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Rank transform of the sequence floor(5n/4); complement of A187231.
2

%I #10 Feb 10 2014 21:27:46

%S 1,3,4,7,8,9,11,14,15,17,18,20,22,23,25,27,29,30,32,34,36,37,39,41,43,

%T 44,46,48,49,51,53,55,56,58,59,62,63,65,66,69,70,72,73,75,77,79,80,83,

%U 84,85,87,89,91,93,94,96,98,99,101,103,105,106,108,110,112,113,114,117,119,120,122,124,125,127,128,131,132,134,135,138,139,141,142,144,146,148,149,151,153,154,156,159,160,161,163,165

%N Rank transform of the sequence floor(5n/4); complement of A187231.

%C See A187224.

%t seqA=Table[Floor[5n/4],{n,1,220}] (*A001068*)

%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]] (*A187230*)

%t Complement[Range[Length[seqA]],limseqU] (*A187231*)

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

%Y Cf. A187224, A187231.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 07 2011