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%I #7 Jan 30 2013 10:50:47
%S -1,2,2,-1,1,-2,-2,4,1,4,-2,-2,-2,-2,4,1,4,-2,-2,1,-1,2,2,-4,-1,-4,2,
%T 2,2,2,-4,-4,-1,-4,-4,8,8,2,2,8,8,-4,-4,-4,-1,-4,-4,-4,2,2,2,2,2,2,2,
%U 2,-4,-4,-4,-1,-4,-4,-4,-4,8,8,2,2,8,8,-4,-4,-1,-4,-4,2,2,2,2,-4,-1,-4,2,2
%N Infinite set of Petoukhov 2^n x 2^n Petoukhov matrices by antidiagonals, generated from w = (-.5 + sqrt(-3)/2)
%C Sergei Petoukhov has pioneered the investigation of certain matrices whose square roots are irrational numbers; and in recognition his discoveries such matrices and their accompanying sequences may be termed "Petoukhov matrices/sequences".
%C Refer to A119633 for a related sequence.
%D Sergey Petoukhov & Matthew He, "symmetrical Analysis Techniques for Genetics systems and Bioinformatics, Advanced Patterns & Applications", IGI Global, 978-1-60566-127-9, October, 2009, Chapters 2, 4, and 6.
%F Given w = (-.5 + sqrt(-3)/2), use the exponent codes of A164092 to create alternating circulant matrices such that a row with 2^n terms generates 2^n x 2^n matrices. Terms in these matrices = exponents for w, then square the matrices. Sequence A164516 = antidiagonals of the infinite set of 2^n x 2^n matrices, exhausting terms in the n-th matrix before using the terms of the next matrix.
%e The exponent codes of A164092 are:
%e .
%e 0; (skip as trivial);
%e 1, -1; (creates the 2x2 matrix [w,1/w; 1/w,w](exponents of w = 1 & -1).
%e 2, 0, -2, 0;
%e 3, 1, -1, 1, -1, -3, -1, 1;
%e 4, 3, .0, 2, .0, -2, .0, 2, 0, -2, -4, -2, 0, -2, 0, 2;
%e ...
%e Exponent codes (above) are generated by adding "1" to each term in n-th row bringing down that subset as the first half of the next row. Second half of the next (n+1)-th) row is created by reversing the terms of n-th row and subtracting "1" from each term. (2, 0, -2, 0) becomes (3, 1, -1, 1) as the first half of the next row. Then append (-1, -3, -1, 1), getting (3, 1, -1, 1, -1, -3, -1, 1) as row 3. Let these rows = "A" for each matrix
%e .
%e In a 2^n * 2^n matrix with a conventional upper left term of (1,1), place A as the top row and left column. Put leftmost term of A into every (n,n) (i.e. diagonal position). Then, odd columns are circulated from position (n,n) downwards while even columns circulate upwards starting from (n,n). Using A with 8 terms we obtain the following 8x8 matrix:
%e .
%e 3, 1, -1, 1, -1, -3, -1, 1;
%e 1, 3, 1, -1, -3, -1, 1, -1;
%e -1, 1, 3, 1, -1, 1, -1, -3;
%e 1, -1, 1, 3, 1, -1, -3, -1;
%e -1, -3, -1, 1, 3, 1, -1, 1;
%e -3, -1, 1, -1, 1, 3, 1, -1;
%e -1, 1, -1, -3, -1, 1, 3, 1;
%e 1, -1, -3, -1, 1, -1, 1, 3;
%e .
%e The foregoing terms are exponents to w, so our new matrix becomes:
%e .
%e 1, w, 1/w, w, 1/w, 1, 1/w, w;
%e w, 1, w, 1/w, 1, 1/w, w, 1/w;
%e 1/w, w, 1, w, 1/w, w, 1/w, 1;
%e w, 1/w, w, 1, w, 1/w, 1, 1/w;
%e 1/w, 1, 1/w, w, 1, w, 1/w, w;
%e 1, 1/w, w, 1/w, w, 1, w, 1/w;
%e 1/w, w, 1/w, 1, 1/w, w, 1, w;
%e w, 1/w, 1, 1/w, w, 1/w, w, 1;
%e .
%e Let the foregoing matrix = Q, then take Q^2 =
%e .
%e -1, 2, -4, 2, -4, 8, -4, 2;
%e 2, -1, 2, -4, 8, -4, 2, -4;
%e -4, 2, -1, 2, -4, 2, -4, 8;
%e 2, -4, 2, -1, 2, -4, 8, -4;
%e -4, 8, -4, 2, -1, 2, -4, 2;
%e 8, -4, 2, -4, 2, -1, 2, -4;
%e -4, 2, -4, 8, -4, 2, -1, 2;
%e 2, -4, 8, -4, 2, -4, 2, -1;
%e .
%e Following analogous procedures for the 2x2 and 4x4 matrices, those are [ -1, 2; 2,-1], and
%e .
%e 1, -2, 4, -2;
%e -2, 1, -2, 4;
%e 4, -2, 1, -2;
%e -2, 4, -2, 1;
%e .
%e Take antidiagonals of the matrices until all terms in each matrix are used.
%Y Cf. A164092, A119633.
%K tabl,sign
%O 1,2
%A _Gary W. Adamson_, Aug 14 2009
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