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%I #12 Sep 08 2022 08:45:56
%S 5,11,18,23,30,36,42,48,54,61,67,72,79,85,91,97,104,110,115,122,128,
%T 135,140,146,153,159,165,171,178,183,189,196,202,208,214,220,227,232,
%U 239,245,251,257,263,270,275,282,288,294,300,306,313,319,324,331,337,343,349,356,362,367,374,380,387,392,398,405,411,417,423,429,435,441,448,454,460,466,472,479,484,491,497,503
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1+sqrt(2), t=1+sqrt(3).
%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=1+sqrt(2), t=1+sqrt(3) gives f=A189469, g=A189470, h=A189471.
%H Ivan Panchenko, <a href="/A189469/b189469.txt">Table of n, a(n) for n = 1..10000</a>
%t r=1; s=1+Sqrt[2]; t=1+Sqrt[3];
%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}] (* A189469 *)
%t Table[g[n], {n, 1, 120}] (* A189470 *)
%t Table[h[n], {n, 1, 120}] (* A189471 *)
%o (PARI) for(n=1,100, print1(n + floor(n*(1+sqrt(2))) + floor(n*(1+sqrt(3))), ", ")) \\ _G. C. Greubel_, Apr 13 2018
%o (Magma) [n + Floor(n*(1+Sqrt(2))) + Floor(n*(1+Sqrt(3))): n in [1..100]]; // _G. C. Greubel_, Apr 13 2018
%Y Cf. A189470, A189471.
%K nonn
%O 1,1
%A _Clark Kimberling_, Apr 22 2011
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