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ISRN Geometry
Volume 2012 (2012), Article ID 348618, 19 pages
Research Article

On a Diophantine Equation That Generates All Integral Apollonian Gaskets

Department of Mathematics, Southern Illinois University Carbondale, Carbondale, IL 62901, USA

Received 1 December 2011; Accepted 20 December 2011

Academic Editors: L. V. Bogdanov and G. Martin

Copyright © 2012 Jerzy Kocik. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


A remarkably simple Diophantine quadratic equation is known to generate all, Apollonian integral gaskets (disk packings). A new derivation of this formula is presented here based on inversive geometry. Also, occurrence of Pythagorean triples in such gaskets is discussed.

1. Introduction

Apollonian disk packing (or Apollonian gasket) is a pattern obtained by starting with three mutually tangent circles of which one contains the other two and then recursively inscribing new circles (disks) in the curvilinear triangular regions (called “ideal triangles’’) formed between the circles. Figure 1 shows a few examples, including (a) a special case of the noncompact “Apollonian strip,” (b) the Apollonian window, which is the only case that has symmetry 𝐷2, (c) the regular threefold gasket, which has symmetry 𝐷3, and (d) a general gasket that may have no mirror symmetry.

Figure 1: Various Apollonian disk packings.

It turns out that there exist an infinite number of integral Apollonian packings, where the curvature of every circle/disk is an integer.

Recall that four pairwise tangent circles are said to be in the Descartes configuration. Suppose 𝑎, 𝑏, 𝑐, and 𝑑 are curvatures of four such circles. Then, the Descartes circle formula states2𝑎2+𝑏2+𝑐2+𝑑2=(𝑎+𝑏+𝑐+𝑑)2.(1.1) One may use this formula to obtain the curvature of the fourth circle 𝑑, given the first three. Being a quadratic equation, it yields two solutions, say 𝑑1 and 𝑑2, such that𝑑1+𝑑2=2(𝑎+𝑏+𝑐).(1.2) Disks 𝑑1 and 𝑑2 will be called the Boyd dual. As observed by the Boyd [1], this “linearized’’ version of (1.1) provides a handy tool for determining the curvatures in a particular Apollonian gasket starting with the first four values. In particular, it follows that if the first four values are integers, so are all.

Remark 1.1. For consistency, the curvature of the circle that contains all of the remaining ones must be given a negative value for (1.1) and (1.2) to hold. It is customary to replace the term “curvature’’ by “bend’’ to account for this convention. Another way of looking at this is to think of disks rather than circles, where the greatest circle is the boundary of an exterior, unbounded, region. This way no two disks in an Apollonian gasket overlap.

In this note, Apollonian gaskets will be labeled by the bends of the five greatest circles, that is, by the five least bends. Why five will become clear when we consider symmetries, the appendix.

Examples of integral Apollonian gaskets include (0,0,1,1,1): Apollonian belt (Figure 1(a)),(1,2,2,3,3): Apollonian window (Figure 1(c)),(2,3,6,7,7): less regular gasket, but with 𝐷1 symmetry,(6,11,14,15,23): quite irregular case.

Note that the regular gasket (see Figure 1) cannot have integer bends as its quintet of curvatures is, up to scale, (13,2,2,2,10+23), and hence its curvatures are populated by elements of [3]. Due to (1.2), the integrality of the first four circles determines integrality of all disks in the packing. An integral Apollonian packing is irreducible if the bends have no common factor except 1.

The problem is to classify and determine all irreducible integral Apollonian gaskets.

2. Integer Disk Packing: The Formula

All integral Apollonian disk packings may be determined using a simple Diophantine quadratic equation with constraints. The derivation of this formula is a much simpler alternative to that of “super-Apollonian packing’’ [24] and is based on inversive geometry.

Theorem 2.1. There is a one-to-one correspondence between the irreducible integral Apollonian gaskets and the irreducible quadruples of nonnegative integers 𝐵,𝑘,𝑛,𝜇 that are solutions to quadratic equation 𝐵2+𝜇2=𝑘𝑛(2.1) with constrains (i)0𝜇𝐵3,(ii)2𝜇𝑘𝑛.(2.2) Every solution to (2.1) corresponds to an Apollonian gasket with the following quintet of the major bends (curvatures): 𝐵0,𝐵1,𝐵2,𝐵3,𝐵4=(𝐵,𝐵+𝑘,𝐵+𝑛,𝐵+𝑘+𝑛2𝜇,𝐵+𝑘+𝑛+2𝜇).(2.3)

Corollary 2.2. The Apollonian gasket is integral if and only if gcd (𝐵,𝑘,𝑛)=1.

Figure 2 locates the curvatures of the theorem in the Apollonian gasket. Note that the triple of integers (𝐵,𝜇,𝑘) is also a good candidate for a label that uniquely identifies an Apollonian gasket (since 𝑛 is determined by 𝑛=(𝐵2+𝜇2)/𝑘).

Figure 2: An Apollonian gasket and its four greatest circles (smallest curvatures).

Equation (2.1) leads to an algorithm producing all integral Apollonian gaskets, ordered by the curvatures, presented in Algorithm 1.

Algorithm 1: An algorithm that produces all integral disk packings.

The appendix shows the first 183 entries for the main bend varying from 0 to 32.

Example 2.3. Figure 3 shows that there are only 3 Apollonian gaskets for 𝐵=6.

Figure 3

Proof of Theorem 2.1 (We denote a circle and its curvature by the same symbol). Consider an Apollonian gasket of disks inscribed inside a circle of curvature 𝐵 (bend equal to 𝐵). Draw an axis through the center of this circle and the center of the next largest circle 𝐵1 (the horizontal axis 𝐴 in Figure 4). Inverting the gasket through a circle 𝐾 of radius 2/𝐵 will produce an Apollonian belt, shown in the figure on the right side of the gasket. Denote its width by 2𝜌. Lines 𝐿0 and 𝐿1 are the images of 𝐵 and 𝐵1, respectively.
The next largest circle, 𝐵2, will show up in the strip as a circle 𝐵2. It will intersect the axis 𝐴. Various Apollonian gaskets will result by varying the strip width (i.e., the parameter 𝜌) and the height of the center of the circle 𝐵2 above the axis. Clearly, the range of is the interval [0,𝜌), as going beyond would result in repetitions of the Apollonian arrangements.
Recall the basic formulas of inversive geometry: inversion through a circle of radius 𝐾 centered at 𝑂 carries a point at distance 𝑑 from 𝑂 to a point at distance 𝑑 according to𝑑𝑑=𝐾2,(2.4) and a disk of radius 𝑟 with center distance 𝑑 from 𝑂 is mapped to a disk of radius 𝑟 with its center located at distance 𝑑 according to (i)𝑟=𝑟𝑅2𝑑2𝑟2,(ii)𝑑=𝑑𝑅2𝑑2𝑟2.(2.5) We will analyze the Apollonian gasket as the inversive image of the Apollonian strip, disk by disk.Step 1. Recall that 𝐵𝑁 represents the curvature of the encompassing circle of the gasket.Step 2. To calculate the curvature 𝐵1 of the greatest circle inside (the image of 𝐿1), apply (2.5)(i) to the point of intersection of 𝐿1 and axis 𝐴 to get: 𝐵1=𝐵+𝐵2𝜌𝑘.(2.6) Since 𝐵 is an integer, so is the last term; denote it by 𝑘=𝐵2𝜌. Clearly, 𝑘 can be any nonnegative integer (to make 𝐵1𝐵).Step 3. For 𝐵2, the image of 𝐵2, we use (2.5)(ii) to get 𝐵2=1𝑟=1𝜌𝑑2𝜌2(2/𝐵)2=𝐵24𝜌𝑑2𝜌2(simplication)=𝐵24𝜌2𝐵+𝜌2+2𝜌2(Pythagoreanthm)=𝐵+4+2𝐵24𝜌𝑛(2.7) As before, we conclude that the last term must be an integer; denote it by 𝑛. Clearly, 𝑛𝑘 (to make 𝐵2𝐵1).Step 4. Similarly, we get a formula for the next largest circle 𝐵3 located below 𝐵2, the image of 𝐵3. Simply use the above formula with =2𝜌 instead of to get 𝐵3=1𝜌𝑑2𝜌2(2/𝐵)2=𝐵24𝜌2𝐵+𝜌2+2𝜌22𝜌2=𝐵+𝜌𝐵2𝑘+4+2𝐵4𝜌𝑛𝐵2𝑚.(2.8) Quite pleasantly, the first three terms coincide with terms from previous steps. Since we have already established that they must be integers, so is the last one; denote it by 𝑚. Thus we have three integers defined by the geometry of the construction: 𝑛=4+2𝐵24𝜌,𝑘=𝜌𝐵2,𝑚=𝐵2.(2.9) Integers 𝑘,𝑛, and 𝑚 are not independent; take the definition for 𝑛 and eliminate and 𝑟 from it to get 4𝑛𝑘=4𝐵2+𝑚2,(2.10) from which it follows immediately that 𝑚 must be even, say 𝑚=2𝜇. Reduce the common factor of 4 to get the “master equation’’ (2.1).As to the constraints, the order of the curvatures gives three inequalities: 𝐵1𝐵𝑘0,𝐵2𝐵1𝑛0,𝐵3𝐵2𝑘2𝜇.(2.11) The additional upper bound for 𝜇 comes from the fact that 𝑘 takes its greatest value at 𝑘=𝐵2+𝜇2. Thus, the last inequality of (2.11), 𝑘2𝜇, implies: 𝐵2+𝜇22𝜇,(2.12) and therefore (after squaring) 𝐵2>3𝜇2.(2.13) This ends the proof.

Figure 4: The method of inversion.

3. Symbols of the Circles in an Apollonian Gasket

The symbol of a circle [5, 6] is a formal fraction ̇𝑥,̇𝑦𝑏,(3.1) where 𝑏=1/𝑟 denotes the bend (signed curvature) of the circle and the position of the center is (𝑥,𝑦)=̇𝑥𝑏,̇𝑦𝑏.(3.2) By reduced coordinates we mean the pair (̇𝑥,̇𝑦). In the case of the Apollonian window (packing with the major curvatures (1,2,2,3,3)), the reduced coordinates and the bend of each circle are integers, see Figure 5.

Figure 5: Apollonian window with labels.

The Boyd linearization (1.2) for bends in the Descartes configurations holds for the reduced coordinates since they satisfy a quadratic equation due to the extended the Descartes circle theorem [24]. Thus’ ̇𝑥4+̇𝑥4=2̇𝑥1+̇𝑥2+̇𝑥3,̇𝑦4+̇𝑦4=2̇𝑦1+̇𝑦2+̇𝑦3.(3.3)

Remark 3.1 (the geometric interpretation of the integrality of the labels). Every circle in the Apollonian window is an integer multiple of its radius above the horizontal axis and from the vertical axis. This is a generalization of Pappus’ arbelos theorem [7] for a single chain in an arbelos-like figure, illustrated in Figure 6.
The question is whether the same may be expected for other integer Apollonian disk packings, that is, are the three numbers in the label all integers?

Figure 6: Classical Pappus’ arbelos theorem: the base is divided randomly, yet every circle in the chain is an even multiple of its radius above the axis.

Proposition 3.2. In the case of the coordinate system with the center located at the center of circle B (see Figure 4), the labels for the Apollonian gasket generated from (𝐵,𝑛,𝑘,𝜇) are as presented in Figure 7.

Figure 7: Labels of circles in the Apollonian packing [𝐵,𝜇,𝑛,𝑘].

Proof. Direct calculations.

One can check that the circles of curvatures 𝐵+𝑘+𝑛±𝜇 are the Boyd dual with respect to the triple of mutually tangent circles of bends (𝐵,𝐵+𝑛,𝐵+𝑘).

The above proposition implies that the reduced coordinates for circles are fractional unless 𝐵𝑘 (see the symbol for 𝐵1 in the figure). Since we have also 2𝜇𝑘 (constraints) and the need for 𝑘2𝜇 (see the symbols for the reduced 𝑦-coordinate), we conclude that we would need 𝜇=0 or 𝑘=2𝜇.

This happens in two cases: Apollonian window (1,2,2,3,3) and Apollonian strip (0,0,1,1,1), that is, for [𝐵,𝜇,𝑘,𝑛] equal to [1,0,1,1] or [0,0,0,1], respectively. However, if 𝑘𝜇 (which happens “often,’’ whenever 𝜇=0), all reduced 𝑦-coordinates are integer. This is a special (integral) case of Pappus’ observation since 𝜇=0 corresponds to the pattern shown in Figure 6, (see the appendix).

4. Pythagorean Triples in Apollonian Gaskets

Given two tangent circles 𝐶1 and 𝐶2, we construct a triangle whose hypotenuse joins the centers and the other sides of which are horizontal or vertical with respect to some fixed axes.


We will associate with this triangle a triple of numbers, namely,(Δ,Γ,𝐻)=𝑏1̇𝑥2𝑏2̇𝑥1,𝑏1̇𝑦2𝑏2̇𝑦1,𝑏1+𝑏2.(4.1) The actual size of the triangle sides is the above triple scaled down by the factor of 𝑏1𝑏2. Hence, the symbol for such triangles isΔ,Γ,𝐻𝑏1𝑏2.(4.2) Clearly, Δ2+Γ2=𝐻2. If the reduced coordinates are integers, then the triples (4.1) are Pythagorean triples. The Apollonian window thus contains infinitely many Pythagorean triples [5, 6]. A few are displayed in Figure 8.

Figure 8: Pythagorean triples in the Apollonian window.

Now, for the arbitrary integer packing. Consider the four major circles 𝐵0, 𝐵1, 𝐵2, and 𝐵3 in an Apollonian gasket (Figure 6). Pairwise, they determine six right triangles. Each of the thick segments in Figure 9 represents the hypotenuse of one of them. This sextet will be called the principal frame for the gasket.

Figure 9: Labels of circles in the Apollonian packing [𝐵,𝜇,𝑛,𝑘].

We will prove later that if the corresponding triples Δ, Γ, 𝐻 are integers, so are all triples in the gasket, But first we have the following.

Proposition 4.1. The labels of the six right triangles in an Apollonian gasket whose hypotenuses join the pairs of the first four greatest circles are, in notation as in Figures 7 and 9, as follows:𝑎=𝑘,0,𝑘𝐵(𝐵+𝑘),𝑏=𝑛2𝐵2/𝑘,2𝜇𝐵/𝑘,𝑛𝐵(𝐵+𝑛),𝑐=𝑘+𝑛2𝜇2𝐵2/𝑘,2𝐵2𝜇𝐵/𝑘,𝑘+𝑛2𝜇𝐵(𝐵+𝑘+𝑛2𝜇),𝐴=2𝑏𝑘+2𝜇+4𝐵𝜇/𝑘,2𝐵𝑛+𝜇+2𝐵2/𝑘,2𝐵+2𝑛+𝑘2𝜇(𝐵+𝑛)(𝐵+𝑘+𝑛2𝜇),𝐵=2𝐵+𝑛2𝜇2𝐵2/𝑘,2𝐵+𝑘𝜇2𝐵2/𝑘,2𝐵+𝑛+2𝑘2𝜇𝐵(𝐵+𝑘+𝑛2𝜇),𝐶=2𝐵+𝑘𝑛+2𝐵2/𝑘,2𝜇+2𝜇𝐵/𝑘,2𝐵+𝑘+𝑛(𝐵+𝑘)(𝐵+𝑛).(4.3)

Proof. Direct calculation from the data of Proposition 3.2.

The master equation is used to bring the triples to a form where the only fractional terms contain 𝐵/𝑘 as a factor. Note that, in general, each of the entries for the reduced coordinates is a linear combination with integer coefficients of 𝐵,𝑘,𝑛, and 𝜇 plus the only, possibly, noninteger term, a multiple of 2𝐵𝜇/𝑘 or 2𝐵2/𝑘. The fact that it can be done proves the following theorem.

Theorem 4.2. If 𝑘2𝐵2 (or, equivalently 𝑘2𝐵𝜇), then all triples (4.1) in the integer Apollonian gasket (B, 𝜇, k) are integers (form Pythagorean triples).

Proof. The following implication is direct: 𝑘2𝐵2𝑘2𝜇𝑘2𝜇2𝑘2𝜇2.(4.4) Combining the premise with the result, 𝑘22𝐵22𝜇2𝑘24(𝐵𝜇)2𝑘2𝜇𝐵. Every entry of (4.3) may be expressed as a linear integral combination of 𝐵, 𝑘, 𝑛, and 𝜇 and an extra term of either 2𝐵𝜇/𝑘 or 2𝐵𝜇/𝑘. This proves the claim.

Integer Pythagorean triples happen frequently. For instance, each of the conditions: 𝜇=0, or 𝑘=1, 𝑘=2 is sufficient.

5. Linear Recurrence for Pythagorean Triples in an Apollonian Gasket

Consider four circles in the Descartes configuration, 𝐶1,,𝐶4 (see Figure 10). The four centers determine six segments that we view as hypotenuses of triangles. They give, after rescaling, six Pythagorean triangles with sides denoted as follows:horizontalsideΔ𝑖𝑗=𝑏𝑖𝑏𝑗𝑥𝑖𝑥𝑗=̇𝑥𝑖𝑏𝑗̇𝑥𝑗𝑏𝑖,verticalsideΓ𝑖𝑗=𝑏𝑖𝑏𝑗𝑦𝑖𝑦𝑗=̇𝑦𝑖𝑏𝑗̇𝑦𝑗𝑏𝑖,hypotenuse𝐵𝑖𝑗=𝑏𝑖𝑏𝑗1𝑏𝑖+1𝑏𝑗=𝑏𝑖+𝑏𝑗(5.1) with 𝑖,𝑗=1,,4. Note that Δ𝑖𝑗=Δ𝑗𝑖 and Γ𝑖𝑗=Γ𝑗𝑖 but 𝐵𝑖𝑗=𝐵𝑗𝑖. They form the frame of the Descartes configuration.

Figure 10: (a) A frame (b) and an adjacent frame. Circle 4 is shaded.

Now, complement the picture with a circle 𝐶4, the Boyd dual to 𝐶1 (see Figure 10). This leads to a new frame, the frame of the Descartes configuration 𝐶4,𝐶2,𝐶3,𝐶4. Quite interestingly, the elements of the new frame are linear combinations of the elements of the initial frame.

Theorem 5.1. Following the notation of (5.1) and Figure 7, the transition of frames is given byΔ41=Δ41+2Δ212Δ13,Γ41=Γ41+2Γ212Γ13,𝐵41=𝐵41+2𝐵12+2𝐵13.(5.2)

Proof. Using the linear relations (1.2) and (3.3), we can express the position and the curvature of the new, fifth, circle in terms of the initial four: ̇𝑥4=2̇𝑥1+2̇𝑥2+2̇𝑥3̇𝑥4,̇𝑦4=2̇𝑦1+2̇𝑦2+2̇𝑦3̇𝑦4,𝑏4=2𝑏1+2𝑏2+2𝑏3𝑏4.(5.3) We can calculate the Pythagorean vectors for each pair 𝐶4𝐶𝑖, 𝑖=1,2,3. By some magic, due to the following regroupings and adding zeros, these can be expressed in terms of the initial four Pythagorean vectors only, as shown here for the pair 𝐶4𝐶1: Δ41=̇𝑥4𝑏1̇𝑥1𝑏4=2̇𝑥1+2̇𝑥2+2̇𝑥3̇𝑥4𝑏1̇𝑥12𝑏1+2𝑏2+2𝑏3𝑏4=̇𝑥1𝑏4̇𝑥4𝑏1+2̇𝑥2𝑏1̇𝑥1𝑏21+2̇𝑥3𝑏1̇𝑥1𝑏3=Δ41+2Δ212Δ13.(5.4) The Γ’s follow the same argument. Only 𝐵s are slightly different: 𝐵41=𝑏4+𝑏1=𝑏4+3𝑏1+2𝑏2+2𝑏3=4𝑏1+2𝑏2+2𝑏3𝑏1+𝑏4=𝐵41+2𝐵12+2𝐵13.(5.5) Thus, the transformation is linear in the entries of Δ, Γ, and, 𝐵, with integer coefficients, as stated.

Actually, we have proved a little more.

Corollary 5.2. If the (scaled) triangles formed by the initial Descartes configuration are integer, so are all of the triangles of the Apollonian window.

Matrix Description
Now we may rephrase our findings on transitions between frames in terms of matrices. The data for the Pythagorean triangles in each frame may be expressed as three vectors (columns), for example, for the initial frame, Δ=[Δ41,Δ42,Δ43,Δ12,Δ23,Δ31]𝑇 and similarly for Γ and 𝐵. Similarly for the second frame, we have Δ, Γ, and 𝐵.

Here are the three matrices of the transition from a frame to the corresponding subsequent frame:𝐴=100000000001000010000122201020210002,𝐵=000001010000000100021200000212120002,(5.6)𝐶=000010000100001000012200102020000221.(5.7) (vertical and horizontal lines in the matrices are drawn for easier inspection). Now acting with them on the initial vectors Δ and Γ, one may reconstruct all the interior triangles. Vectors Δ and Γ may be combined into a single matrix, 𝑇=[ΔΓ]: 𝑇=Δ41Γ41Δ42Γ42Δ43Γ43Δ12Γ12Δ23Γ23Δ31Γ31,𝐵=𝐵41𝐵42𝐵43𝐵12𝐵23𝐵31(5.8) The vector 𝐵 is transformed to 𝐵 by matrices like 𝐴,𝐵,and𝐶, where all of the entries are to be replaced by their absolute values.

We conclude with two examples of Apollonian packings in which every triangle for adjacent circles determines a Pythagorean triple, see Figures 11 and 12. Only some of the triangles are displayed.

Figure 11: Arrangement (𝑏0,𝑏1,𝑏2,𝑏3)=(2,3,6,7) or (𝐵,𝜇,𝑘,𝑛)=(2,4,4,5). Left side shows symbols of the principal and other circles, and right side shows Pythagorean triples.
Figure 12: Arrangement (𝑏0,𝑏1,𝑏2,𝑏3)=(3,5,8,8) or (𝐵,𝜇,𝑘,𝑛)=(3,4,5,5). Both symbols (left) and Pythagorean triples (right). The triangles are reported with absolute values of sides.

6. Concluding Remarks

Integral Apollonian disk packings have a topic of much interest for a while [24, 8, 9] and their occurrence has been analyzed. Yet it is not clear whether the integral Apollonian disk packings admit coordinate systems in which the reduced coordinates are integral, except the known cases of Apollonian strip and Apollonian window. We have, however, a frequent occurrence of gaskets that admit integral Pythagorean triples, constructed in Sections 4 and 5.

Yet another intriguing remark concerns the master equation 𝐵2+𝜇2𝑛𝑘=0.(6.1) If we see it as 𝐵2+𝜇2=hypotenuse2—one may identify the corresponding (scaled) triangle in Figure 4 as the one with legs being the diameter 2/𝐵 of the encompassing circle and altitude of the center of circle 𝐵2—the inversive image of circle 𝐵2. Each solution to the master equation may also be viewed as an integral isotropic vector [𝐵,𝜇,𝑛,𝑘]𝑇 in the Minkowski space 3,1  a la Pedoe map from circles to vectors, see [6]. Such vectors represent points in this Minkowski space—degenerated circles of radius 0.


List of Integral Packings

Here, we provide a list of the first 200 integer Apollonian packings. First, however, we give a few words on symmetry of these arrangements.

Only two integral Apollonian packings have rich symmetry: that of Apollonian Strip, and of Apollonian window (𝐷2). Otherwise, we have two cases, that of symmetry 𝐶1 and 𝐷1, which will be labeled as follows:Skew: means no nontrivial symmetry (thus symmetry of type 𝐶1).

For cases with single axis symmetry (𝐷1), we will distinguish two cases:odd: mirror symmetry 𝐷1, with exactly three circles cut by the axis of symmetry,even: mirror symmetry with infinitely many circles on the axis,even*: even symmetry that is obtained “accidentally” by Algorithm 1, where the two greatest congruent circles are not images of the congruent disks in the Apollonian strip.

Consult Figure 13 for visualization.

Figure 13: Types of symmetries and arrangements.

Yet another attribute of an Apollonian packing is a shift, which we define as the degree the disk on the horizontal axis in the Apollonian strip is raised above this axis in terms of the fraction of its radius: shift=𝜌=2𝜇𝑘.(A.1) See Figure 4. It has values in the interval [0,1], and it measures the degree the gasket is off the axial symmetry: 0 for odd symmetry 𝐷1 and 1 for even symmetry 𝐷1. The fractional value indicates lack of axial symmetry, except the cases of “accidental” even* symmetry, when 𝑛=𝑘 (this gives 𝐵1=𝐵2).

Table 1 displays first 24 integral Apollonian gaskets including symmetry type, principal disk quintet, the parameters 𝐵, 𝑘, 𝑛, and 𝜇, and the shift factor. The first even* case is also included.

Table 1: Bend quintets for Apollonian integer disk packings for principal curvatures 1 through 15, including four indices and shifts.

Table 2 contains a longer list of integer Apollonian gaskets for the principal curvatures from 1 through 32. Curvature quintets (in brackets) are followed by shift factor.

Table 2: Bend quintets for Apollonian integer disk packings for principal curvatures 1 through 32.


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