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a(n) = n + floor(ns/r) + floor(nt/r), where r = e - 2, s = e - 1, t = e.
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%I #4 Jan 05 2020 12:59:37

%S 6,13,21,28,34,42,49,57,64,70,78,85,93,99,106,114,121,129,135,142,150,

%T 157,165,171,178,186,193,199,207,214,222,229,235,243,250,258,265,271,

%U 279,286,294,300,307,315,322,330,336,343,351,358,366,372,379,387,394

%N a(n) = n + floor(ns/r) + floor(nt/r), where r = e - 2, s = e - 1, t = e.

%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 = e - 2, s = e - 1, t = e yields

%C a=A330177, b=A016789, c=A330178.

%F a(n) = n + floor(ns/r) + floor(nt/r), where r = e - 2, s = e - 1, t = e.

%t r = E - 2; s = E - 1; t = E;

%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}] (* A330177 *)

%t Table[b[n], {n, 1, 120}] (* A016789 *)

%t Table[c[n], {n, 1, 120}] (* A330178 *)

%Y Cf. A016789, A330178.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 05 2020