[seqfan] Re: Rewriting squares
jean-paul.allouche at imj-prg.fr
Thu May 16 22:26:05 CEST 2019
Thanks for this point
Le 16/05/2019 à 21:35, Neil Sloane a écrit :
>> 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 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
>> 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
>>>> (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)
>>> never occur.
>>> Similarly, when starting with 5 (A308171), we have to amend the above
>>> 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
>>> - of the digits that occur infinitely often ?
>>> - where finite subsequence a(m..n), n>m>2, will occur again in the
>>> The sequence clearly is not squarefree (we have the cube 1, 6, 3, 1, 6,
>>> 1, 6, 3 quite early)
>>> but can one make some other statement concerning squares, cubes... of
>>> / 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:
>>>>> 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,
>>>>> 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
>>> reached, except for the fixed point "1") ?
>>> Just the initial character? (This wouldn't be much interesting IMHO.)
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