[seqfan] Re: Problem reproducing A019989, A019990, and A019991

[Peter Munn]

Hi Kevin,

On Sat, April 27, 2019 3:14 am, Kevin Ryde via SeqFan wrote:

> bradklee at gmail.com (Brad Klee) writes:
>> Taking Mod[-A307672,3] gives this sequence
>> Gosper calls "magicd" (see G4G link therein),

> Ah, oops, beaut.  Could have that among the definitions.  Does
> an "alternating balanced ternary" representation have a name?
> Davis and Knuth flip every second bit of binary for "folded"
> representation, though balanced ternary doesn't look much like
> folding. And yet maybe "alternating" suggests every second digit
> by position, instead of every second non-zero.

I have been working with Brad and others on this. It means every second
nonzero digit.

I came via a different route but also ended up viewing the sequences using
balanced ternary, which I found easiest to visualise using the complete
ternary tree:
http://mathworld.wolfram.com/CompleteTernaryTree.html
Start at node 1 and write down "1" and take a route down the tree. When
taking the left branch append a "-1" ternary digit (or trit), when taking
the middle branch append a "0" trit, and for the right branch append a
"1". So you would reach node 6 having written down "1" "-1" "0".

The sequences can be seen to have symmetries corresponding to an
equilateral triangle, or a triangular prism. Imagine an oversized Scrabble
piece, but triangular. On one side, write "a", "b", "c" clockwise with the
top of each letter towards a vertex. On the opposite side, write "A", "B",
"C" similarly so that corresponding upper case and lower case letters are
back-to-back. Place the piece at node 1 of the complete ternary tree with
"C" vertex at the top. When moving the piece down the left branch from a
node, (eg from 1 to 2) flip it over along the axis that runs between its
left vertices; similarly for the right branch, flip it over along the axis
that runs between its right vertices; and for the middle branch (eg from
node 2 to node 6) keep its orientation unchanged. A019989 lists n
corresponding to nodes numbered 2n where the piece ends up with vertex "a"
face-up at the top.

The above was my route into analysing these sequences, and it uses Bill
Gosper's letters. We have recently discovered their relationship to
Sierpinki's triangle and arrowhead become neater if we map "a" "b" "c" to
"0" "2" "4" and "A" "B" "C" to "3" "5" "1".

There is much more we are uncovering about the sequences relationship to
Sierpinki's triangle and arrowhead that will appear in the sequences in
due course.

Best Regards,

Peter

>> otherwise I
>> think we are stuck with needing six symbols.
>> Does it factor well enough to produce the parity
>> separate from the offset?  I don't know.
> For non-interest, here's a picture of the ordinary ternary I tried,
which is 9 states.  Not very good except it can count in 3^k blocks.
>  http://user42.tuxfamily.org/temporary/ternary-A019989.pdf  (about 38k)
http://user42.tuxfamily.org/temporary/ternary-A019989.tex

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