Binomial transform, part II

Let's go "back" to the binomial transform. The encyclopedias such as Mathworld and Wikipedia are missing some important points. Here's my blog on the subject:

http://oeisfan.blogspot.com/2009/08/binomial-transform-introduction.html

Since in one way or another, the binomial transform is present or implied as being an important entity in your papers.

Let's connect it to a. Lengths of continued fraction representations of Infinite Farey tree fractions and
b. the Dragon Curve. You asked about other curves. Nope. the "Hex" curve would be a whole new ballgame. This Gray-code system is isomorphic with only SQUARE - type maps (arrays of square cells).; not triangular or hexagonal.

From the other e mail the lengths of CF fractions are:
1
1, 2
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...
This system can be reproduced as follows, with any sequence S(n), which here is (1, 2, 3,..). Ok, to get the left half of row n, we bring down row (n-1) as shown.
Right half, we take row n say row 3 = (1, 2, 3, 2), We Reverse the terms getting (2, 3, 2, 1) and INCREMENT (replace each with the NEXT term in the series (1, 2, 3, 4,....), = (3, 4, 3, 2). Then we APPEND TO (1, 2, 3, 2), getting row 4:
1, 2, 3, 2, 3, 4, 3, 2.

. This is a Gray code linear map since the neighbors of any term are the next or previous term in all cases.
This is a binomial frequency distribution map since we have in row 3 a frequency of (1, 2, 1); one 1, two 2's, and one 3.
In row four we have a binomial frequency of (1, 3, 3, 1) as to (, 1, 2, 3, 4) since there's one 1, three 2's, three 3's, and one 4.

Now we introduce the concept of the binomial transform (bt).
We take row sums of the above triangle =
1, 3, 8, 20,...
but easier, we store Pascal's triangle in my pocket calculator = P:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1
...
= P, and also store the vector [1, 2, 3, 4, 5,....]. = V, then take P * V getting [1, 3, 8, 20, 48, 112,....]
Thus since after 20 we have a 48, this means that the 16 term string would have a binomial frequency of (1, 4, 6, 4, 1).
as to (1, 2, 3, 4, and 5).

Now lets's take the rows of the CF lengths triangle:
1
1, 2
1, 2, 3, 2
...We can create a complete infinite string of "what these rows tend to" as follows:
We take a 2^n bit string and simply revere and INCREMENT, getting a 2^(n+1) bit string. Example:
we have (1, 2, 3, 2, 3, 4, 3, 2) so we append: (3, 4, 5, 4, 3, 4, 3, 2) Then we do the same with that 16 term string to get the 32 bit string, and so on. So our 16 bit string is:
(1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2).
Next, we let the leftmost term (a 1) = "0". When going to the right, put a 1 if next term is higher. If next term is lower, put an 0.
This gets us:
(0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0...) which is IDENTICAL to the L-system "code" for the Dragon curve!
(as shown in Mathworld). That is, the sequence of 0's and 1's may be interpreted as "Right" or "Left" at right angles.
You asked about other Fractals. Nope. this would only be a fractal with tiny squares, not triangules or hexes.

However, from the previous, it's obvious that our initial sequence = (1, 2, 3....); but we can use ANY sequence, say (1, 3, 5,..).
Using the "bring down n-th row" = left half, then right half = "reverse and increment', using (1, 3, 5, 7,....) we obtain.:
1
1, 3
1, 3, 5, 3
1, 3, 5, 3, 5, 7, 5, 3
...
and so on, then recording row sums = (1, 4, 12, 32, 80, 192,....) or "Binomial transform of (1, 3, 5, 7,...) = (1, 4, 12, 32, 80,...).
.
INTERLUDE....: At this point, the "point" of all this is simply to introduce a Gray-code mapping strategy in which the frequency of terms complies with the Binomial frequency of terms selected for the string.
In my other e mail, I introduced the 2-Dimensional map.
We just take the 2^n x 2^n bit string, say (1, 3, 5, 3), and put this as the top row and left column, with (1, 1, 1, 1) as the diagonal: Then for ODD rows, we start at position (n,n) (along the diagonal), and go DOWN if the column is odd. Go UP from (n,n) if the column is odd. For the 4x4 array using (1, 3, 5,...) we obtain:
1, 3, 5, 3
3, 1, 3, 5
5, 3, 1, 3
3, 5, 3, 1
....with a binomial frequence of (1, 2, 1) as to (1, 3, 5) in each row/column.
Note that in any Knights move, we add or subtract "2" to get a neighbor.
The whole point behind introducing these 1and 2 dimensional maps is simply to state that the maps "exist" and I "believe" that there are possibly uknown QM "entities" mapable on such Gray code formats.
That's where you come in.
Next, we will try to map an irrational term, the cyclotomic third root of Unity. = (-.5 + sqrt(-3)/2)
...