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%I #21 Sep 08 2022 08:45:56
%S 1,3,6,8,11,13,16,18,20,22,25,27,30,32,35,37,39,41,44,46,49,51,54,56,
%T 57,60,62,65,67,70,72,75,76,79,81,84,86,89,91,94,95,98,100,103,105,
%U 108,110,113,115,116,119,121,124,126,129,131,134,135,138,140,143,145,148,150,153,154,157,159,162,164,167,169,171,173,175,178,180
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(1/2), t=cos(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 pairwise disjoint. Let f(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define g(n) and h(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=sin(1/2), t=cos(1/2) gives f=A190085, g=A190086, h=A190087.
%H G. C. Greubel, <a href="/A190085/b190085.txt">Table of n, a(n) for n = 1..10000</a>
%F A190085: f(n) = n + [n*sin(1/2)] + [n*cos(1/2)], this sequence.
%F A190086: g(n) = n + [n*csc(1/2)] + [n*cot(1/2)].
%F A190087: h(n) = n + [n*sec(1/2)] + [n*tan(1/2)].
%t r=1; s=Sin[1/2]; t=Cos[1/2];
%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}] (* A190085 *)
%t Table[g[n], {n, 1, 120}] (* A190086 *)
%t Table[h[n], {n, 1, 120}] (* A190087 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sin(1/2)) + floor(n*cos(1/2)), ", ")) \\ _G. C. Greubel_, Mar 04 2018
%o (Magma) [n + Floor(n*Sin(1/2)) + Floor(n*Cos(1/2)): n in [1..100]]; // _G. C. Greubel_, Mar 04 2018
%Y Cf. A190086, A190087.
%K nonn
%O 1,2
%A _Clark Kimberling_, May 04 2011
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