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

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, 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, 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

Offset: 0

Author

Ctibor O. Zizka, Jan 16 2018

Keywords

• nonn

Comments

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.

Examples

			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).
                                 Sequence b after b(k) is
n b(n) k=floor(4*(n+b(n))/3)         incremented by 1
= ==== =====================  ===============================
                              {1,0,0,0,0,0,0,0,0,0,0,0,0,...}
0   1  floor(4*(0+1)/3) =  1  {1,1,0,0,0,0,0,0,0,0,0,0,0,...}
1   1  floor(4*(1+1)/3) =  2  {1,1,1,0,0,0,0,0,0,0,0,0,0,...}
2   1  floor(4*(2+1)/3) =  4  {1,1,1,0,1,0,0,0,0,0,0,0,0,...}
3   0  floor(4*(3+0)/3) =  4  {1,1,1,0,2,0,0,0,0,0,0,0,0,...}
4   2  floor(4*(4+2)/3) =  8  {1,1,1,0,2,0,0,0,1,0,0,0,0,...}
5   0  floor(4*(5+0)/3) =  6  {1,1,1,0,2,0,1,0,1,0,0,0,0,...}
6   1  floor(4*(6+1)/3) =  9  {1,1,1,0,2,0,1,0,1,1,0,0,0,...}
7   0  floor(4*(7+0)/3) =  9  {1,1,1,0,2,0,1,0,1,2,0,0,0,...}
8   1  floor(4*(8+1)/3) = 12  {1,1,1,0,2,0,1,0,1,2,0,0,1,...}
		

Crossrefs

Cf. A136119.

Programs

Mathematica

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 *)