%I #4 Jan 05 2020 13:01:12
%S 1,4,6,9,12,14,17,20,22,25,27,29,32,34,37,40,42,45,47,50,53,55,57,60,
%T 62,65,68,70,73,75,78,81,83,85,88,90,93,95,98,101,103,106,109,111,113,
%U 116,118,121,123,126,129,131,134,137,139,141,143,146,149,151
%N a(n) = n + floor(nr/t) + floor(ns/t), where r = log(2), s = 1, t = log(3).
%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 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 = log(2), s = 1, t = log(3) yields
%C a=A330213, b=A330214, c=A330215.
%F a(n) = n + floor(nr/t) + floor(ns/t), where r = log(2), s = 1, t = log(3).
%t r = Log[2]; s = 1; t = Log[3];
%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}] (* A330213 *)
%t Table[b[n], {n, 1, 120}] (* A330214 *)
%t Table[c[n], {n, 1, 120}] (* A330215 *)
%Y Cf. A330213, A330214.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jan 05 2020