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A190053
a(n) = n + [n*s/r] + [n*t/r]; r=2, s=sin(Pi/3), t=csc(Pi/3).
4
1, 3, 5, 7, 9, 11, 14, 15, 17, 19, 21, 23, 25, 28, 29, 31, 33, 35, 37, 39, 42, 43, 45, 47, 49, 52, 53, 56, 57, 59, 61, 63, 66, 67, 70, 71, 74, 75, 77, 80, 81, 84, 85, 88, 89, 91, 94, 95, 98, 99, 102, 104, 105, 108, 109, 112, 113, 116, 118, 119, 122, 123, 126, 127, 130, 132, 134, 136, 137, 140, 141, 144, 146, 148, 150, 151, 154, 156, 158, 160
OFFSET
1,2
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=2, s=sin(Pi/3), t=csc(Pi/3) gives a=A190053, b=A190054, c=A190055.
LINKS
FORMULA
A190053: a(n) = n + [(n/2)*sin(Pi/3)] + [(n/2)*csc(Pi/3)].
A190054: b(n) = n + [2n*csc(Pi/3)] + [n*(csc(Pi/3))^2].
A190055: c(n) = n + [2n*sin(Pi/3)] + [n*(sin(Pi/3))^2].
MATHEMATICA
r=2; s=Sin[Pi/3]; t=Csc[Pi/3];
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}] (* A190053 *)
Table[b[n], {n, 1, 120}] (* A190054 *)
Table[c[n], {n, 1, 120}] (* A190055 *)
PROG
(PARI) for(n=1, 100, print1(n + floor(n*sin(Pi/3)/2) + floor(n/(2*sin(Pi/3))), ", ")) \\ G. C. Greubel, Jan 10 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/3)/2) + Floor(n/(2*Sin(Pi(C)/3))): n in [1..100]]; // G. C. Greubel, Jan 10 2018
CROSSREFS
Sequence in context: A118820 A117521 A376602 * A327209 A285590 A376957
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 04 2011
STATUS
approved