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Search: id:A123508
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| A123508 |
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1-dimensional quasiperiodic heptagonal sequence. |
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+0 2
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| 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; listen)
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OFFSET
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0,2
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COMMENT
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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.
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REFERENCES
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Peter Steinbach, "Golden Fields: A Case for the Heptagon", Mathematics Magazine, Vol. 70, No. 1, 1997, p. 29.
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FORMULA
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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.
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EXAMPLE
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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.
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CROSSREFS
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Cf. A077998.
Sequence in context: A016571 A055189 A106824 this_sequence A117621 A059660 A035456
Adjacent sequences: A123505 A123506 A123507 this_sequence A123509 A123510 A123511
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KEYWORD
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nonn
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AUTHOR
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Gary W. Adamson and Roger L. Bagula (qntmpkt(AT)yahoo.com), Oct 01 2006
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