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Rank transform of the sequence floor(n/sqrt(3)); complement of A187410.
2

%I #9 Feb 10 2014 21:30:54

%S 1,2,3,5,6,8,10,11,12,13,15,16,18,19,20,22,23,24,25,27,29,30,32,33,35,

%T 36,37,39,40,42,43,44,46,47,49,50,52,53,54,56,57,59,60,62,63,65,66,67,

%U 69,70,71,73,74,76,77,78,79,81,83,84,85,86,88,89,91,93,94,95,96,98,99,101,102,103,105,106,108,110,111,112,113,115,116,118,119,120,122,123,125,126,127,129

%N Rank transform of the sequence floor(n/sqrt(3)); complement of A187410.

%C See A187224.

%t m = 3^(-1/2);

%t seqA = Table[Floor[m*n], {n, 1, 180}] (* A097337 *)

%t seqB = Table[n, {n, 1, 80}]; (* A000027 *)

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

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

%t {#1, 2} & /@ seqB}, 1]];

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

%t [{seqA, seqB}]][[1]] (* A187319 *)

%t Complement[Range[Length[seqA]], limseqU] (* A187410 *)

%t (* by _Peter J. C. Moses_, Mar 09 2011 *)

%Y Cf. A187224, A187410.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 08 2011