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Rank transform of the sequence floor(n/3); complement of A187682.
2

%I #6 Dec 04 2016 19:46:24

%S 1,2,3,4,5,7,8,9,11,12,13,15,16,17,19,20,21,23,24,25,26,27,28,30,31,

%T 32,34,35,36,38,39,40,41,42,43,45,46,47,49,50,51,53,54,55,56,57,58,60,

%U 61,62,64,65,66,68,69,70,71,72,73,75,76,77,79,80,81,83,84,85,86,87,88,90,91,92,94,95,96,98,99,100,102,103,104,106,107,108,110,111,112,113,114,115,117,118,119,121,122

%N Rank transform of the sequence floor(n/3); complement of A187682.

%C See A187224.

%t seqA = Table[Floor[n/3], {n, 1, 220}]

%t seqB = Table[n, {n, 1, 220}];(*A000027*)

%t jointRank[{seqA_,

%t seqB_}] := {Flatten@Position[#1, {_, 1}],

%t Flatten@Position[#1, {_, 2}]} &[

%t Sort@Flatten[{{#1, 1} & /@ seqA, {#1, 2} & /@ seqB}, 1]];

%t limseqU =

%t FixedPoint[jointRank[{seqA, #1[[1]]}] &,

%t jointRank[{seqA, seqB}]][[1]] (*A187681*)

%t Complement[Range[Length[seqA]], limseqU] (*A187682*)

%t (*by _Peter J. C. Moses_, Mar 12 2011*)

%Y Cf. A187224, A187682, A187683, A187685, A187687.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 12 2011