Image:ApollonianGasket-1_2_2_3-Labels.png|Integral Apollonian circle packing defined by circle
curvatures of (−1, 2, 2, 3) Image:ApollonianGasket-3_5_8_8-Labels.png|Integral Apollonian circle packing defined by circle curvatures of (−3, 5, 8, 8) Image:ApollonianGasket-12_25_25_28-Labels.png|Integral Apollonian circle packing defined by circle curvatures of (−12, 25, 25, 28) Image:ApollonianGasket-6_10_15_19-Labels.png|Integral Apollonian circle packing defined by circle curvatures of (−6, 10, 15, 19) Image:ApollonianGasket-10_18_23_27-Labels.png|Integral Apollonian circle packing defined by circle curvatures of (−10, 18, 23, 27) If any four mutually tangent circles in an Apollonian gasket all have integer
curvature (the inverse of their radius) then all circles in the gasket will have integer curvature. Since the equation relating curvatures in an Apollonian gasket, integral or not, is a^2 + b^2 + c^2 + d^2 = 2ab + 2 a c + 2 a d + 2 bc+2bd+2cd,\, it follows that one may move from one quadruple of curvatures to another by
Vieta jumping, just as when finding a new
Markov number. The first few of these integral Apollonian gaskets are listed in the following table. The table lists the curvatures of the largest circles in the gasket. Only the first three curvatures (of the five displayed in the table) are needed to completely describe each gasket – all other curvatures can be derived from these three.
Enumerating integral Apollonian circle packings The curvatures (a, b, c, d) are a root quadruple (the smallest in some integral circle packing) if a . They are primitive when \gcd(a, b, c, d)=1. Defining a new set of variables (x, d_1, d_2, m) by the matrix equation \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ -1 & 1 & 0 & 0\\ -1 & 0 & 1 & 0\\ -1 & 1 & 1 &-2 \end{bmatrix} \begin{bmatrix} x \\ d_1 \\ d_2 \\ m \end{bmatrix} gives a system where (a, b, c, d) satisfies the Descartes equation precisely when x^2+m^2=d_1 d_2. Furthermore, (a, b, c, d) is primitive precisely when \gcd(x, d_1, d_2)=1, and (a, b, c, d) is a root quadruple precisely when x. This relationship can be used to find all the primitive root quadruples with a given negative bend x. It follows from 2m\leq d_1 and 2m\leq d_2 that 4m^2\leq d_1d_2, and hence that 3m^2\leq d_1d_2-m^2=x^2. Therefore, any root quadruple will satisfy 0\leq m \leq |x|/\sqrt{3}. By iterating over all the possible values of m, d_1, and d_2 one can find all the primitive root quadruples. The following
Python code demonstrates this algorithm, producing the primitive root quadruples listed above. import math def get_primitive_bends(n: int) -> tuple[int, int, int, int]: if n == 0: yield 0, 0, 1, 1 return for m in range(math.ceil(n / math.sqrt(3))): s = m**2 + n**2 for d1 in range(max(2 * m, 1), math.floor(math.sqrt(s)) + 1): d2, remainder = divmod(s, d1) if remainder == 0 and math.gcd(n, d1, d2) == 1: yield -n, d1 + n, d2 + n, d1 + d2 + n - 2 * m for n in range(15): for bends in get_primitive_bends(n): print(bends)
The Local-Global Conjecture The curvatures appearing in a primitive integral Apollonian circle packing must belong to a set of six or eight possible residues classes modulo 24, and theoretical results and numerical evidence supported that any sufficiently large integer from these residue classes would also be present as a curvature within the packing. This conjecture, known as the local-global conjecture, was proved to be false in 2023.
Symmetry of integral Apollonian circle packings There are multiple types of
dihedral symmetry that can occur with a gasket depending on the curvature of the circles.
No symmetry If none of the curvatures are repeated within the first five, the gasket contains no symmetry, which is represented by symmetry group
C1; the gasket described by curvatures (−10, 18, 23, 27) is an example.
D1 symmetry Whenever two of the largest five circles in the gasket have the same curvature, that gasket will have
D1 symmetry, which corresponds to a reflection along a diameter of the bounding circle, with no rotational symmetry.
D2 symmetry If two different curvatures are repeated within the first five, the gasket will have D2 symmetry; such a symmetry consists of two reflections (perpendicular to each other) along diameters of the bounding circle, with a two-fold rotational symmetry of 180°. The gasket described by curvatures (−1, 2, 2, 3) is the only Apollonian gasket (up to a scaling factor) to possess D2 symmetry.
D3 symmetry There are no integer gaskets with
D3 symmetry. If the three circles with smallest positive curvature have the same curvature, the gasket will have
D3 symmetry, which corresponds to three reflections along diameters of the bounding circle (spaced 120° apart), along with three-fold rotational symmetry of 120°. In this case the ratio of the curvature of the bounding circle to the three inner circles is 2 − 3. As this ratio is not rational, no integral Apollonian circle packings possess this
D3 symmetry, although many packings come close.
Almost-D3 symmetry The figure at left is an integral Apollonian gasket that appears to have
D3 symmetry. The same figure is displayed at right, with labels indicating the curvatures of the interior circles, illustrating that the gasket actually possesses only the
D1 symmetry common to many other integral Apollonian gaskets. The following table lists more of these
almost-
D3 integral Apollonian gaskets. The sequence has some interesting properties, and the table lists a factorization of the curvatures, along with the multiplier needed to go from the previous set to the current one. The absolute values of the curvatures of the "a" disks obey the
recurrence relation , from which it follows that the multiplier converges to + 2 ≈ 3.732050807.
Sequential curvatures For any integer
n > 0, there exists an Apollonian gasket defined by the following curvatures: (−
n,
n + 1,
n(
n + 1),
n(
n + 1) + 1). For example, the gaskets defined by (−2, 3, 6, 7), (−3, 4, 12, 13), (−8, 9, 72, 73), and (−9, 10, 90, 91) all follow this pattern. Because every interior circle that is defined by
n + 1 can become the bounding circle (defined by −
n) in another gasket, these gaskets can be
nested. This is demonstrated in the figure at right, which contains these sequential gaskets with
n running from 2 through 20. ==History==