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a(n) = n + floor(nr/t) + floor(ns/t), where r = Pi - 1, s = Pi, t = Pi + 1.
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%I #9 Apr 09 2021 22:11:39

%S 1,4,6,9,10,13,15,18,19,22,24,27,28,31,33,36,37,40,42,45,46,49,51,54,

%T 55,58,60,63,64,67,70,72,75,76,79,81,84,85,88,90,93,94,97,99,102,103,

%U 106,108,111,112,115,117,120,121,124,126,129,130,133,136,138

%N a(n) = n + floor(nr/t) + floor(ns/t), 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(nr/t) + floor(ns/t), 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 *)

%Y Cf. A016789, A330182.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 05 2020