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A123508 1-dimensional quasiperiodic heptagonal sequence. 1
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified June 19 05:17 EDT 2019. Contains 324217 sequences. (Running on oeis4.)