%I #12 Mar 17 2014 01:18:07
%S 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,
%T 0,2,5,3,1,3,1,-1,1,3,1,-1,-3,-1,1,-1,1,3
%N Triangle by 2^n term rows, codes used for generating Petoukhov matrices in a Gray code format.
%F 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).
%e First few rows of the triangle =
%e 0;
%e 1, -1;
%e 2, .0, -2, 0;
%e 3, .1, -1, 1, -1, -3, -1, 1;
%e 4, .2, .0, 2, .0, -2, .0, 2, 0, -2, -4, -2, 0, -2, 0, 2;
%e ...
%e We present examples of Petoukhov matrices (Cf. A164091) using rows 2 and 3.
%e .
%e 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:
%e .
%e [2, 0, -2, 0;
%e .0, 2, 0, -2;
%e -2, 0, 2, .0;
%e .0,-2, 0, .2]
%e .
%e 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.
%e .
%e Then square matrix: M^2 =
%e 9, 6, 4, 6;
%e 6, 9, 6, 4;
%e 4, 6, 9, 6;
%e 6, 4, 6, 9;
%e ...
%e The terms (4, 6, 9) may be obtained from a 2 X 3 multiplication table, (Cf. A036561, A164057):
%e .
%e 1,..3,..9,..27,...
%e 2,..6,.18,..54,...
%e 4,.12,.36.........
%e 8..24.............
%e 16................
%e .
%e 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):
%e .
%e 27...18...12...18...12...08...12...18;
%e 18...27...18...12...08...12...18...12;
%e 12...18...27...18...12...18...12...08;
%e 18...12...18...27...18...12...08...12;
%e 12...08...12...18...27...18...12...18;
%e 08...12...18...12...18...27...18...12;
%e 12...18...12...08...12...18...27...18;
%e 18...12...08...12...18...12...18...27;
%e .
%e 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.
%Y Cf. A164057, A036561, A164056, A164057, A147995.
%K tabf,sign
%O 0,4
%A _Gary W. Adamson_, Aug 09 2009