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%I #8 Sep 08 2022 08:45:56
%S 3,7,11,15,20,23,27,31,35,40,43,47,52,55,60,63,68,72,75,80,83,88,92,
%T 95,100,104,108,112,116,120,124,128,132,137,140,144,148,152,157,160,
%U 164,168,172,177,180,185,189,192,197,200,205,209,212,217,220,225,229,233,237,241,245,249,253,257,261,265,269,274,277,281,285,289,294,297,302,305,309,314,317,322,326,329,334
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(2), t=(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 pairwise 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 f(n) = n + [n*s/r] + [n*t/r],
%C g(n) = n + [n*r/s] + [n*t/s],
%C h(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C Taking r=1, s=sqrt(2), t=(1+sqrt(5))/2 gives f=A189364, g=A189365, h=A189366.
%H G. C. Greubel, <a href="/A189364/b189364.txt">Table of n, a(n) for n = 1..10000</a>
%t r = 1; s = Sqrt[2]; t = (1 + Sqrt[5])/2;
%t f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t h[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t Table[f[n], {n, 1, 120}] (* A189364 *)
%t Table[g[n], {n, 1, 120}] (* A189365 *)
%t Table[h[n], {n, 1, 120}] (* A189366 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sqrt(2)) + floor(n*(1+sqrt(5))/2), ", ")) \\ _G. C. Greubel_, Apr 20 2018
%o (Magma) [n + Floor(n*Sqrt(2)) + Floor(n*(1+Sqrt(5))/2): n in [1..100]]; // _G. C. Greubel_, Apr 20 2018
%Y Cf. A189365, A189366.
%K nonn
%O 1,1
%A _Clark Kimberling_, Apr 20 2011
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