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A184871 n+floor(ns/r)+floor(nt/r), where r=log(2), s=log(3), t=log(5).

%I A184871 #9 Sep 04 2015 10:32:10
%S A184871 4,9,13,19,23,28,34,38,43,48,53,58,63,68,72,78,82,87,93,97,102,107,
%T A184871 112,117,122,127,131,137,141,146,151,156,161,165,171,176,180,186,190,
%U A184871 195,200,205,210,215,220,224,230,235,239,245,249,254,260,264,269,274,279,283,288,294,298,303,308,313,318,323,328,332,338,342,347,353,357,362,367,372,377,382,387,391,397,401,406,412,416,421,426,431,436,440,446,450,455,460,465,470,475,480,484,490,495,499,505,509,514,520,524,529,534,539,543,549,554,558,564,568,573,578,583,588
%N A184871 n+floor(ns/r)+floor(nt/r), where r=log(2), s=log(3), t=log(5).
%C A184871 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
%C A184871 {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are disjoint.
%C A184871 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 A184871 a(n)=n+[ns/r]+[nt/r],
%C A184871 b(n)=n+[nr/s]+[nt/s],
%C A184871 c(n)=n+[nr/t]+[ns/t], where []=floor.
%C A184871 Taking r=log(2), s=log(3), t=log(5) yields
%C A184871 a=A184871, b=A184872, c=A184873.
%t A184871 r=Log[2]; s=Log[3]; t=Log[5];
%t A184871 a[n_]:=n+Floor [n*s/r]+Floor[n*t/r];
%t A184871 b[n_]:=n+Floor [n*r/s]+Floor[n*t/s];
%t A184871 c[n_]:=n+Floor[n*r/t]+Floor[n*s/t];
%t A184871 Table[a[n],{n,1,120}]  (* A184871 *)
%t A184871 Table[b[n],{n,1,120}]  (* A184872 *)
%t A184871 Table[c[n],{n,1,120}]  (* A184873 *)
%Y A184871 Cf. A184812, A184872, A184873, A184874 (primes in A184872).
%K A184871 nonn
%O A184871 1,1
%A A184871 _Clark Kimberling_, Jan 23 2011
%E A184871 Name corrected by _Charles R Greathouse IV_, Sep 04 2015