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

Author

Rémy Sigrist, Aug 17 2022

Keywords

• nonn

Comments

See A329919 for further details about the "Square Multiscale" substitution.

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.

Crossrefs

Cf. A022337, A329919, A354535, A356624.

Programs

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).