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Oh, neat, I didn't know about Descartes' theorem!
So you're calculating
k_4 = k_1 + k_2 + k_3 + 2*\sqrt{k_1*k_2 + k_2*k_3 + k_3*k_1}
where curvature k_n is \frac{1}{r_n} (the inverse of the radius).
If I pick O, O' and O_1 to calculate O_2
1 + 1 + 4 + 2*\sqrt{1*1 + 1*4 + 4*1} = 12
Now we can simply repeat with O, O' and O_2 to calculate O_3
1 + 1 + 12 + 2*\sqrt{1*1 + 1*12 + 12*1} = 24
So radius \frac{1}{24} or diameter \frac{1}{12}
And I guess a direct formula for circle O_n could be
d(n) = \frac{1}{n(n+1)}.
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Yes, exactly.
a fractal generated by starting with a triple of circles, each tangent to the other two, and successively filling in more circles, each tangent to another three.
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