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 A189937 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/8), t=cos(Pi/8). 3

%I #13 Sep 08 2022 08:45:56

%S 1,3,6,8,10,13,15,18,20,22,25,27,29,31,33,36,38,40,43,45,48,50,52,55,

%T 57,59,61,63,66,68,70,73,75,78,80,82,85,87,89,91,93,96,98,100,103,105,

%U 107,110,112,115,117,119,121,123,126,128,130,133,135,137,140,142,145,147,149,151,153,156,158,160,163,165,167,170,172,175,177

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/8), t=cos(Pi/8).

%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/8), t=cos(Pi/8) gives a=A189937, b=A189938, c=A189939.

%H G. C. Greubel, <a href="/A189937/b189937.txt">Table of n, a(n) for n = 1..10000</a>

%F A189937: a(n) = n + [n*sin(Pi/8)] + [n*cos(Pi/8)].

%F A189938: b(n) = n + [n*csc(Pi/8)] + [n*cot(Pi/8)].

%F A189939: c(n) = n + [n*sec(Pi/8)] + [n*tan(Pi/8)].

%t r=1; s=Sin[Pi/8]; t=Cos[Pi/8];

%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}] (*A189937*)

%t Table[b[n], {n, 1, 120}] (*A189938*)

%t Table[c[n], {n, 1, 120}] (*A189939*)

%o (PARI) for(n=1,100, print1(n + floor(n*sin(Pi/8)) + floor(n*cos(Pi/8)), ", ")) \\ _G. C. Greubel_, Jan 13 2018

%o (Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/8)) + Floor(n*Cos(Pi(C)/8)): n in [1..100]]; // _G. C. Greubel_, Jan 13 2018

%Y Cf. A189938, A189939.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 01 2011

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