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a(n) is the ratio of the sum of the squares of the bends (curvatures) of the n-th generation of an Apollonian packing to the sum of the squares of the bends of the initial four-circle configuration.
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%I #47 Sep 08 2022 08:45:32

%S 1,17,339,6729,133563,2651073,52620771,1044462201,20731381707,

%T 411494247537,8167690805619,162119333369769,3217883594978523,

%U 63871313899461153,1267772627204287491,25163838602387366361,499473454166134464747,9913977567515527195857

%N a(n) is the ratio of the sum of the squares of the bends (curvatures) of the n-th generation of an Apollonian packing to the sum of the squares of the bends of the initial four-circle configuration.

%C These ratios are independent of the starting configuration. Similar ratios of third and higher moments are not so independent.

%C See A189226 for additional comments, references and links.

%H Vincenzo Librandi, <a href="/A137246/b137246.txt">Table of n, a(n) for n = 1..200</a> [a(188) corrected by _Georg Fischer_, May 24 2019]

%H J. C. Lagarias, C. L. Mallows, and Allan Wilks, <a href="http://arxiv.org/abs/math/0101066"> Beyond the Descartes Circle Theorem</a>, arXiv:math/0101066 [math.MG], 2001.

%H J. C. Lagarias, C. L. Mallows, and Allan Wilks, <a href="http://www.jstor.org/stable/2695498"> Beyond the Descartes Circle Theorem</a>, Amer. Math Monthly, 109 (2002), 338-361.

%H C. L. Mallows, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Mallows/mallows8.html">Growing Apollonian Packings</a>, J. Integer Sequences, 12 (2009), article 09.2.1, page 3.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-3).

%F For n >= 4, a(n) = 20*a(n-1) - 3*a(n-2).

%F O.g.f.: x*(1-x)*(1-2*x)/(1-20*x+3*x^2). - _R. J. Mathar_, Mar 31 2008

%F a(n) = ((41+sqrt(97))*(10+sqrt(97))^(n-1) - (41-sqrt(97))*(10-sqrt(97))^(n-1))/(6*sqrt(97)) for n>1. - _Bruno Berselli_, Jul 04 2011

%e Starting with the configuration with bends (-1,2,2,3) with sum(bends^2) = 18, the next generation contains four circles with bends 3,6,6,15. The sum of their squares is 306 = 18*a(2). The third generation has 12 circles with sum(bends^2) = 6102 = 18*a(3).

%t CoefficientList[Series[(2z^2-3z+1)/(3z^2-20z+1), {z, 0, 30}], z] (* and *) LinearRecurrence[{20, -3}, {1, 17, 339}, 30] (* _Vladimir Joseph Stephan Orlovsky_, Jul 03 2011 *)

%o (PARI) Vec(x*(1-2*x)*(1-x)/(1-20*x+3*x^2)+O(x^30)) \\ _Charles R Greathouse IV_, Jul 03 2011

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(x*(1-x)*(1-2*x)/(1-20*x+3*x^2))); // _Bruno Berselli_, Jul 04 2011

%o (Sage) a=(x*(1-x)*(1-2*x)/(1-20*x+3*x^2)).series(x, 30).coefficients(x, sparse=False); a[1:] # _G. C. Greubel_, May 24 2019

%o (GAP) a:=[1,17,339];; for n in [4..30] do a[n]:=20*a[n-1]-3*a[n-2]; od; a; # _G. C. Greubel_, May 24 2019

%Y Cf. A135849, A105970, A189226, A189227.

%K easy,nonn

%O 1,2

%A _Colin Mallows_, Mar 09 2008