Graphics by P. Bourke, relating to the work of Bagula and Adamson
http://ozviz.wasp.uwa.edu.au/~pbourke/fractals/pascaltriangle/
Monday, August 31, 2009
Sunday, August 30, 2009
A164909 generated from the Dragon Curve
The Dragon Curve is A014577: 1, 1, 0, 1, 1,....
Sequence A164909 begins 1, 2, 3, 2, 3, 4, 3, 2, where the dragon curve terms are considered
codes: A164909 begins with "1", then using (1, 1, 0, 1, 1,...) as coding: 1 = add 1, 0 = subtract 1;
we obtain:
1...2...3...2...3...4...
.....1...1...0...1....1...
where rows of A088696 rows tend to A164909.
Sequence A164909 begins 1, 2, 3, 2, 3, 4, 3, 2, where the dragon curve terms are considered
codes: A164909 begins with "1", then using (1, 1, 0, 1, 1,...) as coding: 1 = add 1, 0 = subtract 1;
we obtain:
1...2...3...2...3...4...
.....1...1...0...1....1...
where rows of A088696 rows tend to A164909.
Binomial transform, an introduction
By operations, we use Pascal's triangle as an infinite lower triangular matrix: A007318 at OEIS,
http://www.tinyurl.com/4zq4q:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1;
...
= P, then multiply P * [any sequence as a vector....]. Say the sequence = (1, 2, 4, 8,...)
Then A007318 * [1,2,4,8,....] = [1, 3, 9, 27,....]; i.e. powers of 3.
The "reason" behind this is is that taking 1,3,3,...items with distinct labels, we create a binomial frequency of items such that given 2 items, (say, 1 and 2); we have one of the 1 and one of the 2 (as in row 1 of Pascal's triangle = (1, 1). Next row = (1, 2, 1), so with 3 items: (1, 2, 4), we have one 1, two 2's, and one 4, = (1, 2, 1) dot (1, 2, 4) = (1 + 4 + 4) = 9, a power of 3.
To create strings with these binomial frequencies of terms, we access the "Lengths of Infinite Farey Tree Continued fractions", in A088696, as follows:
1
1, 2,
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...; where these terms are based on a Gray code reflection principle: To get the next row, we bring down current row as the left half. ; say (1, 2, 3, 2), Now the "reflected" version of this =
(2, 3, 2, 1); but we add "1" to each term since each successive row introduces a new term. Then we append to (1, 2, 3, 2). That is, append (3, 4, 3, 2) to the right of (1, 2, 3, 2); getting:
(1, 2, 3, 2, 3, 4, 3, 2); representing a binomial frequency of terms labeled (1, 2, 3,...) since referring to Pascal's triangle, row 3 applies to 2^3 = 8 terms with a frequency of 4 distinct terms (1, 2, 3, 4); thus one 1, three 2's, three 3's, and one 4.
That's the "meaning" of the Binomial transform since the sums of these terms (1, 2, 3, 2, 3, 4, 3, 2) = 20, where the Binomial transform of (1, 3, 3...) = (1, 3, 8, 20,...).
However, our terms are (1, 2, 4, 8,...), not (1, 2, 3,...) so we index the (1, 2, 4, 8,..) terms with (1, 2, 3, 4,....), replacing the latter with the corresponding terms in (1, 2, 4, 8,...) getting:
1
1, 2
1, 2, 4, 2
1, 2, 4, 2, 4, 8, 4, 2
...where in the latter row we have one 1, three 2's, three 4's, and one 8, a binomial frequency., with sums of the rows = (1, 3, 9, 27,...); in other words, the binomial transform of
(1, 2, 4, 8,...) = (1, 3, 9, 27,...).
http://www.tinyurl.com/4zq4q:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1;
...
= P, then multiply P * [any sequence as a vector....]. Say the sequence = (1, 2, 4, 8,...)
Then A007318 * [1,2,4,8,....] = [1, 3, 9, 27,....]; i.e. powers of 3.
The "reason" behind this is is that taking 1,3,3,...items with distinct labels, we create a binomial frequency of items such that given 2 items, (say, 1 and 2); we have one of the 1 and one of the 2 (as in row 1 of Pascal's triangle = (1, 1). Next row = (1, 2, 1), so with 3 items: (1, 2, 4), we have one 1, two 2's, and one 4, = (1, 2, 1) dot (1, 2, 4) = (1 + 4 + 4) = 9, a power of 3.
To create strings with these binomial frequencies of terms, we access the "Lengths of Infinite Farey Tree Continued fractions", in A088696, as follows:
1
1, 2,
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...; where these terms are based on a Gray code reflection principle: To get the next row, we bring down current row as the left half. ; say (1, 2, 3, 2), Now the "reflected" version of this =
(2, 3, 2, 1); but we add "1" to each term since each successive row introduces a new term. Then we append to (1, 2, 3, 2). That is, append (3, 4, 3, 2) to the right of (1, 2, 3, 2); getting:
(1, 2, 3, 2, 3, 4, 3, 2); representing a binomial frequency of terms labeled (1, 2, 3,...) since referring to Pascal's triangle, row 3 applies to 2^3 = 8 terms with a frequency of 4 distinct terms (1, 2, 3, 4); thus one 1, three 2's, three 3's, and one 4.
That's the "meaning" of the Binomial transform since the sums of these terms (1, 2, 3, 2, 3, 4, 3, 2) = 20, where the Binomial transform of (1, 3, 3...) = (1, 3, 8, 20,...).
However, our terms are (1, 2, 4, 8,...), not (1, 2, 3,...) so we index the (1, 2, 4, 8,..) terms with (1, 2, 3, 4,....), replacing the latter with the corresponding terms in (1, 2, 4, 8,...) getting:
1
1, 2
1, 2, 4, 2
1, 2, 4, 2, 4, 8, 4, 2
...where in the latter row we have one 1, three 2's, three 4's, and one 8, a binomial frequency., with sums of the rows = (1, 3, 9, 27,...); in other words, the binomial transform of
(1, 2, 4, 8,...) = (1, 3, 9, 27,...).
Natural numbers, Gray code string
Refer to "Lengths of Infinite Farey tree continued fractions" where we have
1;
1, 2;
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2;
...Then consider as an infinite string: (1, 1, 2, 1, 2, 3, 2,....) as shown in sequence A088696, at http://www.tinyurl.com/4zq4q
These numbers may be considered codes for generating the binomial transform of sequences.
Refer to Binomial transform, an introduction
1;
1, 2;
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2;
...Then consider as an infinite string: (1, 1, 2, 1, 2, 3, 2,....) as shown in sequence A088696, at http://www.tinyurl.com/4zq4q
These numbers may be considered codes for generating the binomial transform of sequences.
Refer to Binomial transform, an introduction
Lengths of Infinite Farey Tree continued fractions
One half of the Infinite Farey tree, refer to http://www.tinyurl.com/4zq4q, then enter sequence
A088696:
A088696:
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their 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]
...
with the number of terms in each continued fraction representation generating A088696:
1
1 2
1 2 3 2
1 2 3 2 3 4 3 2
where the latter as a sequence may be considered the infinite linear string of the natural numbers based on the Gray code reflection principle. Refer to the entry, "Natural numbers, Gray code string"
Reflected Gray Codes
Refer to the post, "Gray code conversion rules" The following illustrates the reflection principle.
Take base 4, Gray code, using bits 0, 1, 2, 3:
0...000
1...011
2...002
3...003
...
4...013
5...010
6...011
7...012
.
8...022
9...023
10..020
11..021
. ...
Where the reflection rules for Gray code base N uses N terms at a time, refer to rightmost column. Bits are 0,1,2,3. Then the next string of 4 terms repeats last term of previous string (a 3), and continues with the same cycle: 3 ->0 -> 1 -> 2.
However, the next column uses 4 terms with 0, the next 4 with 1, the next 4 with 2's and the next with 3's. Next column going to the left uses 16 0's, 16 1's, etc.
Now refer to the blog, "Lengths of Infinite Farey tree continued fractions".
Take base 4, Gray code, using bits 0, 1, 2, 3:
0...000
1...011
2...002
3...003
...
4...013
5...010
6...011
7...012
.
8...022
9...023
10..020
11..021
. ...
Where the reflection rules for Gray code base N uses N terms at a time, refer to rightmost column. Bits are 0,1,2,3. Then the next string of 4 terms repeats last term of previous string (a 3), and continues with the same cycle: 3 ->0 -> 1 -> 2.
However, the next column uses 4 terms with 0, the next 4 with 1, the next 4 with 2's and the next with 3's. Next column going to the left uses 16 0's, 16 1's, etc.
Now refer to the blog, "Lengths of Infinite Farey tree continued fractions".
Cyclotomic third root of Unity, Gray code map, A164516.
http://www.research.att.com/~njas/sequences/index.html?q=a164516&language=english&go=Search
at http://www.tinyurl.com/4zq4q
then enter sequence A164516
at http://www.tinyurl.com/4zq4q
then enter sequence A164516
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