|
%I #26 Sep 08 2022 08:45:55
%S 2,4,6,8,11,13,15,17,20,22,24,26,28,31,33,35,37,40,42,44,46,48,51,53,
%T 55,57,60,62,64,66,68,71,73,75,77,80,82,84,86,88,91,93,95,97,100,102,
%U 104,106,109,111,113,115,117,120,122,124,126,129,131,133,135
%N a(n) = n + floor(sqrt(3/2)*n).
%C This is the Beatty sequence for 1 + sqrt(3/2).
%C Also, a(n) is the position of 3*n^2 in the sequence obtained by arranging all the numbers in the sets {2*h^2, h >= 1} and {3*k^2, k >= 1} in increasing order. - _Clark Kimberling_, Oct 20 2014
%C Also, numbers n such that floor((n+1)*sqrt(6)) - floor(n*sqrt(6)) = 3. - _Clark Kimberling_, Jul 15 2015
%H Clark Kimberling, <a href="/A184809/b184809.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = n + floor(r*n), where r = sqrt(3/2).
%t (See A184808.)
%o (Magma) [n + Floor(n*Sqrt(3/2)): n in [1..70]]; // _Vincenzo Librandi_, Oct 23 2014
%o (PARI) main(size)={return(vector(size,n,n+floor(sqrt(3/2)*n)))}/* _Anders Hellström_, Jul 15 2015 */
%Y Complement of A184808.
%Y Cf. A184811.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 22 2011
|
