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Positions of 0 in A188011; complement of A188013.
6

%I #19 Sep 08 2022 08:45:56

%S 3,8,16,21,29,37,42,50,55,63,71,76,84,92,97,105,110,118,126,131,139,

%T 144,152,160,165,173,181,186,194,199,207,215,220,228,236,241,249,254,

%U 262,270,275,283,288,296,304,309,317,325,330,338,343,351,359,364,372,377,385,393,398,406,414,419,427,432,440

%N Positions of 0 in A188011; complement of A188013.

%C See A188014 and A188011.

%H G. C. Greubel, <a href="/A188012/b188012.txt">Table of n, a(n) for n = 1..5000</a>

%F a(n+1) = 3*floor(n*phi)+2*n+3 for n>=0, where phi = (1+sqrt(5))/2 (see A188011). - _Michel Dekking_, Sep 28 2017

%t r=(1+5^(1/2))/2; k=3;

%t t=Table[Floor[n*r]-Floor[(n-k)*r]-Floor[k*r],{n,1,220}] (* A188011 *)

%t Flatten[Position[t,0]] (* A188012 *)

%t Flatten[Position[t,1]] (* A188013 *)

%t Table[3*Floor[(n-1)*GoldenRatio] + 2*n + 1, {n, 1, 65}] (* _G. C. Greubel_, Nov 22 2018 *)

%o (PARI) vector(65, n, 3*floor((n-1)*(1+sqrt(5))/2)+2*n+1) \\ _G. C. Greubel_, Nov 22 2018

%o (Magma) [3*Floor((n-1)*(1+Sqrt(5))/2)+2*n+1: n in [1..65]]; // _G. C. Greubel_, Nov 22 2018

%o (Sage) [3*floor((n-1)*(1+sqrt(5))/2)+2*n+1 for n in (1..65)] # _G. C. Greubel_, Nov 22 2018

%Y Cf. A188011, A188013, A188014.

%K nonn

%O 1,1

%A _Clark Kimberling_, Mar 19 2011