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 A182760 Beatty sequence for (3 + 5^(-1/2))/2. 19
 1, 3, 5, 6, 8, 10, 12, 13, 15, 17, 18, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 49, 51, 53, 55, 56, 58, 60, 62, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 81, 82, 84, 86, 87, 89, 91, 93, 94, 96, 98, 99, 101, 103, 105, 106, 108, 110 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A182761 (the complement of A182760), A242671 Sequence in context: A082977 A000210 A329829 * A292646 A022838 A329841 Adjacent sequences:  A182757 A182758 A182759 * A182761 A182762 A182763 KEYWORD nonn AUTHOR Clark Kimberling, Nov 28 2010 STATUS approved

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Last modified January 22 01:28 EST 2022. Contains 350481 sequences. (Running on oeis4.)