A182760 - OEIS

%I #19 Sep 08 2022 08:45:55

%S 1,3,5,6,8,10,12,13,15,17,18,20,22,24,25,27,29,31,32,34,36,37,39,41,

%T 43,44,46,48,49,51,53,55,56,58,60,62,63,65,67,68,70,72,74,75,77,79,81,

%U 82,84,86,87,89,91,93,94,96,98,99,101,103,105,106,108,110

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

%C 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.

%H Vincenzo Librandi, <a href="/A182760/b182760.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = floor(r*n), where r = (3 + 5^(-1/2))/2 = 1.72360...

%e 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.

%t Table[Floor[Sqrt[n/20]+3*n/2], {n,1,100}] (* _G. C. Greubel_, Jan 11 2018 *)

%o (Magma) [Floor(n*(3+5^(-1/2))/2): n in [1..70]]; // _Vincenzo Librandi_, Oct 25 2011

%o (PARI) a(n)=floor(sqrt(n/20)+3*n/2) \\ _Charles R Greathouse IV_, Jul 02 2013

%Y Cf. A182761 (the complement of A182760), A242671

%K nonn

%O 1,2

%A _Clark Kimberling_, Nov 28 2010