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%I #9 Sep 21 2017 03:38:55
%S 1,3,1,3,2,3,1,3,2,2,3,1,3,2,3,1,3,2,2,3,2,3,1,3,2,3,1,3,2,2,3,1,3,2,
%T 3,1,3,2,2,3,2,3,1,3,2,2,3,1,3,2,3,1,3,2,2,3
%N 1-dimensional quasiperiodic heptagonal sequence.
%C Each iterative subset can be parsed into secondary subsets relating to A077998, a sequence generated from the Heptagonal matrix M, [1, 1, 1; 1, 1, 0; 1, 0, 0]: 1, 1, 3, 6, 14, 31, 70, ...as follows (Cf. Steinbach): Performing M^n * [1,0,0] we get 3 sets of vectors, read by rows: 1, 0, 0 1, 1, 1 3, 2, 1 6, 5, 3 .. where the n-th row pertains to the n-th iterative subset of the sequence. E.g. (3, 2, 1) is the distribution of 3's, 2's and 1's in (3,1,3,2,2,3). Furthermore, the vectors generated from M relate to the Heptagon diagonals as follows: (E.g.: given the Heptagon diagonals a = 2.24697960...(1 + 2*Cos 2Pi/7); b = 1,80193773...(2*Cos Pi/7) and c = 1 (the edge); then select any 3-termed row in the vectors, such as row 4, (6, 5, 3). Then a^4 = 6*a + 5*b + 3*1.
%H P. Steinbach, <a href="http://www.jstor.org/stable/2691048">Golden fields: a case for the heptagon</a>, Math. Mag. 70 (1997), no. 1, 22-31, p. 29.
%F Let a(n) = 1; then iterate using the rules 1=>3; 2=>2,3; 3=>1,3,2; Append each successive iterate to the right, creating an infinite string.
%e 1=>3; 3=>1,3,2; then the previous subset generates 3,1,3,2,2,3. The resulting subsets are (1), (1,3,2), (3,1,3,2,2,3)...which we combine to form a continuous sequence.
%Y Cf. A077998.
%K nonn
%O 0,2
%A _Gary W. Adamson_ and _Roger L. Bagula_, Oct 01 2006
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