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A123609
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Quasiperiodic 9-gonal (nonagonal) sequence as a 1-dimensional tiling.
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1
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4, 1, 2, 3, 4, 4, 3, 4, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 2, 3, 4, 1, 2, 3, 4, 3, 4, 2, 3, 4, 1, 2, 3, 4, 4, 3, 4, 2, 3, 4, 1, 2, 3, 4
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OFFSET
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1,1
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COMMENTS
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The frequency of each distinct term (1, 2, 3, or 4) tends to converge to the ratio of each diagonal (a, b, c, or d) to the sum of the 4 diagonal lengths. The four 9-gon (nonagon) diagonals are a=1, b=1.87938524..., c=2.53208888... and d=2.87938524..., with the sum a+b+c+d = d^2 = 8.29085936.... Converting these terms to fractions of the sum, a=0.12061475..., b=0.22668159..., c=0.30540728..., and d=0.34729635.... Through n = 45, we can thus expect sixteen 4's (correct), since round(45*.34729635...) = 16. The numbers of terms in each subset strung together is found in A006357: (1, 4, 10, 30, 85...), thus: (4), (1,2,3,4), (4,3,4,2,3,4,1,2,3,4), ..., while the distributive breakdown of numbers of 1's, 2's, 3's, and 4's may be found in the 4-termed set of vectors in A076264: 1 1 1 1 4 3 2 1 10 9 7 4 30 26 19 10 ... where the sum of 4 terms in a row = the left term in the next row. For example, the frequency distribution of 30 includes ten 4's, nine 3's, seven 2's, and four 1's. Check: the subset of 30 terms generated from the previous subset of 10: (1,2,3,4,2,3,4,1,2,3,4,3,4,2,3,4,1,2,3,4,4,3,4,2,3,4,1,2,3,4).
A fractal structure is suggested by parsing each subset into groups: (1,2,3,4), (2,3,4), (1,2,3,4), (3,4), (2,3,4), (1,2,3,4), (4), (3,4), (2,3,4), (1,2,3,4). That is, 10 groups: four with four terms, three with three terms, two with two terms, and one with one term. Replacing the terms (4,3,2,1) with the diagonal lengths (d,c,b,a) and referring to the set of vectors: (1,1,1,1; 4,3,2,1; 10,9,7,4; ...), label these rows 2,3,4,... and consider (2,3,4,...) exponents to diagonal d=2.87938524..., such that, for example, "4" corresponds to (10,9,7,4), and (Cf. Steinbach) d^4 = 68.738349... = (10*d + 9*c + 7*b + 4*a). Such relationships are a consequence of the "Diagonal Product Formulas" mentioned on p. 23.
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LINKS
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FORMULA
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Using the seed "1", we use the recurrence rules 1=>4; 2=>3,4; 3=>2,3,4; 4=>1,2,3,4; to form iterative subsets which are appended in succession to form a continuous string.
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EXAMPLE
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1=>4, then 4=>1,2,3,4, which, in turn, generates 4,3,4,2,3,4,1,2,3,4 (append next result to right of previous result, getting an infinite aperiodic sequence).
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MATHEMATICA
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Drop[SubstitutionSystem[{1->{4}, 2->{3, 4}, 3->{2, 3, 4}, 4->{1, 2, 3, 4}}, {1}, {5}][[1]], 5] (* Harvey P. Dale, Mar 02 2022 *)
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CROSSREFS
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KEYWORD
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nonn,uned,obsc
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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