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Rank transform of the sequence floor(4n/3); complement of A187686.
3

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

%S 1,3,5,6,8,10,12,13,15,17,19,21,22,24,26,27,29,31,33,34,37,38,40,42,

%T 43,45,47,48,50,52,54,55,58,59,61,63,65,66,68,70,71,74,75,76,79,80,82,

%U 84,85,87,89,91,92,95,96,97,100,102,103,105,107,108,110,112,114,116,117,119,121,123,124,126,128,130,131,133,134,137,138,140,142,144

%N Rank transform of the sequence floor(4n/3); complement of A187686.

%C See A187224.

%t seqA = Table[Floor[4n/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]] (*A187685*)

%t Complement[Range[Length[seqA]], limseqU] (*A187686*)

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

%Y Cf. A187224, A187686.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 12 2011