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a(n) = n + floor(ns/r) + floor(nt/r), where r = log(2), s = 1, t = log(3).
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%I #5 Jan 05 2020 13:01:00

%S 3,7,11,15,19,23,28,31,35,39,43,48,51,56,59,64,67,71,76,79,84,87,92,

%T 96,100,104,107,112,115,120,124,128,132,136,140,144,148,152,156,160,

%U 164,168,173,176,180,184,188,193,196,201,204,209,213,216,221,224,229

%N a(n) = n + floor(ns/r) + floor(nt/r), 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(ns/r) + floor(nt/r), 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. A330214, A330215.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 05 2020