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Positions of 1 in A188395; complement of A188396.
3

%I #12 Nov 22 2018 18:18:30

%S 1,2,4,5,6,7,8,9,11,12,13,14,15,16,18,19,21,22,23,24,25,26,28,29,30,

%T 31,32,33,35,36,38,39,40,41,42,43,45,46,47,48,49,50,52,53,54,55,56,57,

%U 59,60,62,63,64,65,66,67,69,70,71,72,73,74,76,77,79,80,81,82,83,84,86,87,88,89,90,91,93,94,95,96,97,98,100,101,103,104,105,106,107,108,110,111,112

%N Positions of 1 in A188395; complement of A188396.

%C See A187950, A188395.

%H G. C. Greubel, <a href="/A188397/b188397.txt">Table of n, a(n) for n = 1..10000</a>

%t r=2^(-1/2); k=4;

%t t=Table[Floor[n*r+k*r]-Floor[n*r]-Floor[k*r], {n,1,220}] (* A188395 *)

%t Flatten[Position[t,0] ] (* A188396 *)

%t Flatten[Position[t,1] ] (* A188397 *)

%o (PARI) lista(nn) = Vec(select(x->x==1, vector(nn, n, floor((n+4)/sqrt(2)) - floor(n/sqrt(2)) - floor(4/sqrt(2))), 1)); \\ _Michel Marcus_, Apr 26 2018

%Y Cf. A187950, A188395, A188396.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 30 2011