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Rank transform of the sequence 3*floor(n/3); complement of A187579.
2

%I #5 Dec 04 2016 19:46:24

%S 1,2,5,6,7,9,10,11,14,15,16,20,21,22,24,25,26,29,30,31,33,34,35,38,39,

%T 40,44,45,46,48,49,50,53,54,55,59,60,61,63,64,65,68,69,70,72,73,74,77,

%U 78,79,83,84,85,87,88,89,92,93,94,96,97,98,101,102,103,107,108,109,111,112,113,116,117,118,122,123,124,126,127,128,131,132,133,135,136,137,140,141,142,146

%N Rank transform of the sequence 3*floor(n/3); complement of A187579.

%C See A187224.

%t seqA = Table[3*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]] (*A187578*)

%t Complement[Range[Length[seqA]], limseqU] (*A187579*)

%t (*by _Peter J. C. Moses_, Mar 11 2011*)

%Y Cf. A187224, A187579.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 11 2011