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a(n) = n + floor(nr/t) + floor(ns/t), where r = sqrt(2) - 1/2, s = sqrt(2), t = sqrt(2) + 1/2.
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%I #5 Jan 17 2020 23:07:58

%S 1,3,6,7,10,12,15,16,19,21,24,25,28,30,33,34,37,39,42,43,46,48,49,52,

%T 54,57,58,61,63,66,67,70,72,75,76,79,81,84,85,88,90,93,94,97,99,100,

%U 103,105,108,109,112,114,117,118,121,123,126,127,130,132,135

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

%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 = sqrt(2) - 1/2, s = sqrt(2), t = sqrt(2) + 1/2 yields

%C a=A330183, b=A016789, c=A330184.

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

%t r = Sqrt[2] - 1/2; s = Sqrt[2]; t = Sqrt[2] + 1/2;

%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}] (* A330183 *)

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

%t Table[c[n], {n, 1, 120}] (* A330184 *)

%Y Cf. A016789, A330183.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 05 2020