|
%I #12 Sep 08 2022 08:45:56
%S 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,60,62,65,68,71,
%T 74,77,80,83,86,89,92,95,98,101,104,107,110,113,117,120,122,125,128,
%U 131,134,137,140,143,146,149,152,155,158,161,164,167,170,173,177,180,183,185,188,191,194,197,200,203,206,209,212,215,218
%N n + [n*s/r] + [n*t/r]; r=1, s=sin(2*Pi/5), t=csc(2*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(2*Pi/5), t=csc(2*Pi/5) gives
%C a=A190082, b=A190083, c=A190084.
%H G. C. Greubel, <a href="/A190082/b190082.txt">Table of n, a(n) for n = 1..10000</a>
%F A190082: a(n) = n + [n*sin(2*Pi/5)] + [n*csc(2*Pi/5)].
%F A190083: b(n) = n + [n*csc(2*Pi/5)] + [n*(csc(2*Pi/5))^2].
%F A190084: c(n) = n + [n*sin(2*Pi/5)] + [n*(sin(2*Pi/5))^2].
%t r=1; s=Sin[2*Pi/5]; t=Csc[2*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}] (* A190082 *)
%t Table[b[n], {n, 1, 120}] (* A190083 *)
%t Table[c[n], {n, 1, 120}] (* A190084 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sin(2*Pi/5)) + floor(n/sin(2*Pi/5)), ", ")) \\ _G. C. Greubel_, Mar 04 2018
%o (Magma) R:= RealField(); [n + Floor(n*Sin(2*Pi(R)/5)) + Floor(n/Sin(2*Pi(R)/5)): n in [1..100]]; // _G. C. Greubel_, Mar 04 2018
%Y Cf. A190083, A190084.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 04 2011
|
