
COMMENTS

To built the sequence, start from:
1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
At nth step use the rule: " fill a(n)th hole with a(n) " (holes are numbered from 1 at each step)
So step 1 is "fill first hole with 1" giving:
1,1,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
Since a(2)=1 step 2 is still "fill first hole with 1" giving:
1,1,2,1,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
Since a(3)=2 step 3 is "fill second hole with 2" giving:
1,1,2,1,3,_,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
Since a(4)=1 step 4 is "fill first hole with 1" giving:
1,1,2,1,3,1,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
Since a(5)=3 step 5 is "fill third hole with 3" giving:
1,1,2,1,3,1,4,2,5,_,6,_,7,3,8,_,9,_,10,_,11,_,12,_,...
Iterating the process indefinitely yields:
1,1,2,1,3,1,4,2,5,1,6,1,7,3,8,2,9,1,10,4,11,1,12,2,13,5,..
Indices where 1's occur are given by n=1,2,4,6,10,... which are the smallest number of stones in Mancala solitaire which make use of nth hole. If f(k) denotes this sequence k^2/f(k)>pi as k>infty.
Ordinal transform of A028920  Benoit Cloitre, Aug 03 2007
Although A028920 and A130747 are not fractal sequences (according to Kimberling's definition) we say they are "mutual fractal sequences" since the ordinal transform of one gives the other.  Benoit Cloitre, Aug 03 2007
a(A002491(n)) = 1. [From Reinhard Zumkeller, Jun 23 2009]
A082447(n) = number of ones <= n. [From Reinhard Zumkeller, Jul 01 2009]
