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

%I A182760 #27 Feb 11 2025 01:31:00
%S A182760 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 A182760 43,44,46,48,49,51,53,55,56,58,60,62,63,65,67,68,70,72,74,75,77,79,81,
%U A182760 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
%N A182760 Beatty sequence for (3 + 5^(-1/2))/2.
%C A182760 Suppose that u and v are positive real numbers for which the sets S(u)={j*u} and S(v)={k*v}, for j>=1 and k>=1, are disjoint. Let a(n) be the position of n*u 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-sqrt(5))/2.
%H A182760 Vincenzo Librandi, <a href="/A182760/b182760.txt">Table of n, a(n) for n = 1..10000</a>
%F A182760 a(n) = floor(r*n), where r = (3 + 5^(-1/2))/2 = 1.72360...
%e A182760 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 A182760 Table[Floor[Sqrt[n/20]+3*n/2], {n,1,100}] (* _G. C. Greubel_, Jan 11 2018 *)
%o A182760 (Magma) [Floor(n*(3+5^(-1/2))/2): n in [1..70]]; // _Vincenzo Librandi_, Oct 25 2011
%o A182760 (PARI) a(n)=floor(sqrt(n/20)+3*n/2) \\ _Charles R Greathouse IV_, Jul 02 2013
%Y A182760 Cf. A182761 (the complement of A182760), A242671
%K A182760 nonn,easy
%O A182760 1,2
%A A182760 _Clark Kimberling_, Nov 28 2010
%E A182760 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