A356625 After n iterations of the "Square Multiscale" substitution, the largest tiles have side length 3^t / 5^f; a(n) = f (A356624 gives corresponding t's).
0, 1, 2, 3, 1, 4, 2, 5, 3, 6, 4, 2, 7, 5, 3, 8, 6, 4, 9, 7, 5, 3, 10, 8, 6, 4, 11, 9, 7, 5, 12, 10, 8, 6, 4, 13, 11, 9, 7, 5, 14, 12, 10, 8, 6, 15, 13, 11, 9, 7, 5, 16, 14, 12, 10, 8, 6, 17, 15, 13, 11, 9, 7, 18, 16, 14, 12, 10, 8, 6, 19, 17, 15, 13, 11, 9, 7
Offset: 0
Keywords
Examples
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 1 1/5 5 4 81/625 6 2 3/25 7 5 243/3125 8 3 9/125 9 6 729/15625 10 4 27/625
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
- Yotam Smilansky and Yaar Solomon, Multiscale Substitution Tilings, arXiv:2003.11735 [math.DS], 2020.
Programs
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PARI
{ sc = [1]; for (n=0, 76, s = vecmax(sc); print1 (-valuation(s,5)", "); sc = setunion(setminus(sc,[s]), Set([3*s/5, s/5]))) }
Formula
5^a(n) >= 3^A356624(n).
Comments