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%I #5 Mar 03 2013 13:48:24
%S 1,1,2,1,6,2,1,14,8,4,1,30,22,24,4,1,62,52,92,28,8,1,126,114,288,120,
%T 72,8,1,254,240,804,408,384,80,16,1,510,494,2088,1212,1584,46,192,16
%N Triangle, row sums = powers of 3.
%C In A124731, we switch the diagonals. In both triangles, row sums = powers of 3.
%F Let M = the infinite bidiagonal matrix with (1,2,1,2...) in the main diagonal and (2,1,2,1...) in the subdiagonal. The n-th row of the triangle (extracting the zeros) = M^n * [1,0,0,0...].
%e Row 2 = (1, 6, 2) since [1,0,0; 2,2,0; 0,1,1]^2 * [1,0,0] = [1,6,2].
%e First few rows of the triangle are:
%e 1;
%e 1, 2;
%e 1, 6, 2;
%e 1, 14, 8, 4;
%e 1, 30, 22, 24, 4;
%e 1, 62, 52, 92, 28, 8;
%e 1, 126, 114, 288, 120, 72, 8;
%e ...
%Y Cf. A124731, A124732.
%K nonn,tabl
%O 0,3
%A _Gary W. Adamson_ & _Roger L. Bagula_, Nov 05 2006
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