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%I #8 Dec 26 2023 10:11:49
%S 1,2,3,6,8,9,11,13,15,17,18,20,22,24,26,28,29,31,33,35,36,39,40,42,44,
%T 46,48,49,51,53,55,57,58,60,62,64,66,68,69,71,73,75,77,79,80,82,84,86,
%U 88,90,91,93,95,97,98,101,102,104,106,108,109,111,113,115,117,119,120
%N Rank transform of the sequence floor(3(n-2)/2); complement of A187479.
%C See A187224. Although the first term of floor(3(n-2)/2) is negative, we can replace it by 0 without affecting the same joint rankings; thus, the procedure described at A187224 applies.
%t seqA = Table[Floor[3(n-2)/2], {n, 1, 180}]
%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[{seqA, seqB}]][[1]] (* A187478 *)
%t Complement[Range[Length[seqA]], limseqU] (* A187479 *)
%t (* by _Peter J. C. Moses_, Mar 10 2011 *)
%Y Cf. A187422, A187476, A187479.
%K nonn
%O 1,2
%A _Clark Kimberling_, Mar 10 2011
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