A189999 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sinh(1), t=cosh(1).

3, 7, 10, 14, 17, 22, 25, 29, 32, 36, 39, 44, 48, 51, 55, 58, 62, 66, 70, 73, 77, 80, 85, 89, 92, 96, 99, 103, 107, 111, 114, 118, 121, 125, 130, 133, 137, 140, 144, 148, 152, 155, 159, 162, 166, 170, 174, 178, 181, 185, 188, 193, 196, 200, 203, 207, 210, 215, 219, 222, 226, 229, 234, 237, 241, 244, 248, 251, 256, 260, 263, 267, 270

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), t=cosh(1) gives

a=A189999, b=A190000, c=A190001.

Links

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

Formula

A189999: a(n) = n + [n*sinh(1)] + [n*cosh(1)].

A190000: b(n) = n + [n*csch(1)] + [n*coth(1)].

A190001: c(n) = n + [n*sech(1)] + [n*tanh(1)].

Mathematica

r=1; s=Sinh[1]; t=Cosh[1];
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}]  (*A189999*)
Table[b[n], {n, 1, 120}]  (*A190000*)
Table[c[n], {n, 1, 120}]  (*A190001*)
Table[n+Floor[n Cosh[1]]+Floor[n Sinh[1]], {n, 80}] (* Harvey P. Dale, Jan 04 2026 *)

Prog

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

Crossrefs

Cf. A190000, A190001, A190002.

Sequence in context: A248522 A098005 A322533 * A310188 A393700 A171983

Adjacent sequences: A189996 A189997 A189998 * A190000 A190001 A190002

Keyword

nonn

Author

Clark Kimberling, May 03 2011

Status

approved