[seqfan] Re: Rewriting squares

[Neil Sloane]

> or with 6 (639181...). This last sequence might be worth adding.
Me: I think it is enough just to have A308170 and -1

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Thu, May 16, 2019 at 1:54 PM jean-paul allouche <
jean-paul.allouche at imj-prg.fr> wrote:

> Dear Maximilian
>
> I think that the morphism in A308171
> is not quite correct: the image of 9 should be 18
> and the image of 8 should be 46. This is the same
> morphism as for A308170. This morphism has
> three infinite fixed points respectively beginning
> with 1 (A308170), or with 5 (A308171) or with 6
> (639181...). This last sequence might be worth adding.
>
> best wishes
> jean-paul
>
>
>
> Le 16/05/2019 à 01:11, M. F. Hasler a écrit :
> > 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
> >
> > --
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>
>
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