OFFSET
1,2
COMMENTS
Suppose that u and v are positive real numbers for which the sets S(u)={ju} and S(v)={kv}, for j>=1 and k>=1, are disjoint. Let a(n) be the position of nu when the numbers in S(u) and S(v) are jointly ranked. Then, as is easy to prove, a is the Beatty sequence of the number r=1+u/v, and the complement of a is the Beatty sequence of s=1+v/u. For A182760, take u = golden ratio = (1+sqrt(5))/2 and v=sqrt(5), so that r=(3+5^(-1/2))/2 and s=(7-5^(-1/2)/2.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = floor(r*n), where r = (3 + 5^(-1/2))/2 = 1.72360...
EXAMPLE
Let u=(1+sqrt(5))/2 and v=sqrt(5). When the numbers ju and kv are jointly ranked, we write U for numbers of the form ju and V for the others. Then the ordering of the ranked numbers is given by U V U V U U V U V U V U U .. The positions of U are given by A182760.
MATHEMATICA
Table[Floor[Sqrt[n/20]+3*n/2], {n, 1, 100}] (* G. C. Greubel, Jan 11 2018 *)
PROG
(Magma) [Floor(n*(3+5^(-1/2))/2): n in [1..70]]; // Vincenzo Librandi, Oct 25 2011
(PARI) a(n)=floor(sqrt(n/20)+3*n/2) \\ Charles R Greathouse IV, Jul 02 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 28 2010
STATUS
approved