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 A190059 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/5), t=csc(Pi/5). 3

%I #12 Sep 08 2022 08:45:56

%S 2,6,9,12,15,19,22,25,29,32,35,39,42,45,48,52,54,58,62,65,68,71,75,78,

%T 81,85,87,91,95,98,101,104,108,110,114,118,120,124,127,131,134,137,

%U 141,143,147,151,153,157,160,164,166,170,174,176,180,183,186,190,193,197,199,203,207,209,213,216,219,222,226,230

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/5), t=csc(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=sin(Pi/5), t=csc(Pi/5) gives

%C a=A190059, b=A190060, c=A190061.

%H G. C. Greubel, <a href="/A190059/b190059.txt">Table of n, a(n) for n = 1..10000</a>

%F A190059: a(n) = n + [n*sin(Pi/5)] + [n*csc(Pi/5].

%F A190060: b(n) = n + [n*csc(Pi/5)] + [n*(csc(Pi/5))^2].

%F A190061: c(n) = n + [n*sin(Pi/5)] + [n*(sin(Pi/5))^2].

%t r=1; s=Sin[Pi/5]; t=Csc[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}] (*A190059*)

%t Table[b[n], {n, 1, 120}] (*A190060*)

%t Table[c[n], {n, 1, 120}] (*A190061*)

%o (PARI) for(n=1,30, print1(n + floor(n*sin(Pi/5)) + floor(n/sin(Pi/5)), ", ")) \\ _G. C. Greubel_, Jan 10 2018

%o (Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/5)) + Floor(n/Sin(Pi(C)/5)): n in [1..30]]; // _G. C. Greubel_, Jan 10 2018

%Y Cf. A190060, A190061.

%K nonn

%O 1,1

%A _Clark Kimberling_, May 04 2011

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Last modified June 10 13:47 EDT 2023. Contains 363205 sequences. (Running on oeis4.)