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A190082 n+[ns/r]+[nt/r]; r=1, s=cos(2pi/5), t=sec(2pi/5). 3
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 60, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 117, 120, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 177, 180, 183, 185, 188, 191, 194, 197, 200, 203, 206, 209, 212, 215, 218 (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=cos(2pi/5), t=sec(2pi/5) gives

a=A190082, b=A190083, c=A190084.

LINKS

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

FORMULA

A190082:  a(n)=n+[n*cos(2pi/5)]+[n*sec(2pi/5].

A190083:  b(n)=n+[n*sec(2pi/5)]+[n*(sec(2pi/5))^2].

A190084:  c(n)=n+[n*cos(2pi/5)]+[n*(cos(2pi/5))^2].

MATHEMATICA

r=1; s=Cos[2Pi/5]; t=Sec[2Pi/5];

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}]  (*A190082*)

Table[b[n], {n, 1, 120}]  (*A190083*)

Table[c[n], {n, 1, 120}]  (*A190084*)

CROSSREFS

Cf. A190083, A190084.

Sequence in context: A189934 A189386 A016789 * A165334 A189512 A190361

Adjacent sequences:  A190079 A190080 A190081 * A190083 A190084 A190085

KEYWORD

nonn

AUTHOR

Clark Kimberling, May 04 2011

STATUS

approved

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Last modified August 20 06:32 EDT 2017. Contains 290824 sequences.