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%I #13 Sep 08 2022 08:45:57
%S 5,11,16,23,28,34,40,46,51,58,63,69,74,81,86,92,98,104,109,116,121,
%T 127,132,139,144,150,156,162,167,174,179,185,190,197,202,208,214,220,
%U 225,232,237,243,248,255,260,266,272,278,283,290,295,301,306,313,319,324,331,336,342,348,354,359,365,371,377,382,389,394,400
%N n + [n*s/r] + [n*t/r]; r=1, s=sinh(Pi/2), t=cosh(Pi/2).
%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 f(n) = n + [n*s/r] + [n*t/r],
%C g(n) = n + [n*r/s] + [n*t/s],
%C h(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C Taking r=1, s=sinh(Pi/2), t=cosh(Pi/2) gives
%C f=A190323, g=A190324, h=A190325.
%H G. C. Greubel, <a href="/A190323/b190323.txt">Table of n, a(n) for n = 1..10000</a>
%F A190323: f(n) = n + [n*sinh(Pi/2)] + [n*cosh(Pi/2)].
%F A190324: g(n) = n + [n*csch(Pi/2)] + [n*coth(Pi/2)].
%F A190325: h(n) = n + [n*sech(Pi/2)] + [n*tanh(Pi/2)].
%t r=1; s=Sinh[Pi/2]; t=Cosh[Pi/2];
%t f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t Table[f[n], {n, 1, 120}] (* A190323 *)
%t Table[g[n], {n, 1, 120}] (* A190324 *)
%t Table[h[n], {n, 1, 120}] (* A190325 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sinh(Pi/2)) + floor(n*cosh(Pi/2)), ", ")) \\ _G. C. Greubel_, Apr 04 2018
%o (Magma) R:=RealField(); [n + Floor(n*Sinh(Pi(R)/2)) + Floor(n*Cosh(Pi(R)/2)): n in [1..100]]; // _G. C. Greubel_, Apr 04 2018
%Y Cf. A190324, A190325.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 08 2011
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