

A026179


a(n) = nth number k such that s(j) < s(k) for all j < k, where s = A026177.


7



1, 2, 5, 6, 8, 11, 14, 15, 17, 18, 20, 23, 24, 26, 29, 32, 33, 35, 38, 41, 42, 44, 45, 47, 50, 51, 53, 54, 56, 59, 60, 62, 65, 68, 69, 71, 72, 74, 77, 78, 80, 83, 86, 87, 89, 92, 95, 96, 98, 99, 101, 104, 105, 107, 110, 113, 114, 116, 119
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OFFSET

1,2


COMMENTS

Conjecture: (a(n)) = complement of A026225 after removal of the initial 1 in A026179.


LINKS

Table of n, a(n) for n=1..59.


FORMULA

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 Deléham, Jun 27 2006
After first term, these are the numbers of the form (3i+2)*3^j, where i >= 0, j >= 0.  Clark Kimberling, Oct 19 2016


MATHEMATICA

a[b_] := Table[Mod[n/b^IntegerExponent[n, b], b], {n, 1, 160}]
p[b_, d_] := Flatten[Position[a[b], d]]
p[3, 1] (* A026225 *)
p[3, 2] (* A026179 without initial 1 *)
(* Clark Kimberling, Oct 19 2016 *)


CROSSREFS

Sequence in context: A247062 A059009 A214642 * A300063 A230902 A243680
Adjacent sequences: A026176 A026177 A026178 * A026180 A026181 A026182


KEYWORD

nonn


AUTHOR

Clark Kimberling


STATUS

approved



