%I
%S 0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,
%T 0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,
%U 0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0
%N a(n) = floor(nr)  1  floor((n1)r), where r = sqrt(2).
%C Is this A159684 with an additional 0 in front?  _R. J. Mathar_, Mar 20 2011
%C The answer is yes, since it follows right from the definitions of the sequences that (a(n)) is equal to A159684 with a different offset.  _Michel Dekking_, Jan 31 2017
%H Vincenzo Librandi, <a href="/A188037/b188037.txt">Table of n, a(n) for n = 1..5000</a>
%H Heinz H. Bauschke, Minh N. Dao, Scott B. Lindstrom, <a href="https://arxiv.org/abs/1804.08880">The DouglasRachford algorithm for a hyperplane and a doubleton</a>, arXiv:1804.08880 [math.OC], 2018.
%F a(n) = floor(nr)  floor(r)  floor(nr  r), where r = sqrt(2).
%t r=2^(1/2)); k=1;
%t t=Table[Floor[n*r]Floor[(nk)*r]Floor[k*r],{n,1,220}]
%t Table[Floor[n Sqrt[2]]  Floor[Sqrt[2]]  Floor[n Sqrt[2]  Sqrt[2]], {n, 100}] (* _Vincenzo Librandi_, Jan 31 2017 *)
%o (MAGMA) [Floor(n*Sqrt(2))Floor(Sqrt(2))Floor(n*Sqrt(2) Sqrt(2)): n in [1..100]]; // _Vincenzo Librandi_, Jan 31 2017
%o (PARI) a(n) = floor(n*sqrt(2))1floor((n1)*sqrt(2)) \\ _Felix FrÃ¶hlich_, Jan 31 2017
%Y Cf. A080754, A083088, A188014, A188037, A188038.
%Y A159684 is an essentially identical sequence.
%K nonn
%O 1
%A _Clark Kimberling_, Mar 19 2011
