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%I #16 Sep 08 2022 08:45:56
%S 1,3,5,7,9,11,14,15,17,19,21,23,25,28,29,31,33,35,37,39,42,43,45,47,
%T 49,52,53,56,57,59,61,63,66,67,70,71,74,75,77,80,81,84,85,88,89,91,94,
%U 95,98,99,102,104,105,108,109,112,113,116,118,119,122,123,126,127,130,132,134,136,137,140,141,144,146,148,150,151,154,156,158,160
%N a(n) = n + [n*s/r] + [n*t/r]; r=2, s=sin(Pi/3), t=csc(Pi/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 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=2, s=sin(Pi/3), t=csc(Pi/3) gives a=A190053, b=A190054, c=A190055.
%H G. C. Greubel, <a href="/A190053/b190053.txt">Table of n, a(n) for n = 1..10000</a>
%F A190053: a(n) = n + [(n/2)*sin(Pi/3)] + [(n/2)*csc(Pi/3)].
%F A190054: b(n) = n + [2n*csc(Pi/3)] + [n*(csc(Pi/3))^2].
%F A190055: c(n) = n + [2n*sin(Pi/3)] + [n*(sin(Pi/3))^2].
%t r=2; s=Sin[Pi/3]; t=Csc[Pi/3];
%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}] (* A190053 *)
%t Table[b[n], {n, 1, 120}] (* A190054 *)
%t Table[c[n], {n, 1, 120}] (* A190055 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sin(Pi/3)/2) + floor(n/(2*sin(Pi/3))), ", ")) \\ _G. C. Greubel_, Jan 10 2018
%o (Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/3)/2) + Floor(n/(2*Sin(Pi(C)/3))): n in [1..100]]; // _G. C. Greubel_, Jan 10 2018
%Y Cf. A190054, A190055.
%K nonn
%O 1,2
%A _Clark Kimberling_, May 04 2011
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