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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/3), t=csc(Pi/3).
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%I #16 Oct 20 2024 01:42:04

%S 2,5,8,11,14,17,21,23,26,29,32,35,39,42,44,47,50,53,56,60,63,66,68,71,

%T 74,78,81,84,87,89,92,95,99,102,105,108,111,113,117,120,123,126,129,

%U 132,134,138,141,144,147,150,153,157,159,162,165,168,171,174,178,180,183,186,189,192,196,199,202,204,207,210,213,217,220

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, 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=1, s=sin(Pi/3), t=csc(Pi/3) gives a=A190006, b=A190007, c=A190008.

%H G. C. Greubel, <a href="/A190006/b190006.txt">Table of n, a(n) for n = 1..10000</a>

%F A190006: a(n) = n + [n*sin(Pi/3)] + [n*csc(Pi/3)].

%F A190007: b(n) = n + [n*csc(Pi/3)] + [n*(csc(Pi/3))^2].

%F A190008: c(n) = n + [n*sin(Pi/3)] + [n*(sin(Pi/3))^2].

%t r=1; 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}] (* A190006 *)

%t Table[b[n], {n, 1, 120}] (* A190007 *)

%t Table[c[n], {n, 1, 120}] (* A190008 *)

%o (PARI) for(n=1,100, print1(n + floor(n*sin(Pi/3)) + floor(n/sin(Pi/3)), ", ")) \\ _G. C. Greubel_, Jan 11 2018

%o (Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/3)) + Floor(n/Sin(Pi(C)/3)): n in [1..100]]; // _G. C. Greubel_, Jan 11 2018

%Y Cf. A180007, A190008.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 03 2011