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a(n) = n + floor(ns/r) + floor(nt/r), where r = tau - 1/2, s = tau, t = tau + 1/2, tau = golden ratio = (1+sqrt(5))/2.
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%I #4 Jan 05 2020 13:00:45

%S 3,7,12,16,21,25,30,34,39,42,46,51,55,60,64,69,73,78,81,85,90,94,99,

%T 103,108,112,117,121,124,129,133,138,142,147,151,156,160,163,168,172,

%U 177,181,186,190,195,199,204,207,211,216,220,225,229,234,238,243,246

%N a(n) = n + floor(ns/r) + floor(nt/r), where r = tau - 1/2, s = tau, t = tau + 1/2, tau = golden ratio = (1+sqrt(5))/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 = tau - 1/2, s = tau, t = tau + 1/2 yields

%C a=A330185, b=A016789, c=A330186.

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

%t tau = GoldenRatio; r = tau - 1/2; s = tau; t = tau + 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}] (* A330185 *)

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

%t Table[c[n], {n, 1, 120}] (* A330186 *)

%Y Cf. A016789, A330186.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 05 2020