A123508 1-dimensional quasiperiodic heptagonal sequence.

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, 3, 1, 3, 2, 2, 3, 2, 3, 1, 3, 2, 2, 3, 1, 3, 2, 3, 1, 3, 2, 2, 3

Offset: 0

Author

Gary W. Adamson and Roger L. Bagula, Oct 01 2006

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31, p. 29.

Crossrefs

Cf. A077998.

Formula

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.