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A189929
a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(2pi/5), t=cos(2pi/5).
4
1, 3, 5, 8, 10, 12, 15, 17, 19, 22, 24, 26, 29, 31, 33, 35, 38, 40, 42, 45, 46, 48, 51, 53, 55, 58, 60, 62, 64, 67, 69, 71, 74, 76, 78, 81, 83, 85, 88, 90, 91, 93, 96, 98, 100, 103, 105, 107, 110, 112, 114, 117, 119, 121, 123, 126, 128, 130, 133, 135, 137, 139, 141, 143, 146, 148, 150, 153, 155, 157, 159, 162, 164, 166, 169, 171, 173, 176
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=1, s=sin(2pi/5), t=cos(2pi/5) gives
sin(2pi/5)=sqrt((5+sqrt(5))/8); cos(2pi/5)=(-1+sqrt(5))/4.
LINKS
FORMULA
A189929: a(n) = n + [n*sin(2*pi/5)] + [n*cos(2*pi/5)].
A189930: b(n) = n + [n*csc(2*pi/5)] + [n*cot(2*pi/5)].
A189931: c(n) = n + [n*sec(2*pi/5)] + [n*tan(2*pi/5)].
MATHEMATICA
r=1; s=Sin[2Pi/5]; t=Cos[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}] (*A189929*)
Table[b[n], {n, 1, 120}] (*A189930*)
Table[c[n], {n, 1, 120}] (*A189931*)
PROG
(PARI) for(n=1, 100, print1(n + floor(n*sin(2*Pi/5)) + floor(n*cos(2*Pi/5)), ", ")) \\ G. C. Greubel, Jan 13 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(2*Pi(C)/5)) + Floor(n*Cos(2*Pi(C)/5)): n in [1..100]]; // G. C. Greubel, Jan 13 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 01 2011
STATUS
approved