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A189999 n+[ns/r]+[nt/r]; r=1, s=sinh(1), t=cosh(1). 4
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 (list; graph; refs; listen; history; text; internal format)
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+[ns/r]+[nt/r],

b(n)=n+[nr/s]+[nt/s],

c(n)=n+[nr/t]+[ns/t], where []=floor.

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

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

LINKS

Table of n, a(n) for n=1..73.

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*)

CROSSREFS

Cf. A190000, A190001, A190002.

Sequence in context: A047355 A248522 A098005 * A171983 A003231 A189460

Adjacent sequences:  A189996 A189997 A189998 * A190000 A190001 A190002

KEYWORD

nonn

AUTHOR

Clark Kimberling, May 03 2011

STATUS

approved

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Last modified October 21 12:41 EDT 2014. Contains 248377 sequences.