A137246 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.
1, 17, 339, 6729, 133563, 2651073, 52620771, 1044462201, 20731381707, 411494247537, 8167690805619, 162119333369769, 3217883594978523, 63871313899461153, 1267772627204287491, 25163838602387366361, 499473454166134464747, 9913977567515527195857
Offset: 1
Examples
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).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200 [a(188) corrected by _Georg Fischer_, May 24 2019]
- J. C. Lagarias, C. L. Mallows, and Allan Wilks, Beyond the Descartes Circle Theorem, arXiv:math/0101066 [math.MG], 2001.
- J. C. Lagarias, C. L. Mallows, and Allan Wilks, Beyond the Descartes Circle Theorem, Amer. Math Monthly, 109 (2002), 338-361.
- C. L. Mallows, Growing Apollonian Packings, J. Integer Sequences, 12 (2009), article 09.2.1, page 3.
- Index entries for linear recurrences with constant coefficients, signature (20,-3).
Programs
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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
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Magma
R
:=PowerSeriesRing(Integers(), 30); Coefficients(R!(x*(1-x)*(1-2*x)/(1-20*x+3*x^2))); // Bruno Berselli, Jul 04 2011 -
Mathematica
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 *) -
PARI
Vec(x*(1-2*x)*(1-x)/(1-20*x+3*x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 03 2011
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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
Formula
For n >= 4, a(n) = 20*a(n-1) - 3*a(n-2).
O.g.f.: x*(1-x)*(1-2*x)/(1-20*x+3*x^2). - R. J. Mathar, Mar 31 2008
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
Comments