A164057 -id:A164057 - cp's OEIS frontend

A164056 Triangle of 2^n terms by rows, derived from A088696 as to length of continued fractions, lengths increase = 1, decrease = 0. A088696 can be generated using the following algorithm: Rows 0 and 1 begin 1; 1,2; then for all further rows, bring down current row then append to the right: (1 added to each term in current row). Row 2 (1, 2, 3, 2) then becomes: (1, 2, 3, 2, 3, 4, 3, 2).

0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0

Offset: 0

Gary W. Adamson, Aug 08 2009

Complement of the sequence = A164057

			A088696 begins:
1;
1, 2;
1, 2, 3, 2;
1, 2, 3, 2, 3, 4, 3, 2;
...
Triangle A164056 =
0;
0, 1;
0, 1, 1, 0;
0, 1, 1, 0, 1, 1, 0, 0;
0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0;
...

Jon Maiga, Table of n, a(n) for n = 0..1022 (Rows 0..9)

Cf. A088696, A164057

A088696[n_]:=A088696[n]=Flatten[NestList[Join[#,Reverse[#]+1]&,{1},15]][[n]]; Table[If[n>1,If[A088696[n+1]>A088696[n], 1, 0], 0], {n,0,100}] (* Jon Maiga, Sep 30 2019 *)

Given number of terms in half of the Stern-Brocot infinite Farey tree (cf. A088696); left the leftmost term in each row = 0, then the next term = 1 if the next corresponding positional term in A088696 increases; otherwise 0.

More terms from Jon Maiga, Sep 30 2019

Keyword tabf from Michel Marcus, Sep 30 2019

A164092 Triangle by 2^n term rows, codes used for generating Petoukhov matrices in a Gray code format.

Original entry on oeis.org

0, 1, -1, 2, 0, -2, 0, 3, 1, -1, 1, -1, -3, -1, 1, 4, 2, 0, 2, 0, -2, 0, 2, 0, -2, -4, -2, 0, -2, 0, 2, 5, 3, 1, 3, 1, -1, 1, 3, 1, -1, -3, -1, 1, -1, 1, 3

Offset: 0

Gary W. Adamson, Aug 09 2009

			First few rows of the triangle =
0;
1, -1;
2, .0, -2, 0;
3, .1, -1, 1, -1, -3, -1, 1;
4, .2, .0, 2, .0, -2, .0, 2, 0, -2, -4, -2, 0, -2, 0, 2;
...
We present examples of Petoukhov matrices (Cf. A164091) using rows 2 and 3.
.
Row 3 = [2, 0, -2, 0] = A. We crease an "alternating column circulant. If by convention such matrices have an upper left term (1,1), then odd rows cycle from term (n,n) downward using A. Even rows circulate from (n,n) upwards (Cf. A164057). Using these rules, we obtain the exponents for constants k in 4 X 4 Petoukhov matrices:
.
[2, 0, -2, 0;
.0, 2, 0, -2;
-2, 0, 2, .0;
.0,-2, 0, .2]
.
Let the Petoukhov constant k = phi, 1.6180339,... then insert k into the matrix using the exponents shown, getting [phi^2, 1, 1/phi^2, 1; 1, phi^2, 1, 1/phi^2; 1/phi^2, 1, phi^2, 1; 1, 1/phi^2, 1, phi^2] = M.
.
Then square matrix: M^2 =
9, 6, 4, 6;
6, 9, 6, 4;
4, 6, 9, 6;
6, 4, 6, 9;
...
The terms (4, 6, 9) may be obtained from a 2 X 3 multiplication table, (Cf. A036561, A164057):
.
1,..3,..9,..27,...
2,..6,.18,..54,...
4,.12,.36.........
8..24.............
16................
.
As antidiagonals of this array, we see the terms (4, 6, 9). Similarly, for the 8 X 8 matrix, we apply exponents to phi in the next row using the same circulant rule. As indicated by the next antidiagonal of the 2 X 3 table, the 8 X 8 matrix uses the terms (8, 12, 18, 27), but with a binomial frequency of (1, 3, 3, 1). The 8 X 8 matrix is likewise a square of the corresponding matrix using the exponents [3, 1, -1, 1, -1, -3, -1, 1], then applying the circulant rule. Let this 8 X 8 phi matrix = Q. Then Q^2 = the 8 X 8 Petoukhov matrix (Cf. A164057):
.
27...18...12...18...12...08...12...18;
18...27...18...12...08...12...18...12;
12...18...27...18...12...18...12...08;
18...12...18...27...18...12...08...12;
12...08...12...18...27...18...12...18;
08...12...18...12...18...27...18...12;
12...18...12...08...12...18...27...18;
18...12...08...12...18...12...18...27;
.
Note the binomial distribution of (by rows and columns) one 27, three 18's three 12's and one 8. A harmonic relationship is preserved by Knight's moves in any direction including wrap arounds; any neighbor = (2/3) or (3/2) * another neighbor.

Cf. A164057, A036561, A164056, A164057, A147995.

Let a(0) = 0. Add "1" to each term in n-th row, then bring down to create the first half of the next row. Reverse terms of n-th row and subtract "1", then append, as the right half of row (n+1).

A164279 Triangle of 2^n terms per row, a Petoukhov sequence generated from (3,2).

Original entry on oeis.org

1, 3, 2, 9, 6, 4, 6, 27, 18, 12, 18, 12, 8, 12, 18, 81, 54, 36, 54, 36, 24, 36, 54, 36, 24, 16, 24, 36, 24, 36, 54

Offset: 0

Gary W. Adamson, Aug 11 2009

Row sums = powers of 5: (1, 5, 25, 125,...).
Petoukhov has pioneered the investigation of a class of matrices that are squares of other matrices composed of entirely irrational terms. A164279 terms = top rows, left columns of the Petoukhov matrices shown in A164092.
The Petoukhov matrices associated with A164279 are shown in A164092 along with their derivation from phi, 1.618033989...
The original Petoukhov matrices were in a binary Karnaugh map format.
I have standardized the matrices and sequences, mapping them on the Gray code format shown in A147995. This allows for a ("1 operation" change from one term to the next. For example, in A164279, the next term is either (3/2)*(current term) or (2/3)*(current term) depending on the corresponding positional code of A164057: (a 1 or 0).
Note the binomial frequence of terms per row: (e.g. one 27, three 18's, three 12's, and one 8) in row 3.

			The distinct terms per row are (Cf. A036561): (1; 2,3; 4,6,9; 8,12,18,27; 16,24,36,54,81;) while the codes of A164057 begin:
.
1;
1, 0;
1, 0, 0, 1;
1, 0, 0, 1, 0, 0, 1, 1;
1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1;
...
Given (1, 3, 9, 27,...) as leftmost row terms and following the operational rules: (multiply current term by (3/2) if the corresponding code = 1; (or by (2/3) if 0). This generates A164279: .
1;
3, 2;
9, 6, 4, 6;
27, 18, 12, 18, 12, 8, 12, 18;
81, 54, 36, 54, 36, 24, 36, 54, 36, 24, 16, 24, 36, 24, 36, 54;
...

Sergei Petoukhov & Matthew He, "Symmetrical Analysis Techniques for Genetic Systems and Bioinformatics - Advanced Patterns and Applications"; IGI Global, 978-1-60566-127-9, October, 2009; Chapters 2, 4, and 6.

Cf. A036561, A164057, A147995.

Using the row terms of A036562 (a 2x3 multiplication table): (1, 3,2; 4,6,9;, 8,12,18,27;...), rows of A164279 have leftmost terms extracting the power of 9 from A036562: (1, 3, 9, 27,...). Then accessing the corresponding row codes from A164057, and starting from the left, first term = a power of 9, then given the codes of A164057 (0 or 1), the next row term of A164279 = (3/2)*current term) if the corresponding term of A164057 = 1, and (2/3)*current term if 0.

A164309 Triangle read by rows, generated from the binomial expansion of (5x + 2).

Original entry on oeis.org

1, 5, 2, 25, 10, 4, 10, 125, 50, 20, 50, 20, 8, 20, 50, 625, 250, 100, 250, 100, 40, 100, 250, 100, 40, 16, 40, 100, 40, 100, 250

Offset: 0

Gary W. Adamson, Aug 12 2009

Row sums = powers of 7: (1, 7, 49, 343,...).

(5x + 2)^3 = 125x^3 + 150x^2 +60x + 8; with reference to row 3, since there is one 125, three 50's = 150, three 20's = 60, and one 8 = total 343 = 7^3.

			First we construct a 5^n * 2^n array:
.
1...2...4...8...16...
5..10..20..40........
25.50................
125..................
.
Extract diagonal terms: (1; 5,2; 25,10,4;...) then use the multiplication rules given in the formulas section.
.
First few rows of triangle A164057:
1;
1, 0;
1, 0, 0, 1;
1, 0, 0, 1, 0, 0, 1, 1;
1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1;
...
Using the rules, we obtain:
.
1;
5, 2;
25, 10, 4, 10;
125, 50, 20, 50, 20, 8, 20, 50;
625, 250, 100, 250, 100, 40, 100, 250, 100, 40, 16, 40, 100, 40, 100, 250;
...
Example: place row 3 of A164056 on top of construction of row 3, A164309:
.
(1,...0,...0,...1,...0,...0,...1,...1):
(125,50,..20,..50,..20,...8,..20,..50)
.
"50) = (2/5)*125, while 50 = (5/2)*20; etc.

Cf. A164057, A164308.

Given terms in a 2^n * 5^n multiplication table (diagonals of the array); map the terms in a triangle with 2^n terms per row using the template of A164057: (1;, 1,0; 1,0,0,1; 1,0,0,1,0,0,1,1;...).
Starting with a power of 5 at left, next term to the right = (5/2)*(current term) if the A164057 term = "1".
If the A164057 term = 0, the next term of A164309 = (2/5)*(current term).

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A164092 Triangle by 2^n term rows, codes used for generating Petoukhov matrices in a Gray code format.

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A164279 Triangle of 2^n terms per row, a Petoukhov sequence generated from (3,2).

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A164309 Triangle read by rows, generated from the binomial expansion of (5x + 2).

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