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%I #11 Sep 08 2022 08:45:56
%S 2,5,8,11,15,17,20,23,27,30,32,35,39,42,45,47,51,54,57,60,62,66,69,72,
%T 75,79,81,84,87,91,94,96,99,103,106,109,111,114,118,121,124,126,130,
%U 133,136,139,143,145,148,151,155,158,160,163,166,170,173,175,178,182,185,188,190,194,197,200,203,207,209,212,215,218,222
%N n + [n*s/r] + [n*t/r]; r=1, s=cos(Pi/5), t=sec(Pi/5).
%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 a(n) = n + [n*s/r] + [n*t/r],
%C b(n) = n + [n*r/s] + [n*t/s],
%C c(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C Taking r=1, s=cos(Pi/5), t=sec(Pi/5) gives
%C a=A190079, b=A190080, c=A190081.
%H G. C. Greubel, <a href="/A190079/b190079.txt">Table of n, a(n) for n = 1..10000</a>
%F A190079: a(n) = n + [n*cos(Pi/5)] + [n*sec(Pi/5)].
%F A190080: b(n) = n + [n*sec(Pi/5)] + [n*(sec(Pi/5))^2].
%F A190081: c(n) = n + [n*cos(Pi/5)] + [n*(cos(Pi/5))^2].
%t r=1; s=Cos[Pi/5]; t=Sec[Pi/5];
%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}] (*A190079*)
%t Table[b[n], {n, 1, 120}] (*A190080*)
%t Table[c[n], {n, 1, 120}] (*A190081*)
%o (PARI) for(n=1,100, print1(n + floor(n*cos(Pi/5)) + floor(n/cos(Pi/5)), ", ")) \\ _G. C. Greubel_, Feb 15 2018
%o (Magma) R:= RealField(); [n + Floor(n*Cos(Pi(R)/5)) + Floor(n/Cos(Pi(R)/5)): n in [1..100]]; // _G. C. Greubel_, Feb 15 2018
%Y Cf. A190080, A190081.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 04 2011
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