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Positions of 1 in A190483.
4

%I #20 May 06 2021 22:12:20

%S 1,4,6,8,9,11,13,16,18,21,23,25,26,28,30,33,35,37,38,40,42,45,47,49,

%T 50,52,54,55,57,59,62,64,66,67,69,71,74,76,78,79,81,83,86,88,91,93,95,

%U 96,98,100,103,105,107,108,110,112,115,117,120,122,124,125,127,129,132,134,136,137,139,141,144,146,148,149,151,153,154,156,158

%N Positions of 1 in A190483.

%C See A190483.

%H G. C. Greubel, <a href="/A190485/b190485.txt">Table of n, a(n) for n = 1..1000</a>

%t r = Sqrt[2]; b = 2; c = 1;

%t f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];

%t t = Table[f[n], {n, 1, 200}] (* A190483 *)

%t Flatten[Position[t, 0]] (* A190484 *)

%t Flatten[Position[t, 1]] (* A190485 *)

%t Flatten[Position[t, 2]] (* A190486 *)

%o (Python)

%o from sympy import sqrt, floor

%o r=sqrt(2)

%o def a190483(n): return floor((2*n + 1)*r) - 2*floor(n*r) - floor(r)

%o print([n for n in range(1, 501) if a190483(n)==1]) # _Indranil Ghosh_, Jul 02 2017

%Y Cf. A190483, A190485, A190486.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 11 2011