A165196

A165196:
1, 2, 4, 5, 7, 12, 14, 15
17, 22, 37, 42, 44, 49, 51, 52
54, 59, 74, 79, 94, 146, 161, 166, 168, 173, 188, 193, 195, 200, 202, 203,
205, 210, 225, 230, 245, 297, 312. 317. 332. 384,. 587, 639, 654, 706, 721, 726, 728, 733, 748, 753, 768, 820, 835, 840, 842,
847, 862 ,867, 869, 874, 876, 877
...
where a(64) = 877.

Dragon curve coding, part II

note here that this is a binomial frequency, since in the 8 bit row we have one 1, three 2's, three 3's, and one 4.Similarly, the next row would have a binomial frequency of (1, 4, 6, 4, 1), and so on.OK, the foregoing are linear maps. We can create a 2-D map using the same terms as follows: Top row and left column we place(1, 2, 3, 2, 3, 4, 3, 2), with (1,1,1,1,....) on the diagaonal. If leftmost col = 1, then odd rows circulate from position (n,n) DOWN, while evens circulate UP from position (n,n). This is tricky but crucial. This is a new type of Gray Code MAP:.1, 2, 3, 2, 3, 4, 3, 22, 1, 2, 3, 4, 3, 2, 33, 2, 1, 2, 3, 2, 3, 42, 3, 2, 1, 2, 3, 4, 33, 4, 3, 2, 1, 2, 3, 24, 3, 2, 3, 2, 1, 2, 33, 2, 3, 4, 3, 2, 1, 22, 3, 4, 3, 2, 3, 2, 1...OK, this is an important type of Gray Code map which in any Knight's move with wrap-arounds, there's one a "1" difference.In any row or column, there's a binomial frequency as to (1, 2, 3, 4) being (1, 3, 3, 1).Let's stop at this point and just say that for such 2^n x 2^n matrics, using the initial sequence S(n) = (1, 2, 3, 4, 5,....).the SUMS of terms in the matrices = (1, 6, 32, 160, 768,...) since in the 8x8 matrix each row has a sum = 20 and there's 8 such rows = 160.Before completing all of the connections, we can introduce the conclusion. We have created a type of 2^n x 2^n Gray Code Karnaugh map in which we can map terms in any sequence (S(n)) such that there will be a binomial frequency of terms in each row and column.Briefly, in your 1974 paper, you have the square of Pascal's triangle =Q. Just multiply that square * a diagonalized vector:V:10, 20, 0, 40, 0, 0, 80, 0, 0, 0, 16,...Getting Triangle A038208. Bring that up and you will find a "CYVIN" reference. I think he's retired and lives in Oslo. There's 2 of them - could be Father and Son.At any rate, any cites to "CYVIN" always indicate important applications to organic chemistry. So triangle A038208 = Q * V =12, 24, 8, 48, 24, 24, 8...which is symmetrical where as the square of Pacal's triangle is not symmetrical:12, 14, 4, 18, 12, 6, 1... = triangle T.Then triangle T * [1, 2, 3, 4, 5,...] = the sequence (1, 6, 32, 160, 768,...) showing the SUMS of terms in the 2^n x 2^n matrices.To conclude for now, we have introduced triangle T (not in your 1974 paper); which can be linked to a Gray Code map derived from the lengths of continued fractions in one half of the Infinite Farey Tree.Triangle T is symmetrical and has important applications to organic chemistry per citation in A038208 (Cyvin).Next, we will show further connections to your Harter-Heighway "Dragon Curve"....to be continued.

Dragon curve coding, part I

. An "L" system for fractal curves is the series of bits that encodes the directions of moves, through which we can recover the entire fractal. First, we will show the connections between the binomial frequency, infinite farey tree continued fraction lengths, and the Dragon curve, then go on from there in next e mail. The "L" system for the Dragon Curve (Cf. the Mathworld entry on "Dragon Curve" - they have it shown right there. Its entryA014577 in OEIS: http://www.tinyurl.com/4zq4q shown as: (1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0,...). b. now, bring up A088696, lengths of continued fraction representations of infinite farey tree terms. (one half of the tree). This is a very important set of terms since your papers touch upon this theme repeatedly. The fractions are:1/21/3, 2/31/4, 2/5, 3/5, 3/41/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5...and the corresponding continued fraction representations are: [2][3), [1,2][4] [2,2], [1,1,2], [1,3][5], [3,2], [2,1,2], [2,3], [1,1,3], [1,1,1,2], [1,2,2], [1,4]...now simply write down the number of terms in each CF representation = 11, 21, 2, 3, 21, 2, 3, 2, 3, 4, 3, 2Now taking this, write down a R for leftmost term = Right = 1 in the Dragon curve if the NEXT term is greater. If less, write down an 0. This gets us the L system for the Dragon curve: (inserting an initial 0) we obtain rows which tend to the Dragon Curve L system mentioned in Mathworld:(0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0,...). ...note here that this is a binomial frequency, since in the 8 bit row we have one 1, three 2's, three 3's, and one 4.Similarly, the next row would have a binomial frequency of (1, 4, 6, 4, 1), and so on.OK, the foregoing are linear maps. We can create a 2-D map using the same terms as follows: Top row and left column we place(1, 2, 3, 2, 3, 4, 3, 2), with (1,1,1,1,....) on the diagaonal. If leftmost col = 1, then odd rows circulate from position (n,n) DOWN, while evens circulate UP from position (n,n). This is tricky but crucial. This is a new type of Gray Code MAP:.

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)
...

The Dragon Curve mapping codes

As shown in Mathworld, the "L" sequence for the Dragaon curve =
(1, 1, 0, 1, 1, 0, 0,....).
This should actually be prefaced with an 0 getting: (0, 1, 1, 0, 1, 1, 0, 0);
where these terms can be derived from the lengths of continued fraction representations of
one-half of the Infinite Farey tree. We use
1
1, 2
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
....= such lengths, then use the substitutions: "1" if next term going to the right is higher; otherwise 0.; but letting leftmost term -= 0; thus the 8 bit string =
0, 1, 1, 0, 1, 1, 0, 0;
now we can use these codes to remap any sequence S(n) say (1, 3, 5, 7,...) onto a 2^n x 2^n Gray Code format.; getting
1, 3, 5, 3
3, 1, 3, 5
5, 3, 1, 3
3, 5, 3, 1
...
as shown in A002697, where the sum of these terms = 48.
And we note that A038208 * [1, 3, 5, 7,...] = (1, 8, 48, 256,....) where A038208 =
1
2, 2
4, 8, 4
8, 24, 24, 8
...a symmetrical triangle with special properties relating to biochemistry per comments shown
in A038208.

A165193

eigensequence of triangle A038208 where A038208 =
1
2, 2
4, 8, 4
8, 24, 24, 8
...
= triangle T, then Triangle T * any sequence S(n) as a vector say [1, 3, 5, 7,...] =
(1, 8, 48, 256, 1280,...); = A002697, where (Cf. comments); for example we have the
array
1, 3, 5, 3
3, 1, 3, 5
5, 3, 1, 3
3, 5, 3, 1
...total of terms = 48, where T * [1, 3, 5, 7,...] = (1, 8, 48, 256,...), thus the "48".