

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
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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=(75^(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



