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%I #5 Mar 30 2012 18:57:24
%S 1,2,4,6,7,9,10,12,14,15,17,19,20,22,24,25,27,28,30,32,33,35,37,38,40,
%T 41,43,45,46,49,50,51,53,55,56,58,59,61,63,64,67,68,69,71,73,74,76,77,
%U 80,81,82,84,86,87,89,91,92,94,95,98,99,100,102,104,105,107,109,111,112,113,116,117,118,120,122,123,125,126,129,130,131,134,135,136,138
%N n+[ns/r]+[nt/r]; r=1, s=1/e, t=1/(e+1).
%C This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
%C a(n)=n+[ns/r]+[nt/r],
%C b(n)=n+[nr/s]+[nt/s],
%C c(n)=n+[nr/t]+[ns/t], where []=floor.
%C Taking r=1, s=1/e, t=1/(e+1) gives
%C a=A189518, b=A189519, c=A189520.
%F The three sequences a=A189518, b=A189519, c=A189520 are given by
%F a(n)=n+[n/e]+[n/(e+1)],
%F b(n)=n+[ne]+[ne/(e+1)],
%F c(n)=3n+[ne]+[n/e].
%F Is there a simple formula for the complement of a?
%t r=1; s=1/E; t=1/(E+1);
%t a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t Table[a[n], {n, 1, 120}] (*A189518*)
%t Table[b[n], {n, 1, 120}] (*A189519*)
%t Table[c[n], {n, 1, 120}] (*A189520*)
%Y Cf. A189519, A189520.
%K nonn
%O 1,2
%A _Clark Kimberling_, Apr 23 2011
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