|
%I #9 Apr 22 2013 02:37:52
%S 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,
%T 16,24,36,24,36,54
%N Triangle of 2^n terms per row, a Petoukhov sequence generated from (3,2).
%C Row sums = powers of 5: (1, 5, 25, 125,...).
%C 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.
%C The Petoukhov matrices associated with A164279 are shown in A164092 along with their derivation from phi, 1.618033989...
%C The original Petoukhov matrices were in a binary Karnaugh map format.
%C 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).
%C Note the binomial frequence of terms per row: (e.g. one 27, three 18's, three 12's, and one 8) in row 3.
%D 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.
%F 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.
%e 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:
%e .
%e 1;
%e 1, 0;
%e 1, 0, 0, 1;
%e 1, 0, 0, 1, 0, 0, 1, 1;
%e 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1;
%e ...
%e 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: .
%e 1;
%e 3, 2;
%e 9, 6, 4, 6;
%e 27, 18, 12, 18, 12, 8, 12, 18;
%e 81, 54, 36, 54, 36, 24, 36, 54, 36, 24, 16, 24, 36, 24, 36, 54;
%e ...
%Y Cf. A036561, A164057, A147995.
%K nonn,tabf
%O 0,2
%A _Gary W. Adamson_, Aug 11 2009
|
