

A190059


a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(pi/5), t=csc(pi/5).


3



2, 6, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42, 45, 48, 52, 54, 58, 62, 65, 68, 71, 75, 78, 81, 85, 87, 91, 95, 98, 101, 104, 108, 110, 114, 118, 120, 124, 127, 131, 134, 137, 141, 143, 147, 151, 153, 157, 160, 164, 166, 170, 174, 176, 180, 183, 186, 190, 193, 197, 199, 203, 207, 209, 213, 216, 219, 222, 226, 230
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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=sin(pi/5), t=csc(pi/5) gives
a=A190059, b=A190060, c=A190061.


LINKS

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


FORMULA

A190059: a(n) = n + [n*sin(pi/5)] + [n*csc(pi/5].
A190060: b(n) = n + [n*csc(pi/5)] + [n*(csc(pi/5))^2].
A190061: c(n) = n + [n*sin(pi/5)] + [n*(sin(pi/5))^2].


MATHEMATICA

r=1; s=Sin[Pi/5]; t=Csc[Pi/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}] (*A190059*)
Table[b[n], {n, 1, 120}] (*A190060*)
Table[c[n], {n, 1, 120}] (*A190061*)


PROG

(PARI) for(n=1, 30, print1(n + floor(n*sin(Pi/5)) + floor(n/sin(Pi/5)), ", ")) \\ G. C. Greubel, Jan 10 2018
(MAGMA) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/5)) + Floor(n/Sin(Pi(C)/5)): n in [1..30]]; // G. C. Greubel, Jan 10 2018


CROSSREFS

Cf. A190060, A190061.
Sequence in context: A189752 A206813 A189371 * A190332 A187912 A186500
Adjacent sequences: A190056 A190057 A190058 * A190060 A190061 A190062


KEYWORD

nonn


AUTHOR

Clark Kimberling, May 04 2011


STATUS

approved



