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%I #11 Sep 08 2022 08:45:57
%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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