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COMMENTS
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To build the sequence, start from:
1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
At the n-th 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 n=1,2,4,6,10,... which are the smallest number of stones in Mancala solitaire which make use of the n-th hole. If f(k) denotes this sequence then lim_{k->oo} k^2/f(k) = Pi.
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
Another way (less self-referent) to construct the sequence.
Step 1: Let's start from the integers separated by a hole:
1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
Step 2: Put integers in the holes leaving 2 holes between each integer giving:
1,*1*,2,_,3,_,4,*2*,5,_,6,_,7,*3*,8,_,9,_,10,*4*,11,_,12,_,...
Step 3: Put integers in the holes leaving 3 holes between each integer giving:
1,1,2,*1*,3,_,4,2,5,_,6,_,7,3,8,*2*,9,_,10,4,11,_,12,_,...
Step 4: Put integers in the holes leaving 4 holes between each integer giving:
1,1,2,1,3,*1*,4,2,5,_,6,_,7,3,8,2,9,_,10,4,11,_,12,*2*,...
Iterating the process yields the sequence
1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 1, 7, 3, 8, 2, 9, 1, 10, 4, 11, 1, 12, 2,... (End)
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