# [seqfan] Re: morphism in A284940 compared with A080580

jean-paul allouche jean-paul.allouche at imj-prg.fr
Tue Apr 25 21:07:45 CEST 2017

```sure
jean-paul

Le 25/04/17 à 21:06, Andrew Weimholt a écrit :
> Correction
> b=A284939.
>
> Let a=A080580 and A be the set of terms of a
> Let b=A284939
>
> b(n)=0 => n in A => a(n) = a(n-1)+2 => b(a(n-1)+2),b(a(n-1)+3) = 0,1
>
> b(n)=1 => n not in A => a(n) = a(n-1)+4 => b(a(n-1)+2), b(a(n-1)+3),
> b(a(n-1)+4), b(a(n-1)+5) = 1,1,0,1
>
> One should be able to prove this rigorously using induction, since the
> initial terms agree.
>
> Andrew
>
>
>
>> On Tue, Apr 25, 2017 at 3:13 AM, Neil Sloane <njasloane at gmail.com> wrote:
>>
>>> In other words, RJM is saying that it appears that
>>>
>>>   "the positions of 0's in the fixed point of the morphism 0 -> 01, 1 ->
>>> 1101"
>>> are given by
>>> "a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence,
>>> a(n)=a(n-1)+4 otherwise".
>>>
>>> That is a pretty interesting conjecture (A284940 =? A080580)!  It
>>> would be worth checking it for a few million terms, or more.
>>>
>>> Clark Kimberling (if you are on this list), I know you
>>> have recently been studying many similar "Positions of 0's in fixed
>>> point of ..."
>>> sequences.  Have you observed any other apparent coincidences of this
>>> type?
>>> 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.
>>> Email: njasloane at gmail.com
>>>
>>>
>>>
>>> On Tue, Apr 25, 2017 at 3:28 AM, Richard J. Mathar
>>> <mathar at mpia-hd.mpg.de> wrote:
>>>> The first 1000 terms (at least) of A284940 equal the first
>>>> 1000 terms of A080580. Can this be demonstrated for all general values?
>>>>
>>>> Richard
>>>>
>>>> --
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>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
> --
> Seqfan Mailing list - http://list.seqfan.eu/

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