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Rank transform of the sequence floor(3n/4); complement of A187229.
2

%I #13 Jan 29 2016 11:12:47

%S 1,2,4,6,7,8,10,11,12,14,16,18,19,20,22,24,25,27,28,30,31,32,34,35,36,

%T 38,40,42,43,44,46,47,48,50,52,53,54,56,58,59,60,62,64,66,67,68,70,72,

%U 73,75,76,78,79,80,82,83,84,86,88,90,91,92,94,96,97,99,100,102,103,104,106,108,109,111,112,114,115,116,118,120,121,123,124,126,127,128,130,131,132,134,136,138,139,140,142,143,144,146,148,149,150

%N Rank transform of the sequence floor(3n/4); complement of A187229.

%C See A187224.

%t seqA=Table[Floor[3n/4],{n,1,220}] (*A057353*)

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

%t jointRank[{seqA_,seqB_}]:={Flatten@Position[#1,{_,1}],Flatten@Position[#1,{_,2}]}&[Sort@Flatten[{{#1,1}&/@seqA,{#1,2}&/@seqB},1]];

%t limseqU=FixedPoint[jointRank[{seqA,#1[[1]]}]&,jointRank[{seqA,seqB}]][[1]] (*A187228*)

%t Complement[Range[Length[seqA]],limseqU] (*A187229*)

%t (*by _Peter J. C. Moses_, Mar 07 2011*)

%Y Cf. A187224, A187229.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 07 2011