A298307 - OEIS

%I #24 Jan 30 2018 21:27:40

%S 1,1,1,0,2,0,1,0,1,2,0,0,1,1,2,0,0,1,1,0,1,2,0,0,1,2,0,0,1,1,2,0,0,1,

%T 1,0,2,0,1,0,1,1,2,0,0,1,2,0,0,1,1,0,2,0,1,0,1,1,2,0,0,1,1,0,2,0,1,0,

%U 2,0,1,0,1,2,0,0,1,1,1,0,2,0,1,0,2,0,1,0,1,2,0,0,1,1,2,0,0,1,1,0,2,0,1,0,1,2,0,0,1

%N Start with a(0) = 1 and add at step n >= 0 the term 1 at position floor(4*(n+a(n))/3).

%C Sum_{i=0..n} a(i)/n = 3/4. For sequences of the type:  a(0) = 1, in step n >= 0 add the term 1 at position floor(k*(n+a(n)), k rational number > 1 we have Sum_{i=0..n} a(i)/n = 1/k.

%e Define a sequence b whose terms are initially b(0)=1 and, for n > 0, b(n)=0, i.e., b = {1,0,0,0,0,0,0,0,0,...}; then, for n = 0,1,2,..., increment b(floor(4*(n+b(n))/3)) by 1. For n >= 0, a(n) is the final value of b(n).

%e                                  Sequence b after b(k) is

%e n b(n) k=floor(4*(n+b(n))/3)         incremented by 1

%e = ==== =====================  ===============================

%e                               {1,0,0,0,0,0,0,0,0,0,0,0,0,...}

%e 0   1  floor(4*(0+1)/3) =  1  {1,1,0,0,0,0,0,0,0,0,0,0,0,...}

%e 1   1  floor(4*(1+1)/3) =  2  {1,1,1,0,0,0,0,0,0,0,0,0,0,...}

%e 2   1  floor(4*(2+1)/3) =  4  {1,1,1,0,1,0,0,0,0,0,0,0,0,...}

%e 3   0  floor(4*(3+0)/3) =  4  {1,1,1,0,2,0,0,0,0,0,0,0,0,...}

%e 4   2  floor(4*(4+2)/3) =  8  {1,1,1,0,2,0,0,0,1,0,0,0,0,...}

%e 5   0  floor(4*(5+0)/3) =  6  {1,1,1,0,2,0,1,0,1,0,0,0,0,...}

%e 6   1  floor(4*(6+1)/3) =  9  {1,1,1,0,2,0,1,0,1,1,0,0,0,...}

%e 7   0  floor(4*(7+0)/3) =  9  {1,1,1,0,2,0,1,0,1,2,0,0,0,...}

%e 8   1  floor(4*(8+1)/3) = 12  {1,1,1,0,2,0,1,0,1,2,0,0,1,...}

%t mx = 104; t = Join[{1}, 0 Range@mx]; k = 1; While[4 k < 3 (mx + 2), t[[ Floor[ 4(k + t[[k]])/3]]]++; k++]; Join[{1}, t] (* _Robert G. Wilson v_, Jan 18 2018 *)

%Y Cf. A136119.

%K nonn

%O 0,5

%A _Ctibor O. Zizka_, Jan 16 2018