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

Kevin Ryde user42_kevin at yahoo.com.au
Mon Feb 3 12:08:37 CET 2020

```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.)

```