A182760 Beatty sequence for (3 + 5^(-1/2))/2.

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, 112, 113, 115, 117, 118, 120, 122, 124, 125, 127, 129, 130, 132, 134, 136, 137, 139, 141, 143, 144, 146, 148, 149, 151, 153, 155, 156, 158, 160, 162, 163, 165, 167, 168

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 * A379805 A292646 A022838

Adjacent sequences: A182757 A182758 A182759 * A182761 A182762 A182763

Keyword

nonn,changed

Author

Clark Kimberling, Nov 28 2010

Extensions

More than the usual number of terms are shown in order to distinguish this from a very similar sequence. - N. J. A. Sloane, Jan 20 2025

Status

approved