cp's OEIS Frontend

A298307 Start with a(0) = 1 and add at step n >= 0 the term 1 at position floor(4*(n+a(n))/3).

%I A298307 #27 Mar 16 2025 03:19:42
%S A298307 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 A298307 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 A298307 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 A298307 Start with a(0) = 1 and add at step n >= 0 the term 1 at position floor(4*(n+a(n))/3).
%C A298307 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 A298307 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 A298307                                  Sequence b after b(k) is
%e A298307 n b(n) k=floor(4*(n+b(n))/3)         incremented by 1
%e A298307 = ==== =====================  ===============================
%e A298307                               {1,0,0,0,0,0,0,0,0,0,0,0,0,...}
%e A298307 0   1  floor(4*(0+1)/3) =  1  {1,1,0,0,0,0,0,0,0,0,0,0,0,...}
%e A298307 1   1  floor(4*(1+1)/3) =  2  {1,1,1,0,0,0,0,0,0,0,0,0,0,...}
%e A298307 2   1  floor(4*(2+1)/3) =  4  {1,1,1,0,1,0,0,0,0,0,0,0,0,...}
%e A298307 3   0  floor(4*(3+0)/3) =  4  {1,1,1,0,2,0,0,0,0,0,0,0,0,...}
%e A298307 4   2  floor(4*(4+2)/3) =  8  {1,1,1,0,2,0,0,0,1,0,0,0,0,...}
%e A298307 5   0  floor(4*(5+0)/3) =  6  {1,1,1,0,2,0,1,0,1,0,0,0,0,...}
%e A298307 6   1  floor(4*(6+1)/3) =  9  {1,1,1,0,2,0,1,0,1,1,0,0,0,...}
%e A298307 7   0  floor(4*(7+0)/3) =  9  {1,1,1,0,2,0,1,0,1,2,0,0,0,...}
%e A298307 8   1  floor(4*(8+1)/3) = 12  {1,1,1,0,2,0,1,0,1,2,0,0,1,...}
%t A298307 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 A298307 Cf. A136119.
%K A298307 nonn
%O A298307 0,5
%A A298307 _Ctibor O. Zizka_, Jan 16 2018