A190326 - OEIS

%I #10 Sep 08 2022 08:45:57

%S 10,21,31,42,53,63,74,84,95,106,116,127,137,148,159,169,180,190,201,

%T 212,222,233,243,254,265,275,286,296,307,318,328,339,349,360,371,381,

%U 392,402,413,424,434,445,455,466,477,487,498,508,519,530,540,551,561,572

%N n + [n*s/r] + [n*t/r]; r=1/2, 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=1/2, s=sinh(Pi/2), t=cosh(Pi/2) gives

%C f=A190326, g=A190327, h=A190328.

%H G. C. Greubel, <a href="/A190326/b190326.txt">Table of n, a(n) for n = 1..10000</a>

%F A190326:  f(n) = n + [2*n*sinh(Pi/2)] + [2*n*cosh(Pi/2)].

%F A190327:  g(n) = n + [n*csch(Pi/2)/2] + [n*coth(Pi/2)].

%F A190328:  h(n) = n + [n*sech(Pi/2)/2] + [n*tanh(Pi/2)].

%t r=1/2; 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}]  (*A190326*)

%t Table[g[n], {n, 1, 120}]  (*A190327*)

%t Table[h[n], {n, 1, 120}]  (*A190328*)

%o (PARI) for(n=1,100, print1(n + floor(2*n*sinh(Pi/2)) + floor(2*n*cosh(Pi/2)), ", ")) \\ _G. C. Greubel_, Apr 04 2018

%o (Magma) R:=RealField(); [n + Floor(2*n*Sinh(Pi(R)/2)) + Floor(2*n*Cosh(Pi(R)/2)): n in [1..100]]; // _G. C. Greubel_, Apr 04 2018

%Y Cf. A190327, A190328.

%K nonn

%O 1,1

%A _Clark Kimberling_, May 08 2011