

A026367


a(n) = least k such that s(k) = n, where s = A026366.


7



1, 2, 4, 5, 7, 8, 9, 10, 12, 13, 15, 16, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 31, 32, 34, 35, 37, 38, 39, 40, 42, 43, 45, 46, 48, 49, 51, 52, 53, 54, 56, 57, 59, 60, 61, 62, 64, 65, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 80, 81, 83, 84, 86
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OFFSET

1,2


COMMENTS

Complement of A026368; also the rank transform (as at A187224) of A004526 (after removal of the initial two zeros).  Clark Kimberling, Mar 10 2011
Conjecture: 1 < n*r  a(n) < 2 for n>=1, where r = (1 + sqrt(3))/2.  Clark Kimberling, Apr 29 2017


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = a(n1)+2 if n is in the set 2A+1, a(n) = a(n1)+1 if n is not in 2A+1 where 2A+1 = {2a(1)+1,2a(2)+1,2a(3)+1,...} = {3,5,9,11,15,17,19,21,25,27,31,33,...}. a(n) = (1+sqrt(3))/2*n+O(1) For example, a(3) = a(2)+2 = 2+2 = 4 since 3 is in 2A+1. a(10) = a(9)+1 = 12+1 = 13 since 10 is not in 2A+1.  Benoit Cloitre, Apr 23 2008


MATHEMATICA

seqA = Table[Ceiling[n/2], {n, 1, 180}] (* A004526 *)
seqB = Table[n, {n, 1, 80}]; (* A000027 *)
jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
{#1, 2} & /@ seqB}, 1]];
limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]] (* A026367 *)
Complement[Range[Length[seqA]], limseqU] (* A026368 *)
(* Peter J. C. Moses, Mar 10 2011 *)
s = Nest[Flatten[# /. {0 > {1, 1}, 1 > {1, 1, 0}}] &, {0}, 13] (* A285431 *)
Flatten[Position[s, 0]] (* A026368 *)
Flatten[Position[s, 1]] (* A026367 *)
(* Clark Kimberling, Apr 28 2017 *)


CROSSREFS

Cf. A026368, A187224, A004526, A285431.
Sequence in context: A187895 A079726 A047498 * A039069 A285423 A173025
Adjacent sequences: A026364 A026365 A026366 * A026368 A026369 A026370


KEYWORD

nonn


AUTHOR

Clark Kimberling


EXTENSIONS

Complement sequence in first comment corrected by Nathan Fox, Mar 21 2014


STATUS

approved



