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A123609 Quasiperiodic nonagonal sequence as a 1-dimensional tiling. +0
1
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 (list; graph; listen)
OFFSET

1,1

COMMENT

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 nonagon diagonals are a = 1, b = 1.87938524...,

c = 2.5320888862... and 2.87938524157, with the sum a+b+c+d = d^2 = 8.2985936...Converting

these terms to percentages, a = .120614..., b = .226681..., c = .305407...,

and d = .347296. Through n = 45, we can thus expect 16 4's (correct), since

round:(45*.347296...) = 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, 7 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.

REFERENCES

Peter Steinbach, "Golden Fields: A Case for the Heptagon", Mathematics Magazine, Vol. 70, No. 1, 1997, p. 29.

FORMULA

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.

EXAMPLE

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).

CROSSREFS

Cf. A123508, A006357, A076264, A091024.

Sequence in context: A090204 A156915 A072046 this_sequence A075617 A055182 A084454

Adjacent sequences: A123606 A123607 A123608 this_sequence A123610 A123611 A123612

KEYWORD

nonn,uned,obsc

AUTHOR

Gary W. Adamson and Roger L. Bagula (qntmpkt(AT)yahoo.com), Oct 03 2006

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Last modified November 25 20:09 EST 2009. Contains 167514 sequences.


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