OFFSET
1,2
COMMENTS
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".
Refer to A119633 for a related sequence.
REFERENCES
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.
FORMULA
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.
EXAMPLE
The exponent codes of A164092 are:
.
0; (skip as trivial);
1, -1; (creates the 2x2 matrix [w,1/w; 1/w,w](exponents of w = 1 & -1).
2, 0, -2, 0;
3, 1, -1, 1, -1, -3, -1, 1;
4, 3, .0, 2, .0, -2, .0, 2, 0, -2, -4, -2, 0, -2, 0, 2;
...
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
.
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:
.
3, 1, -1, 1, -1, -3, -1, 1;
1, 3, 1, -1, -3, -1, 1, -1;
-1, 1, 3, 1, -1, 1, -1, -3;
1, -1, 1, 3, 1, -1, -3, -1;
-1, -3, -1, 1, 3, 1, -1, 1;
-3, -1, 1, -1, 1, 3, 1, -1;
-1, 1, -1, -3, -1, 1, 3, 1;
1, -1, -3, -1, 1, -1, 1, 3;
.
The foregoing terms are exponents to w, so our new matrix becomes:
.
1, w, 1/w, w, 1/w, 1, 1/w, w;
w, 1, w, 1/w, 1, 1/w, w, 1/w;
1/w, w, 1, w, 1/w, w, 1/w, 1;
w, 1/w, w, 1, w, 1/w, 1, 1/w;
1/w, 1, 1/w, w, 1, w, 1/w, w;
1, 1/w, w, 1/w, w, 1, w, 1/w;
1/w, w, 1/w, 1, 1/w, w, 1, w;
w, 1/w, 1, 1/w, w, 1/w, w, 1;
.
Let the foregoing matrix = Q, then take Q^2 =
.
-1, 2, -4, 2, -4, 8, -4, 2;
2, -1, 2, -4, 8, -4, 2, -4;
-4, 2, -1, 2, -4, 2, -4, 8;
2, -4, 2, -1, 2, -4, 8, -4;
-4, 8, -4, 2, -1, 2, -4, 2;
8, -4, 2, -4, 2, -1, 2, -4;
-4, 2, -4, 8, -4, 2, -1, 2;
2, -4, 8, -4, 2, -4, 2, -1;
.
Following analogous procedures for the 2x2 and 4x4 matrices, those are [ -1, 2; 2,-1], and
.
1, -2, 4, -2;
-2, 1, -2, 4;
4, -2, 1, -2;
-2, 4, -2, 1;
.
Take antidiagonals of the matrices until all terms in each matrix are used.
CROSSREFS
KEYWORD
tabl,sign
AUTHOR
Gary W. Adamson, Aug 14 2009
STATUS
approved