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a(n) = n + floor(ns/r) + floor(nt/r), where r = Pi - 1, s = Pi, t = Pi + 1.
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%I #9 Jan 28 2023 15:37:38

%S 3,7,12,16,21,25,30,34,39,43,48,52,57,61,66,69,73,78,82,87,91,96,100,

%T 105,109,114,118,123,127,132,135,139,144,148,153,157,162,166,171,175,

%U 180,184,189,193,198,201,205,210,214,219,223,228,232,237,241,246

%N a(n) = n + floor(ns/r) + floor(nt/r), where r = Pi - 1, s = Pi, t = Pi + 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 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 = Pi - 1, s = Pi, t = Pi + 1 yields a=A330181, b=A016789, c=A330182.

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

%t r = Pi - 1; s = Pi; t = Pi + 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}] (* A330181 *)

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

%t Table[c[n], {n, 1, 120}] (* A330182 *)

%t Table[n+Floor[n Pi/(Pi-1)]+Floor[n (Pi+1)/(Pi-1)],{n,60}] (* _Harvey P. Dale_, Jan 28 2023 *)

%Y Cf. A016789, A330182.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 05 2020