cp's OEIS Frontend

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

%I A164092 #12 Mar 17 2014 01:18:07
%S A164092 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 A164092 0,2,5,3,1,3,1,-1,1,3,1,-1,-3,-1,1,-1,1,3
%N A164092 Triangle by 2^n term rows, codes used for generating Petoukhov matrices in a Gray code format.
%F A164092 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 A164092 First few rows of the triangle =
%e A164092 0;
%e A164092 1, -1;
%e A164092 2, .0, -2, 0;
%e A164092 3, .1, -1, 1, -1, -3, -1, 1;
%e A164092 4, .2, .0, 2, .0, -2, .0, 2, 0, -2, -4, -2, 0, -2, 0, 2;
%e A164092 ...
%e A164092 We present examples of Petoukhov matrices (Cf. A164091) using rows 2 and 3.
%e A164092 .
%e A164092 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 A164092 .
%e A164092 [2, 0, -2, 0;
%e A164092 .0, 2, 0, -2;
%e A164092 -2, 0, 2, .0;
%e A164092 .0,-2, 0, .2]
%e A164092 .
%e A164092 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 A164092 .
%e A164092 Then square matrix: M^2 =
%e A164092 9, 6, 4, 6;
%e A164092 6, 9, 6, 4;
%e A164092 4, 6, 9, 6;
%e A164092 6, 4, 6, 9;
%e A164092 ...
%e A164092 The terms (4, 6, 9) may be obtained from a 2 X 3 multiplication table, (Cf. A036561, A164057):
%e A164092 .
%e A164092 1,..3,..9,..27,...
%e A164092 2,..6,.18,..54,...
%e A164092 4,.12,.36.........
%e A164092 8..24.............
%e A164092 16................
%e A164092 .
%e A164092 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 A164092 .
%e A164092 27...18...12...18...12...08...12...18;
%e A164092 18...27...18...12...08...12...18...12;
%e A164092 12...18...27...18...12...18...12...08;
%e A164092 18...12...18...27...18...12...08...12;
%e A164092 12...08...12...18...27...18...12...18;
%e A164092 08...12...18...12...18...27...18...12;
%e A164092 12...18...12...08...12...18...27...18;
%e A164092 18...12...08...12...18...12...18...27;
%e A164092 .
%e A164092 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 A164092 Cf. A164057, A036561, A164056, A164057, A147995.
%K A164092 tabf,sign
%O A164092 0,4
%A A164092 _Gary W. Adamson_, Aug 09 2009