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

0,2

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.

Links

Table of n, a(n) for n=0..55.

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

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.

Example

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.

Crossrefs

Cf. A077998.

Sequence in context: A106824 A317203 A229215 * A117621 A178055 A247856

Adjacent sequences: A123505 A123506 A123507 * A123509 A123510 A123511

Keyword

nonn

Author

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

Status

approved