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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).

%I A356625 #13 Aug 21 2022 06:15:28
%S A356625 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,
%T A356625 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,
%U A356625 13,11,9,7,18,16,14,12,10,8,6,19,17,15,13,11,9,7
%N 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).
%C A356625 See A329919 for further details about the "Square Multiscale" substitution.
%H A356625 Rémy Sigrist, <a href="/A356625/b356625.txt">Table of n, a(n) for n = 0..10000</a>
%H A356625 Yotam Smilansky and Yaar Solomon, <a href="https://arxiv.org/abs/2003.11735">Multiscale Substitution Tilings</a>, arXiv:2003.11735 [math.DS], 2020.
%F A356625 5^a(n) >= 3^A356624(n).
%e A356625 The first terms, alongside the corresponding side lengths, are:
%e A356625   n   a(n)  Side length
%e A356625   --  ----  -----------
%e A356625    0     0            1
%e A356625    1     1          3/5
%e A356625    2     2         9/25
%e A356625    3     3       27/125
%e A356625    4     1          1/5
%e A356625    5     4       81/625
%e A356625    6     2         3/25
%e A356625    7     5     243/3125
%e A356625    8     3        9/125
%e A356625    9     6    729/15625
%e A356625   10     4       27/625
%o A356625 (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]))) }
%Y A356625 Cf. A022337, A329919, A354535, A356624.
%K A356625 nonn
%O A356625 0,3
%A A356625 _Rémy Sigrist_, Aug 17 2022