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A190304 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=csch(1), t=sech(1). 3

%I

%S 1,4,6,9,12,14,16,19,21,24,27,29,32,34,36,39,42,44,47,49,51,54,56,59,

%T 62,64,66,69,71,74,77,79,82,84,86,89,91,94,97,99,101,104,106,109,112,

%U 114,116,119,121,124,127,129,132,133,136,139,141,144,147,149,151,154,156,159,162,164,167,169,171,174,177,179,182,183,186,189

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=csch(1), t=sech(1).

%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+[ns/r]+[nt/r],

%C b(n) = n+[nr/s]+[nt/s],

%C c(n) = n+[nr/t]+[ns/t], where []=floor.

%C Taking r=1, s=csch(1), t=sech(1) gives

%C a=A190304, b=A190305, c=A190306.

%H G. C. Greubel, <a href="/A190304/b190304.txt">Table of n, a(n) for n = 1..10000</a>

%F A190304: a(n)=n+[n*csch(1)]+[n*sech(1)].

%F A190305: b(n)=n+[n*sinh(1)]+[n*tanh(1)].

%F A190306: c(n)=n+[n*cosh(1)]+[n*coth(1)].

%t r=1; s=Csch[1]; t=Sech[1];

%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}] (*A190304*)

%t Table[b[n], {n, 1, 120}] (*A190305*)

%t Table[c[n], {n, 1, 120}] (*A190306*)

%o (PARI) for(n=1, 30, print1(n + floor(n/sinh(1)) + floor(n/cosh(1)), ", ")) \\ _G. C. Greubel_, Dec 29 2017

%o (MAGMA) [n + Floor(n/Sinh(1)) + Floor(n/Cosh(1)): n in [1..30]]; // _G. C. Greubel_, Dec 29 2017

%Y Cf. A190305, A190306, A190265.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 07 2011

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Last modified May 8 21:56 EDT 2021. Contains 343668 sequences. (Running on oeis4.)