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Let $r_1,r_2\dots$ be the radii of Apollonian gasket. I would like to know for which values $\alpha$ we have $$\sum_{n=1}^\infty r_n^\alpha<\infty.$$

Integral Apollonian circle packing defined by circle curvatures of (−1, 2, 2, 3)

I know that if three circles $A$, $B$ and $C$ are tangent to two circles $D_1$ and $D_2$ then $$d_1+d_2=2(a+b+c),$$ where $a$, $b$, $c$, $d_1$ and $d_2$ denote the curvatures of the corresponding circle. (For example, on the picture, $3+3=2(2+2-1)$.)

In principle, it gives a recursive formula for $r_n$, but I was not able to figure out how to use it.

Motivation: I would like to know if one can cover whole measure of a square with countable number of disjoint open discs with radii $r_1,r_2\dots$ such that $$\sum_{n=1}^\infty r_n^\alpha<\infty$$ for all $\alpha>1$. By some reason I believe that Apollonian gasket is optimal in this sense; at least it worth to check it.

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For what its worth, while in principle one has a recursive formula for the curvatures, you are not alone in being unable to effectively use it. My advisor in grad school spent some time looking at the question of 'given starting curvatures of a gasket what curvatures can/cannot appear in the gasket?' without doing an exhaustive depth search. As far as I know this question is still open (the best one can do is find some modular restrictions), so short of finding a better answer, you probably wouldn't be able to calculate the critical $\alpha$ without a computer like in Gerald's answer. –  ARupinski Jan 17 '13 at 3:56
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1 Answer

up vote 6 down vote accepted

This critical value is $\alpha_0 \approx 1.3056867$ ... For $\alpha > \alpha_0$ the series converges, for $\alpha < \alpha_0$ it diverges.

Before the inexpensive computer, it was difficult to tell whether the critical value is ${}> 1$ or not.

Boyd, David W. The sequence of radii of the Apollonian packing. Math. Comp. 39 (1982), no. 159, 249–254.



mentioned in the comments... arbitrarily packed disks, not necessarily touching as in Apollonian packings. The critical value (= dimension of the residual set) is shown to be ${}> 1.02$.

Larman, D. G. On the Besicovitch dimension of the residual set of arbitrarily packed disks in the plane. J. London Math. Soc. 42 1967 292–302.


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This number is also the Hausdorff dimension of the complementary set when the disks are removed. It is an open question on what the smallest Hausdorff dimension of a residual set is, but there is a lower bound due to Larman of 1.02? –  Anthony Quas Jan 17 '13 at 3:48
@Anthony: I would be very interested in a citation for both assertions. –  Rodrigo A. Pérez Jan 17 '13 at 3:52
I don't have mathscinet access right now so I can't look in the paper, but is that critical value special to the (1,2,2)-gasket in the question or is it independent of the specific set of curvatures in the gasket? –  ARupinski Jan 17 '13 at 3:59
You get the same critical value with different starting curvatures. Maybe it is the 1.02 I was thinking of when I said it was hard to tell if it was ${}>1$. –  Gerald Edgar Jan 17 '13 at 14:58
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