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%I #22 Nov 17 2016 23:37:40
%S 1,2,66,1314,26082,517698,10275714,203961186,4048396578,80356048002,
%T 1594975770306,31658447262114,628384017931362,12472705016840898,
%U 247568948283023874,4913960850609954786,97536510167350024098,1935988320795170617602,38427156885401362279746,762735172745641733742114
%N a(n) is the ratio of the sum of squares of the bends of the circles that are added in the n-th generation of Apollonian packing, to the sum of squares of the bends of the initial three circles.
%C For more references and links, see A189226.
%H Colin Barker, <a href="/A154637/b154637.txt">Table of n, a(n) for n = 0..750</a>
%H Colin Mallows, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Mallows/mallows8.html">Growing Apollonian packings</a>, J. Integer Sequences v.12, article 09.2.1 (2009).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-3).
%F G.f.: (1-18*x+29*x^2) / (1-20*x+3*x^2).
%F From _Colin Barker_, Nov 16 2016: (Start)
%F a(n) = ((133-13*sqrt(97))*(10+sqrt(97))^n - (10-sqrt(97))^n*(133+13*sqrt(97))) / (3*sqrt(97)) for n>0.
%F a(n) = 20*a(n-1) - 3*a(n-2) for n>2.
%F (End)
%e Starting with three circles with bends -1,2,2, the ssq is 9. The first derived generation has two circles, each with bend 3. So a(1) = (9+9)/9 = 2.
%t CoefficientList[Series[(29 z^2 - 18 z + 1)/(3 z^2 - 20 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{20, -3}, {1, 2, 66}, 100] (* _Vladimir Joseph Stephan Orlovsky_, Jul 03 2011 *)
%o (PARI) Vec((1-18*x+29*x^2)/(1-20*x+3*x^2) + O(x^30)) \\ _Colin Barker_, Nov 16 2016
%Y For starting with four circles, see A137246. For sums of bends, see A135849 and A154636. For three dimensions, see A154638 - A154645.
%Y Cf. also A189226, A189227.
%K easy,nonn
%O 0,2
%A _Colin Mallows_, Jan 13 2009
%E More terms from _N. J. A. Sloane_, Nov 22 2009
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