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A356624
After n iterations of the "Square Multiscale" substitution, the largest tiles have side length 3^t / 5^f; a(n) = t (A356625 gives corresponding f's).
2
0, 1, 2, 3, 0, 4, 1, 5, 2, 6, 3, 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0, 10, 7, 4, 1, 11, 8, 5, 2, 12, 9, 6, 3, 0, 13, 10, 7, 4, 1, 14, 11, 8, 5, 2, 15, 12, 9, 6, 3, 0, 16, 13, 10, 7, 4, 1, 17, 14, 11, 8, 5, 2, 18, 15, 12, 9, 6, 3, 0, 19, 16, 13, 10, 7, 4, 1, 20, 17
OFFSET
0,3
COMMENTS
See A329919 for further details about the "Square Multiscale" substitution.
LINKS
Yotam Smilansky and Yaar Solomon, Multiscale Substitution Tilings, arXiv:2003.11735 [math.DS], 2020.
FORMULA
5^A356625(n) >= 3^a(n).
EXAMPLE
The first terms, alongside the corresponding side lengths, are:
n a(n) Side length
-- ---- -----------
0 0 1
1 1 3/5
2 2 9/25
3 3 27/125
4 0 1/5
5 4 81/625
6 1 3/25
7 5 243/3125
8 2 9/125
9 6 729/15625
10 3 27/625
PROG
(PARI) { sc = [1]; for (n=0, 78, s = vecmax(sc); print1 (valuation(s, 3)", "); sc = setunion(setminus(sc, [s]), Set([3*s/5, s/5]))) }
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Aug 17 2022
STATUS
approved