A190323 n + [n*s/r] + [n*t/r]; r=1, s=sinh(Pi/2), t=cosh(Pi/2).

5, 11, 16, 23, 28, 34, 40, 46, 51, 58, 63, 69, 74, 81, 86, 92, 98, 104, 109, 116, 121, 127, 132, 139, 144, 150, 156, 162, 167, 174, 179, 185, 190, 197, 202, 208, 214, 220, 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

Offset

1,1

Comments

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

f(n) = n + [n*s/r] + [n*t/r],

g(n) = n + [n*r/s] + [n*t/s],

h(n) = n + [n*r/t] + [n*s/t], where []=floor.

Taking r=1, s=sinh(Pi/2), t=cosh(Pi/2) gives

f=A190323, g=A190324, h=A190325.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

A190323: f(n) = n + [n*sinh(Pi/2)] + [n*cosh(Pi/2)].

A190324: g(n) = n + [n*csch(Pi/2)] + [n*coth(Pi/2)].

A190325: h(n) = n + [n*sech(Pi/2)] + [n*tanh(Pi/2)].

Mathematica

r=1; s=Sinh[Pi/2]; t=Cosh[Pi/2];
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}]  (* A190323 *)
Table[g[n], {n, 1, 120}]  (* A190324 *)
Table[h[n], {n, 1, 120}]  (* A190325 *)

Prog

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

Crossrefs

Cf. A190324, A190325.

Sequence in context: A272666 A314174 A314175 * A314176 A314177 A314178

Adjacent sequences: A190320 A190321 A190322 * A190324 A190325 A190326

Keyword

nonn

Author

Clark Kimberling, May 08 2011

Status

approved