%I #32 May 25 2018 09:13:53
%S 0,0,1,1,0,2,1,0,0,1,1,2,0,0,1,1,0,3,1,0,0,1,1,3,0,0,1,1,0,2,1,0,0,1,
%T 1,4,0,0,1,1,0,3,1,0,0,1,1,2,0,0,1,1,0,2,1,0,0,1,1,3,0,0,1,1,0,3,1,0,
%U 0,1,1,4,0,0,1,1,0,2,1,0,0,1,1,2,0,0,1,1,0,3,1,0,0,1,1,3,0,0,1,1,0,2,1,0,0,1,1,5,0,0,1,1,0,3,1,0,0,1,1,2
%N Lexicographically least sequence of nonnegative integers that avoids 3/2-powers.
%C Rowland and Shallit showed that this sequence is 6-regular.
%H Eric Rowland, <a href="/A269518/b269518.txt">Table of n, a(n) for n = 0..20000</a>
%H Lara Pudwell and Eric Rowland, <a href="http://arxiv.org/abs/1510.02807">Avoiding fractional powers over the natural numbers</a>, arXiv:1510.02807 [math.CO] (2015). Also Electronic Journal of Combinatorics, Volume 25(2) (2018), #P2.27
%H Eric Rowland and Jeffrey Shallit, <a href="http://arxiv.org/abs/1101.3535">Avoiding 3/2-powers over the natural numbers</a>, arXiv:1101.3535 [math.CO] (2011).
%H Eric Rowland and Jeffrey Shallit, <a href="http://dx.doi.org/10.1016/j.disc.2011.12.019">Avoiding 3/2-powers over the natural numbers</a>, Discrete Mathematics 312 (2012) 1282-1288.
%F a(6 n + 5) = a(n) + 2. - _Eric Rowland_, Oct 01 2016
%t (* This gives the first 7776 terms. *)
%t Replace[SubstitutionSystem[{n_Integer :> {one, 0, zero, 1, one, n + 2}, zero -> {zero, 0, one, 1, zero, 2}, one -> {zero, 0, one, 1, zero, 3}}, {zero}, {{5}}], {zero -> 0, one -> 1}, {1}] (* _Eric Rowland_, Oct 01 2016 *)
%Y Cf. A269517 (the lexicographically least sequence that avoids a/b-powers for all a/b >= 3/2).
%K nonn
%O 0,6
%A _Eric Rowland_, Feb 28 2016