A137246 -id:A137246 - cp's OEIS frontend

A189226 Curvatures in the nickel-dime-quarter Apollonian circle packing, ordered first by generation and then by size.

-11, 21, 24, 28, 40, 52, 61, 157, 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397, 132, 156, 160, 189, 204, 205, 216, 237, 253, 285, 288, 309, 316, 336, 349, 405, 412, 421, 453, 460, 469, 472, 517, 544, 565, 616, 628, 685, 717, 741, 1084, 1093, 1104, 1125, 1128, 1140

Offset: 1

Jonathan Sondow, Apr 18 2011

For a circle, curvature = 1/radius. The curvatures of a quarter, nickel, and dime are approximately proportional to 21, 24, and 28, respectively. Three mutually tangent circles with curvatures 21, 24, 28 can be inscribed in a circle of curvature 11.
Apollonius's and Descartes's Theorems say that, given three mutually tangent circles of curvatures a, b, c, there are exactly two circles tangent to all three, and their curvatures are a + b + c +- 2*sqrt(ab + ac + bc). (Here negative curvature of one of the two circles means that the three circles are inscribed in it.)
Fuchs (2009) says "An Apollonian circle packing ... is made by repeatedly inscribing circles into the triangular interstices in a Descartes configuration of four mutually tangent circles. Remarkably, if the original four circles have integer curvature, all of the circles in the packing will have integer curvature as well." That is because if a + b + c - 2s*qrt(ab + ac + bc) is an integer, then so is a + b + c + 2*sqrt(ab + ac + bc).
For n > 1, the n-th generation of the packing has 4*3^(n-2) circles.
Infinitely many of the curvatures are prime numbers A189227. In fact, in any integral Apollonian circle packing that is primitive (i.e., the curvatures have no common factor), the prime curvatures constitute a positive fraction of all primes (Bourgain 2012) and there are infinitely many pairs of tangent circles both of whose curvatures are prime (Sarnak 2007, 2011).
Fuchs and Sanden (2012) report on experiments with the nickel-dime-quarter Apollonian circle packing, which they call the coins packing P_C.

			The 1st-generation curvatures are -11, 21, 24, 28, the 2nd are 40, 52, 61, 157, and the 3rd are 76, 85, 96, 117, 120, 132, 181, 213, 237, 376, 388, 397. The 4th generation begins 132, 156, 160, 189, 204, 205, 216, ....
As 21 + 24 + 28 +- 2*sqrt(21*24 + 21*28 + 24*28) = 157 or -11, the sequence begins -11, 21, 24, 28, ... and 157 is in it.
The primes 157 and 397 are the curvatures of two circles that are tangent.

D. Austin, When Kissing Involves Trigonometry, AMS feature column March 2006.
J. Bourgain, Integral Apollonian circle packings and prime curvatures, arXiv:1105.5127 [math.NT], 2011-2012.
J. Bourgain and A. Kontorovich, On the Strong Density Conjecture for Integral Apollonian Circle Packings, arXiv:1205.4416 [math.NT], 2012-2013. See figure 1.
S. Butler, R. Graham, G. Guettler and C. Mallows, Irreducible Apollonian configurations and packings, Discrete & Computational Geometry, 44 (2010), 487-507.
E. Fuchs, Arithmetic Properties of Apollonian Circle Packings, Ph.D. thesis 2009.
E. Fuchs and K. Sanden, Some experiments with integral Apollonian circle packings, Experiment. Math. 20 (2011), 380-399.
R. L. Graham, J. C. Lagarias, C. L. Mallows, Allan Wilks, and C. H. Yan, Apollonian Circle Packings: Number Theory, J. Number Theory, 100 (2003), 1-45.
R. L. Graham, J. C. Lagarias, C. L. Mallows, Allan Wilks, and C. H. Yan, Apollonian Circle Packings: Geometry and Group Theory I. The Apollonian Group., Discrete & Computational Geometry, 34 (2005), no. 4, 547-585.
K. E. Hirst, The Apollonian Packing of Circles, J. London Math. Soc. s1-42(1) (1967), 281-291.
E. Kasner and F. Supnick, The Apollonian packing of circles, Proc. Nat. Acad. Sci. U.S.A. 29 (1943), 378-384.
A. Kontorovich, From Apollonius to Zaremba: Local-global phenomena in thin orbits, Bull. Amer. Math. Soc., 50 (2013), 187-228.
J. C. Lagarias, C. L. Mallows, and Allan Wilks, Beyond the Descartes Circle Theorem, Amer. Math Monthly, 109 (2002), 338-361.
L. Levine, W. Pegden, C. K. Smart, Apollonian Structure in the Abelian Sandpile, arXiv:1208.4839 [math.AP], 2012-2014.
D. Mackenzie, A Tisket, a Tasket, an Apollonian Gasket, American Scientist, 98 (2010).
C. L. Mallows, Growing Apollonian Packings, J. Integer Sequences, 12 (2009), article 09.2.1.
I. Peterson, Circle game, Science News, 4/21/01.
I. Peterson, Temple circles, Math Trek, 4/23/01.
P. Sarnak, Letter to Lagarias on integral Apollonian packings, June, 2007.
P. Sarnak, Integral Apollonian packings, MAA Lecture, Jan 2009.
P. Sarnak, Integral Apollonian packings, Amer. Math. Monthly, 118 (2011), 291-306.
K. E. Stange, The sensual Apollonian circle packing, arXiv:1208.4836 [math.NT], 2012-2014.
Wikipedia, Integral Apollonian circle packings
Wikipedia, Descartes' theorem

Cf. A042944, A042946, A042945, A045506, A045673, A045864, A052483, A060790, A135849, A137246, A154636, A154637, A154638, A171090, A189227, A218155, A248938, A265495, A294510.

root = {-11, 21, 24, 28};
triples = Subsets[root, {3}];
a = {root};
Do[
  ng = Table[Total@t + 2 Sqrt@Total[Times @@@ Subsets[t, {2}]], {t, triples}];
  AppendTo[a, Sort@ng];
  triples = Join @@ Table[{t, r} = tr; Table[Append[p, r], {p, Subsets[t, {2}]}], {tr, Transpose@{triples, ng}}]
  , {k, 3}];
Flatten@a (* Andrey Zabolotskiy, May 29 2022 *)

a(n) == 0, 4, 12, 13, 16, or 21 (mod 24).

Terms a(28) and beyond from Andrey Zabolotskiy, May 29 2022

A135849 a(n) is the ratio of the sum of the bends (curvatures) of the circles in the n-th generation of an Apollonian packing to the sum of the bends in the initial four-circle configuration.

Original entry on oeis.org

1, 5, 39, 297, 2259, 17181, 130671, 993825, 7558587, 57487221, 437222007, 3325314393, 25290849123, 192350849805, 1462934251071, 11126421459153, 84622568920011, 643601286982629, 4894942589100999, 37228736851860105, 283145067047577843, 2153474325825042429

Offset: 1

Colin Mallows, Mar 06 2008

These ratios are independent of the starting configuration.

For more comments, references and links, see A189226.

			Starting with the configuration with bends (-1,2,2,3) with sum(bends) = 6, the next generation contains four circles with bends 3,6,6,15. The sum is 30 = 6*a(2). The third generation has 12 circles with sum(bends) = 234 = 6*a(3).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
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 (8,-3).

Cf. A105970, A137246, A138264, A189226, A189227.

I:=[1, 5, 39]; [n le 3 select I[n] else  8*Self(n-1) - 3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 25 2012

CoefficientList[Series[(2 z^2 - 3 z + 1)/(3 z^2 - 8 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{8, -3}, {1, 5, 39}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)

Vec((2*x^3 - 3*x^2 + x)/(3*x^2 - 8*x + 1)+O(x^99)) \\ Charles R Greathouse IV, Jul 03 2011

For n >= 4, a(n) = 8*a(n-1) - 3*a(n-2).
For n>2, [a(n+2), a(n+3)] = the 2 X 2 matrix [0,1; -3,8]^n * [5,39]. Example: [0,1; -3,8]^3 * [5,39] = [a(5), a(6)] = [2259, 17181]. - Gary W. Adamson, Mar 09 2008 (typo corrected by Jonathan Sondow, Dec 24 2012)
a(n) = floor(C * A138264(n)), where C = 1.057097576... = (1/2)*((1/9) + sqrt((1/81) + 4)). Example: a(7) = 130671 = floor(C * A138264(7)) = floor(C * 123613). A135849(n)/A138264(n) tends to C. - Gary W. Adamson, Mar 09 2008
O.g.f.: 2*x/3 +7/9 +(59*x-7)/(9*(1-8*x+3*x^2)). - R. J. Mathar, Apr 24 2008
a(n) = 31*sqrt(13)*(A^n - B^n)/234 - 7*(A^n + B^n)/18 for n>1 where A=3/(4-sqrt(13)) and B=3/(4+sqrt(13)). - R. J. Mathar, Apr 24 2008

A154636 a(n) is the ratio of the sum of the bends of the circles that are drawn in the n-th generation of Apollonian packing to the sum of the bends of the circles in the initial configuration of 3 circles.

Original entry on oeis.org

1, 2, 18, 138, 1050, 7986, 60738, 461946, 3513354, 26720994, 203227890, 1545660138, 11755597434, 89407799058, 679995600162, 5171741404122, 39333944432490, 299156331247554, 2275248816682962, 17304521539721034, 131610425867719386, 1000969842322591986

Offset: 0

Colin Mallows, Jan 13 2009

For comments and more references and links, see A189226.

			Starting from three circles with bends -1,2,2 summing to 3, the first derived generation consists of two circles, each with bend 3. So a(1) is (3+3)/3 = 2.

Colin Barker, Table of n, a(n) for n = 0..1000
C. L. Mallows, Growing Apollonian Packings, J. Integer Sequences, 12 (2009), article 09.2.1.
Index entries for linear recurrences with constant coefficients, signature (8,-3).

Other sequences relating to the two-dimensional case are A135849, A137246, A154637. For the three-dim. case see A154638 - A154645. Five dimensions: A154635.

Cf. also A189226, A189227.

CoefficientList[Series[(5 z^2 - 6 z + 1)/(3 z^2 - 8 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{8, -3}, {1, 2, 18}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)

Vec((1 - x)*(1 - 5*x) / (1 - 8*x + 3*x^2) + O(x^30)) \\ Colin Barker, Jul 15 2017

G.f.: (1 - x)*(1 - 5*x) / (1 - 8*x + 3*x^2).
From Colin Barker, Jul 15 2017: (Start)
a(n) = ((-(-7+sqrt(13))*(4+sqrt(13))^n - (4-sqrt(13))^n*(7+sqrt(13)))) / (3*sqrt(13)) for n>0.
a(n) = 8*a(n-1) - 3*a(n-2) for n>2.
(End)

More terms from N. J. A. Sloane, Nov 22 2009

A154637 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.

Original entry on oeis.org

1, 2, 66, 1314, 26082, 517698, 10275714, 203961186, 4048396578, 80356048002, 1594975770306, 31658447262114, 628384017931362, 12472705016840898, 247568948283023874, 4913960850609954786, 97536510167350024098, 1935988320795170617602, 38427156885401362279746, 762735172745641733742114

Offset: 0

Colin Mallows, Jan 13 2009

For more references and links, see A189226.

			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.

Colin Barker, Table of n, a(n) for n = 0..750
Colin Mallows, Growing Apollonian packings, J. Integer Sequences v.12, article 09.2.1 (2009).
Index entries for linear recurrences with constant coefficients, signature (20,-3).

For starting with four circles, see A137246. For sums of bends, see A135849 and A154636. For three dimensions, see A154638 - A154645.

Cf. also A189226, A189227.

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 *)

Vec((1-18*x+29*x^2)/(1-20*x+3*x^2) + O(x^30)) \\ Colin Barker, Nov 16 2016

G.f.: (1-18*x+29*x^2) / (1-20*x+3*x^2).
From Colin Barker, Nov 16 2016: (Start)
a(n) = ((133-13*sqrt(97))*(10+sqrt(97))^n - (10-sqrt(97))^n*(133+13*sqrt(97))) / (3*sqrt(97)) for n>0.
a(n) = 20*a(n-1) - 3*a(n-2) for n>2.
(End)

More terms from N. J. A. Sloane, Nov 22 2009

A265189 Soddy circles: the two circles tangent to each of three mutually tangent circles.

Original entry on oeis.org

69, 46, 23, 6, 138, 70, 30, 21, 5, 105, 132, 33, 11, 4, -132, 138, 92, 46, 12, 276, 140, 60, 42, 10, 210, 153, 136, 72, 17, 306, 207, 138, 69, 18, 414, 210, 90, 63, 15, 315, 216, 135, 24, 10, -135, 238, 119, 102, 21, 357, 252, 63, 28, 9, 0

Offset: 1

Colin Barker, Dec 04 2015

For any three mutually tangent circles (with radii a, b, and c), one can construct a fourth circle (the inner Soddy circle, with radius d) that is mutually tangent internally to the three circles, and a fifth circle (the outer Soddy circle, with radius e) that is mutually tangent externally to the three circles. For this sequence all five radii have integral lengths.
The sequence is an array of 5-tuples (a,b,c,d,e) ordered by increasing values of a, with a > b > c.
A positive value for the outer Soddy circle indicates that it contains the three circles; a negative value indicates that it is exterior to the three circles; a value of 0 indicates that it has an infinite radius, that is, it is a straight line.

Colin Barker, Table of n, a(n) for n = 1..1000
Kival Ngaokrajang, Illustration of a(1) - a(5), a(41) - a(45) and a(51) - a(55)
Eric Weisstein's World of Mathematics, Soddy Circles
Wikipedia, Descartes' theorem

Cf. A256694.
See also the many sequences arising from Apollonian circle packing: A135849, A137246, A154636, etc.
Also the sequences related to Soddy's circle packings: A046159, A046160, A062536, etc.

soddy(amax) = {
  my(L=List(), abc, t, u);
  for(a=1, amax,
    for(b=1, a-1,
      for(c=1, b-1,
        abc=a*b*c;
        if(issquare(abc*(a+b+c), &t),
          u=a*b+a*c+b*c;
          if(abc%(u+2*t) == 0,
            if(u-2*t != 0,
              if(abc%(u-2*t) == 0,
                listput(L, [a,b,c,abc\(u+2*t),-abc\(u-2*t)])
              )
            ,
              listput(L, [a,b,c,abc\(u+2*t),0])
            )
          )
        )
      )
    )
  );
  Vec(L)
}
soddy(253)

A256694 The radius of the largest of four circles with different integer radii arranged so that each circle is tangent externally to the other three circles.

Original entry on oeis.org

69, 70, 132, 138, 140, 153, 198, 207, 210, 216, 238, 252, 264, 264, 264, 270, 276, 280, 285, 290, 306, 345, 350, 390, 396, 396, 414, 420, 429, 432, 459, 476, 483, 490, 504, 504, 520, 528, 528, 528, 539, 540, 552, 560, 567, 570, 580, 594, 595, 612, 621, 630

Offset: 1

Colin Barker, Apr 08 2015

nonn

Kival Ngaokrajang, Illustration of the first three solutions
Eric Weisstein's World of Mathematics, Soddy Circles.

See also the many sequences arising from Apollonian circle packing: A135849, A137246, A154636, etc.

Also the sequences related to Soddy's circle packings: A046159, A046160, A062536, etc.

soddy(k) = {
  s=[];
  for(a=1, k,
    for(b=1, a-1,
      for(c=1, b-1,
        if(issquare(a*b*c*(a+b+c), &t),
          if(a*b*c % (a*b+a*c+b*c+2*t) == 0,
            s=concat(s, a)
          )
        )
      )
    )
  );
  s
}
soddy(500)

A136270 a(n) = 20a(n-1) - 3a(n-2).

Original entry on oeis.org

1, 17, 337, 6689, 132769, 2635313, 52307953, 1038253121, 20608138561, 409048011857, 8119135821457, 161155572393569, 3198754040407009, 63491614090959473, 1260236019697968433, 25014245551686490241

Offset: 1

Gary W. Adamson, Mar 19 2008

a(n)/a(n-1) tends to (sqrt(97) + 10), an eigenvalue of the matrix and root of the characteristic polynomial x^2 - 20x + 3.

			a(4) = 20*a(3) - 3*a(2) = 20*337 - 3*17.
[a(3), a(4)] = [0,1; -3,20] ^3 * [1,1] = [337, 6689].

G. C. Greubel, Table of n, a(n) for n = 1..765
Index entries for linear recurrences with constant coefficients, signature (20,-3).

Cf. A137246.

LinearRecurrence[{20, -3}, {1, 17}, 50] (* G. C. Greubel, Feb 23 2017 *)

x='x+O('x^50); Vec((1-3*x)/(1-20*x+3*x^2)) \\ G. C. Greubel, Feb 23 2017

a(n) = 20*a(n-1) - 3*a(n-2), n>2; a(1) = 1, a(2) = 17.
[a(3), a(4)] = the 2 X 2 matrix [0,1; -3,20]^n * [1,1].
A137246(n) = 20*a(n) - 3*a(n-1), n>4.
O.g.f.: (1-3*x)/(1-20*x+3*x^2). - R. J. Mathar and Alexander R. Povolotsky, Mar 31 2008
a(n) = (1/2)*(10 - sqrt(97))^n - (9/194)*sqrt(97)*(10 + sqrt(97))^n + (1/2)*(10 + sqrt(97))^n + (9/194)*(10 - sqrt(97))^n*sqrt(97) - Alexander R. Povolotsky, Mar 31 2008

cp's OEIS Frontend

A189226 Curvatures in the nickel-dime-quarter Apollonian circle packing, ordered first by generation and then by size.

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A135849 a(n) is the ratio of the sum of the bends (curvatures) of the circles in the n-th generation of an Apollonian packing to the sum of the bends in the initial four-circle configuration.

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A154636 a(n) is the ratio of the sum of the bends of the circles that are drawn in the n-th generation of Apollonian packing to the sum of the bends of the circles in the initial configuration of 3 circles.

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A154637 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.

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A265189 Soddy circles: the two circles tangent to each of three mutually tangent circles.

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A256694 The radius of the largest of four circles with different integer radii arranged so that each circle is tangent externally to the other three circles.

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A136270 a(n) = 20*a(n-1) - 3*a(n-2).

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A136270 a(n) = 20a(n-1) - 3a(n-2).