%I
%S 1,2,5,6,8,11,14,15,17,18,20,23,24,26,29,32,33,35,38,41,42,44,45,47,
%T 50,51,53,54,56,59,60,62,65,68,69,71,72,74,77,78,80,83,86,87,89,92,95,
%U 96,98,99,101,104,105,107,110,113,114,116,119
%N a(n) = nth number k such that s(j) < s(k) for all j < k, where s = A026177.
%C Conjecture: (a(n)) = complement of A026225 after removal of the initial 1 in A026179.
%F Let the sequence 1, 0, 1, 1, 0, 0, 1, 0, 1, ... be defined as the fixed point of the morphism 1>101 and 0>100, starting from a(1)=1. The indices of 0 are 2, 5, 6, 8, 11, 14, 17, 18, ... (this sequence with first term omitted).  _Philippe DelĂ©ham_, Jun 27 2006
%F After first term, these are the numbers of the form (3i+2)*3^j, where i >= 0, j >= 0.  _Clark Kimberling_, Oct 19 2016
%t a[b_] := Table[Mod[n/b^IntegerExponent[n, b], b], {n, 1, 160}]
%t p[b_, d_] := Flatten[Position[a[b], d]]
%t p[3, 1] (* A026225 *)
%t p[3, 2] (* A026179 without initial 1 *)
%t (* _Clark Kimberling_, Oct 19 2016 *)
%K nonn
%O 1,2
%A _Clark Kimberling_
