[seqfan] Re: Need proof that Kimberling's A26185 = A198173

[Neil Sloane]

Kevin, great, thanks!  I thought there were probably more of these
duplicates but I didn't have time to do a search.
I'll do the merging over the next couple of days


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 Mon, Feb 3, 2020 at 7:54 PM Kevin Ryde via SeqFan <seqfan at list.seqfan.eu>
wrote:

> njasloane at gmail.com (Neil Sloane) writes:
> >
> > I have now processed all these pairs or triples or even quadruples ...
>
> Also maybe A026177 = A026196 = A026216 ?  A026177 is Michel Dekking's
> Pi_odd in section 2.3.  I had some work-in-progress bits for that
> one, which could be in that section already.  I hadn't thought of
> distinguishing type III and IV but instead went ceil(2n/3).
>
>     A026177
>     Let d=A060236(n) be the lowest non-0 ternary digit of n.
>       If d=1 then a(n)=ceil(2n/3), otherwise a(n)=2n.
>     a(n) = ceil(2n / 3^A137893(n)).
>     a(3n + (1 if n=1 mod 3)) = 3*a(n) is all multiples of 3.
>     a(3n) = 3*a(n) - (1 if n=1 mod 3).
>
> The second last is "sequence is its own multiples of 3" giving
> A026177 = A026216, after suitable explanation.  I don't know how to
> get the further apparent A026177 = A026196; the latter coming from
> Pi_42 of section 2.4.
>
> The inverses of each would be A026178 = A026197 = A026217 which
> R. J. Mathar already contemplates in A026217.
>
> (I arrived at A026177 since its choice of 2n/3 or 2n goes according
> to terdragon left or right turn.  New high values and positions, and
> increments between those, are then turn or segment expansion related.)
>
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