Just the initial character? (This wouldn't be much interesting IMHO.)
--
Maximilian