[seqfan] Re: morphism in A284940 compared with A080580

[Andrew Weimholt]

Let a=A080580, and A be the set of terms of a
Let b=A294940

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

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.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> 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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