A190002 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=(sinh(1))^2, t=(cosh(1))^2.

4, 8, 14, 18, 22, 28, 32, 38, 42, 46, 52, 56, 60, 66, 70, 76, 80, 84, 90, 94, 100, 104, 108, 114, 118, 122, 128, 132, 138, 142, 146, 152, 156, 160, 166, 170, 176, 180, 184, 190, 194, 200, 204, 208, 214, 218, 222, 228, 232, 238, 242, 246, 252, 256, 260, 266, 270, 276, 280, 284, 290, 294, 300, 304, 308, 314, 318, 322, 328, 332, 338

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

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

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

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

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

a=A190002, b=A190003, c=A190004.

Links

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

Formula

A190002: a(n) = n + [n*(sinh(1))^2] + [n*(cosh(1))^2].

A190003: b(n) = n + [n*(csch(1))^2] + [n*(coth(1))^2].

A005408: c(n) = 2*n - 1.

Mathematica

r=1; s=Sinh[1]^2; t=Cosh[1]^2;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}]  (* A190002 *)
Table[b[n], {n, 1, 120}]  (* A190003 *)
Table[c[n], {n, 1, 120}]  (* A005408 *)
Table[a[n]/2, {n, 1, 120}](* A190004 *)
Table[b[n]/2, {n, 1, 120}](* A182760 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*(sinh(1))^2) + floor(n*(cosh(1))^2), ", ")) \\ G. C. Greubel, Jan 11 2018
(Magma) [n + Floor(n*(Sinh(1))^2) + Floor(n*(Cosh(1))^2): n in [1..100]]; // G. C. Greubel, Jan 11 2018

Crossrefs

Cf. A189999, A190003, A005408.

Sequence in context: A312505 A312506 A312507 * A312508 A312509 A312510

Adjacent sequences: A189999 A190000 A190001 * A190003 A190004 A190005

Keyword

nonn

Author

Clark Kimberling, May 03 2011

Status

approved