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).

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

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.

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

Cf. A022336, A329919, A354535, A356625.

Sequence in context: A277141 A021438 A195822 * A025638 A215591 A025639

Adjacent sequences: A356621 A356622 A356623 * A356625 A356626 A356627

Keyword

nonn

Author

Rémy Sigrist, Aug 17 2022

Status

approved