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a(n) = [n*r]-[k*r]-[n*r-k*r], where r=sqrt(2), k=3, [ ]=floor.
3

%I #10 Sep 08 2022 08:45:56

%S 0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,

%T 0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,

%U 0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1

%N a(n) = [n*r]-[k*r]-[n*r-k*r], where r=sqrt(2), k=3, [ ]=floor.

%C See A188014.

%H G. C. Greubel, <a href="/A188041/b188041.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = [n*r]-[3*r]-[n*r-3*r], where r=sqrt(2).

%t r=2^(1/2); k=3;

%t t=Table[Floor[n*r]-Floor[(n-k)*r]-Floor[k*r],{n,1,220}] (*A188041*)

%t Flatten[Position[t,0]] (*A188042*)

%t Flatten[Position[t,1]] (*A188043*)

%o (PARI) for(n=1, 50, print1(floor(n*sqrt(2)) - floor(3*sqrt(2)) - floor((n-3)*sqrt(2)), ", ")) \\ _G. C. Greubel_, Apr 10 2018

%o (Magma) [Floor(n*Sqrt(2)) - Floor(3*Sqrt(2)) - Floor((n-3)*Sqrt(2)): n in [1..50]]; // _G. C. Greubel_, Apr 10 2018

%Y Cf. A188014, A187969.

%K nonn

%O 1

%A _Clark Kimberling_, Mar 19 2011