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The American Mathematical Monthly

ISSN: 0002-9890 (Print) 1930-0972 (Online) Journal homepage: https://maa.tandfonline.com/loi/uamm20

Fibonacci Numbers, Integer Compositions, andNets of Antiprisms

Rick Mabry

To cite this article: Rick Mabry (2019) Fibonacci Numbers, Integer Compositions,and Nets of Antiprisms, The American Mathematical Monthly, 126:9, 786-801, DOI:10.1080/00029890.2019.1644124

To link to this article: https://doi.org/10.1080/00029890.2019.1644124

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Fibonacci Numbers, Integer Compositions,and Nets of Antiprisms

Rick Mabry

Abstract. All geometrically distinct nets (edge-unfoldings) are explicitly constructed for theinfinite family of polyhedra known as uniform antiprisms. The Fibonacci numbers F2n areshown to count the number of nets of the n-antiprism that have point symmetry. (The n-antiprism refers to the antiprism whose surface consists of two regular n-gons separated bya band of 2n equilateral triangles.) Smaller evenly indexed Fibonacci numbers count certainsubfamilies of these nets. Well-known formulas for sums of alternately indexed Fibonaccinumbers motivate and become accessories to our constructions and proofs. For counting andconstructing the symmetric nets, we first use a known result connecting compositions of inte-gers and Fibonacci numbers. In an alternative proof, we illustrate a striking self-similarityamong the counts of certain natural subfamilies of the symmetric nets, which is used to gener-ate an efficient recursive construction of all the symmetric nets of the n-antiprism from thoseof the (n − 1)-antiprism. (This second proof is found in the online supplement to this arti-cle.) Finally, each nonsymmetric net of the n-antiprism is constructed by combining a pair ofdistinct symmetric nets of the n-antiprism.

1. NETS OF ANTIPRISMS. We construct all distinct nets (edge-unfoldings) of theuniform antiprism of order n. In the process we obtain formulas for the total numbertn of all such distinct nets, distinctness meaning geometric noncongruence. The num-bers sn of all of such nets that have point symmetry turn out to be the evenly indexedFibonacci numbers, as do the counts of certain smaller subcollections with variousdistinguishing properties.

The nets we are examining are geometric objects, but our proofs and constructionsrely almost entirely on certain combinatorial properties of the antiprism, which we willencode using some customized graph-like diagrams in Section 2.3.

Figure 1. The antiprism of ordern = 5.

Figure 2. A net of the antiprism. Figure 3. A symmetric net.

Having a look at Figures 1–3 will help us get started; depicted are an antiprism oforder n = 5, one of its t5 = 1540 nets, and one of its s5 = 55 symmetric nets.

doi.org/10.1080/00029890.2019.1644124MSC: Primary 52B10, Secondary 05C30; 05C10; 05C50Color versions of one or more of the figures in the article can be found online at www.tandfonline.com/uamm.Supplemental data for this article can be accessed on the publisher’s website.

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We call our nets symmetric when they have point symmetry; there are no nets ofantiprisms with bilateral symmetry. We let T n and Sn respectively denote the col-lection of nets and the collection of symmetric nets of order n. (Thus tn = |Tn| andsn = |Sn|.)

Our main results are that

sn = F2n (1)

and

tn = 1

2sn(sn + 1) (2)

hold for all n, where Fj is the usual j th Fibonacci number (F1 = 1, F2 = 1, Fj+2 =Fj+1 + Fj for all j ≥ 1). To prove (1) we make use of a known result connectingcompositions of integers and Fibonacci numbers. We will also indicate a self-similarityamong certain subsets of Sn that can be used to prove (1) directly.

For (2), the fact that j (j + 1)/2 is the number of pairs of elements of a finite setof size j will be the basis for constructing the complete set Tn. Each distinct pair ofsymmetric nets will be used to generate a distinct (not necessarily symmetric) net.

2. TERMINOLOGY AND LIGHT BACKGROUND. We need to define what it iswe are counting—these distinct nets of antiprisms. What is an antiprism? What is anet and when are two nets distinct? Our definitions are straightforward and intuitive,but in the literature there are some differing versions of the term net, so we will brieflyaddress those. (Readers who feel familiar with these topics might simply skim thisnext subsection.)

2.1. Nets, dual maps, and spanning trees. The nets we depict are of the uniformantiprisms, which are the polyhedra whose faces form a band (often called a ribbon)of 2n equilateral triangles separating two regular n-gons (which can be consideredcaps above and below the band). The two n-gons are parallel and (when viewed alongan axis through their common centers) are offset by an angle of π/n. We refer to thesesolids simply as the antiprisms of order n or just n-antiprisms, for short.

Our combinatorial considerations are not dependent on the triangular faces beingequilateral; they could just as well be congruent isosceles triangles, all edges joiningthe two regular n-gons being of equal length. And we are not much concerned herewith the antiprisms as solids—we are only interested in the edge-unfoldings of theirsurfaces. More specifically, we mean those plane figures—the nets of the polyhedra—that result from cutting the polyhedron’s surface along some set of edges joining itsfaces so that the remaining edges can be used as hinges to fold the surface into aconnected, plane figure. There are always the same number, 2n + 1, of such hinge-edges (called fold edges in [13]) in a net of an n-antiprism.

We should be even more specific. We referred to the “plane figures” resulting fromsuch unfoldings, but this can be problematic. This is because some nets have faces thatoverlap when unfolded, as in Figure 4, where n = 13 in the first net and two facesproperly overlap, albeit slightly, and n = 96 in the second, where there is much moreoverlap. (When n = 12 the overlap can be at a single boundary point of the triangularfaces.) When such overlapping occurs, one loses the ability to “cut out the net” froma piece of cardboard or paper in order to do some real folding, so it is natural that the

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overlapping cases are often unwanted. The overlapping cases have been called “would-be nets” in [5] and lately one sees the terms “simple unfoldings” and “overlappingunfoldings” used to distinguish the two types.

For us this does not matter, even though the distinction is one of the importantideas regarding polyhedra. Indeed, the terms edge-unfolding and net are often meantto exclude the overlapping cases. For example, the most famous problem in this realmasks: Does every convex polyhedron have a net? This is often called Durer’s problem,but was probably first stated by Shephard [13]. So far, the overwhelmingly predom-inant conjecture is affirmative, as no counterexamples are known or predicted. See,for instance, [4–6, 12]. (The most recent of these intriguingly describes “substantialempirical evidence both for and against” the prevailing conjecture.)

Figure 4. When n > 12 the triangular faces of ann-antiprism’s net can properly overlap.

Figure 5. The first antiprism of Figure 4, jawsslightly agape.

Even in light of such overlapping terminology, we will use the term net to refer tosuch figures, overlapping or not. For our purposes the net can be considered to be apolygonal line in the plane, possibly self-intersecting, and we consider two nets to bedistinct if they are not congruent as such. For a more rigorous and technical approachone can see Sections 1.6–7 of [1], where a net has a more general definition.

Soon we will introduce diagrams that free us from such concerns, reducing thingsto some bare combinatorial essentials. Among other things we need not concern our-selves with here are the many three-dimensional aspects of unfolding of polyhedra.For that the reader can consult one of the papers on the cutting edge, such as [2],where continuous blooming is the primary subject. Figure 5 shows one of our nets inthe process of blooming.

Each of the polygonal faces of a polyhedron is adjacent to other faces, adjacencymeaning abutting along common edges. Thus we can associate these edges withunordered pairs of the adjacent faces. For representing the polyhedron a planar mapof the polyhedron is often used. One means of obtaining such a map is by meansof a stereographic projection (or a Schlegel diagram), which preserves all the adja-cencies of the polyhedron if we agree to allow the unbounded exterior of the mapto represent one of the faces. Figure 6 shows such a map of the antiprism of ordern = 5, the exterior of the projection corresponding to one of the two pentagonal faces,numbered 0.

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Figure 6. A projection of the 5-antiprism. Figure 7. The projection plus G5 (dashed edges).

From the planar map one then defines a graph, called the dual of the planar map, thevertices of this graph representing the faces of the polyhedron, and these vertices beingconsidered adjacent in the graph (connected by an edge in the graph) whenever thecorresponding faces of the polyhedron or map are adjacent (separated by an edge in thepolyhedron or map). Thus the ordered pairs of faces in the map become correspondingordered pairs of vertices in the dual. (The dual use of the term “edge” might seemproblematic, but it should be clear in context.) We denote by Gn the dual graph of then-antiprism. Figure 7 shows G5 superimposed on a planar map of the 5-antiprism.

We may think of our net as being “hinged together” by means of a minimal setof edges joining the polyhedron’s adjacent faces. Every face is connected to everyother face by some path—some sequence of pairwise-hinged faces, but there can beno “circuits” formed—no sequence connecting a face to itself. Translated to the dual,where connections of vertices replace those of faces, the nets correspond to spanningtrees of the dual. In turn, a spanning tree of Gn specifies a net when imposed on theantiprism. (The spanning tree itself has no independent geometric properties, so cannotrepresent a net without being associated with the antiprism.) See Figure 8, which showsa spanning tree of G5 (a subset of the edges of G5, in bold) and the net corresponding tothe spanning tree. For more about nets as spanning trees, see [15] or Chapter 7 of [12]and the references therein.

Figure 8. A spanning tree of G5 and its corresponding net.

2.2. Labeling and taxonomy. We label our polygonal faces according to the picturein Figure 9, in which a sufficiently suggestive quantity of labels are marked. Only the

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band of 2n triangles is shown; the two n-gons are not pictured. We label those topand bottom n-gonal faces with 0 and 2n + 1, respectively. Triangles marked 1 throughn run consecutively along and adjacent to the “top” of the antiprism, while trianglesmarked n + 1 through 2n are connected along the “bottom” of the antiprism.

Figure 9. Our labeling of the band of the n-antiprism.

Notice that none of the top (respectively, bottom) triangles are adjacent to the bot-tom (respectively, top) n-gon. The unordered pairs of labels of pairwise-adjacent tri-angular faces are of the forms {j, n + j} (1 ≤ j ≤ n), {j, n + j − 1} (2 ≤ j ≤ n), and{1, 2n}, while the adjacencies of the triangles to the n-gons are of the form {0, j} and{n + j, 2n + 1} (1 ≤ j ≤ n).

Given a net of an antiprism, we find it convenient to give names to its various bodyparts. Let Figure 10 serve as an example. There we see that the two heads of the net(the two n-gons) are always connected by a neck (given in the example by the path{1, 9, 2, 10}). The neck always consists of pairs of adjacent and oppositely orientedtriangular faces. The pairing is a consequence of the alternating connections to theheads (see Figure 9); an odd number of triangles in the neck would create a circuit offaces from a head to itself. We define the neck size to be the number h of such pairs,which is always at least 1 (h = 2 in Figure 10, h = 1 in Figure 11). There can be onlyone neck—one path from head to head, else there would be a circuit. Attached to andextending the neck can be triangular faces that form the collar, which naturally hastwo lapels.

Figure 10. h = 2, k1 = 2, k2 = 1. Figure 11. h = 1, k1 = 0, k2 = 1.

There are always distinct labeled spanning trees that give rise to congruent nets.Two of these are shown in Figure 12; we consider these to be the same net.

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Figure 12. Two congruent nets, distinct only when labeled.

We therefore standardize the labeling by henceforth insisting that the neck face touch-ing head 0 is always numbered 1, as in the right member of Figure 12. Our labeling ofthe antiprism ensures that the labels of the band can be cycled so that this is so. Havingdone that, we can distinguish the first lapel and second lapel, with lengths denoted byk1 and k2, respectively. The first lapel consists of those band members that form a pathfrom face 1 not intersecting the neck. In Figure 10, k1 = 2 and k2 = 1, the first andsecond collars being, respectively, the paths {3} and {16, 8}. Notice that one or both ofthe lapels might be empty; see the example in Figure 11, where k1 = 0 and k2 = 1.

2.3. Arrow-grams. We now introduce “arrow-grams” as convenient encodings ofantiprism nets that contain all the combinatorial ingredients we need.

The diagram in Figure 13 is essentially the graph Gn of an antiprism (where n = 5),but where arrows take the place of edges to the heads. We can use subgraphs of thisdiagram to more easily represent our nets. Figure 14 shows an example.

Figure 13. The complete arrow-gram of the 5-antiprism.

We standardize our arrow-grams by requiring the top row of vertices to be the facesnumbered 1 through n (the top band members) and the up-arrows indicate edges toface 0 (top n-gon); down-arrows from faces n + 1 through 2n (bottom band) go toface 2n + 1 (bottom n-gon). We further standardize the arrow-grams by forcing themto be “neck-centric”: the edge in the center of the neck is the center edge of the arrow-gram. A result of this is that we are sometimes forced to draw a wrap-around edge—an edge joining the far-right band member with the far-left one (the edge {4, 8} inFigures 13 and 14 is such).

Figure 14. An arrow-gram of a net.

For future reference, the conditions that associate a spanning tree of Gn with a netare next restated in terms of arrow-grams.

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Lemma (Arrow-gram criteria). A sub-diagram of a complete arrow-gram of an n-antiprism is the arrow-gram of a net of the antiprism if and only if (1) every bandmember is connected by some path to an arrow, (2) there is one and only one pathconnecting any up-arrow to a down-arrow, and (3) there is never a path from an up-arrow to an up-arrow or a down-arrow to a down-arrow.

It is important to note that a neck-centered arrow-gram of a symmetric net is itselfsymmetric in the plane with respect to the center of its neck. It follows that a neck-centered symmetric arrow-gram can never have a wrap-around edge. (Suppose other-wise. If that edge is ultimately connected to the neck, then symmetry shows that theentire band is connected, which is impossible for a net. Otherwise, such an edge, bydefinition, connects the vertices at the extreme left and right ends of the arrow-gram.But each of these vertices must be connected by some path to a head, while symmetryimplies that those two heads are different. That leaves the two heads connected by thispath. But they are already connected by the neck, which is in the center, so we nowhave two paths connecting the heads, forming a circuit, which is impossible.)

2.4. Special case: n = 3. Our aim is to collect and count all unique plane figurescorresponding to the nets of our antiprisms, uniqueness being determined by geometricnoncongruence. However, an exception to this uniqueness occurs when n = 3, becauseall the faces of the uniform 3-antiprisms, including the 3-gonal heads, are congruent.Thus if we wish to include n = 3, we need to consider certain congruent nets as beingdistinct. Fortunately, this can be done in a natural way.

In fact, we can make a case for considering the “degenerate” cases of 2-antiprisms,and even the 1-antiprism. We save the details of all three of these special cases forthe online supplement to this article, since they can be omitted here without loss ofcontinuity.

3. SYMMETRIC NETS. In this section we prove a result, Lemma 1, that lets usenumerate Sn, the symmetric members of Tn. (Here is where the Fibonacci numbersfinally appear.) Since we are considering only symmetric nets, we have k1 = k2, so wedenote the lapel lengths simply by k. Note that we must have

1 ≤ h ≤ n, 0 ≤ k ≤ n − 1, 1 ≤ h + k ≤ n, (3)

so these restrictions will be implicitly assumed henceforth. A consequence of thelemma is that each pair (h, k) satisfying (3) is attained by at least one net in Tn.

Lemma 1. For each n ≥ 1, the number sn(h, k) of symmetric nets of the n-antiprismhaving neck size h and lapel size k is given by

sn(h, k) ={

1 if h + k = n

F2(n−h−k) if 1 ≤ h + k < n.(4)

Before proving the lemma, we note that it immediately implies the following, whichwe had set as one of our main goals (equation (1)).

Theorem 1. There are F2n symmetric nets of the n-antiprism. That is, sn = F2n.

Proof. By Lemma 1, the number of symmetric nets with neck size h is

sn(h, ∗) :=n−h∑k=0

sn(h, k)

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= 1 +n−h−1∑

k=0

F2(n−h−k)

= 1 + F2 + F4 + · · · + F2(n−h)

= F1+2(n−h),

where the last equality follows from telescoping the preceding sum after expressingthe even-indexed Fibonacci numbers like so: F2i = F2i+1 − F2i−1 (i ≥ 1). Similarly,we then have

sn =n∑

h=1

sn(h, ∗)

=n∑

h=1

F1+2(n−h)

= F1 + F3 + · · · + F2n−1

= F2n.

Our theorem is now proven as soon as Lemma 1 is proven.

Proof of Lemma 1. We are to show that sn(h, k) = F2(n − h − k) if h + k < n andthat sn(h, k) = 1 if h + k = n. The latter is clear, though, since if h + k = n, thenthere is just one member of Sn(h, k). This follows since the neck and lapels use up allthe band members, as in Figure 15, which shows all such cases for n = 5.

Figure 15. sn(h, k) = 1 when h + k = n, illustrated for n = 5.

So let us assume that h + k < n and let m = n − h − k.Recall that for any integer m > 0, a composition of m is a sequence (x1, x2, . . . , x�)

of positive integers whose sum is m. Note that order matters; (2, 2, 1) and (1, 2, 2) aredistinct compositions of 5, for instance; see [14]. (A composition of m is sometimescalled an ordered partition of m.) Let Cm denote the collection of all compositions ofm.

Consider a “partial” arrow-gram of a member of Sn(h, k) as in the top diagram ofFigure 16, where n = 11, h = 3, and k = 2. The neck and lapels are set, but we havenot yet attached the remaining vertices, of which there are m = n − h − k = 6 on eachside. (Our neck-centered arrow-gram is symmetric, so we focus on just one side.)

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Figure 16. Grouping and connecting the remaining n − h − k vertices.

Each composition of m will correspond to exactly one way to group these m ver-tices. As an example, the second diagram in Figure 16 shows the composition (3, 1, 2)

of 6 and the associated grouping; we connect the vertices within each group.Each group needs to be connected to a head, and to do that we simply select one

vertex from each group, an example being shown in the last arrow-gram of Figure 16.Clearly there are 3 × 1 × 2 = 6 choices for the last step in that example, and in general,if the composition of m is (x1, x2, . . . , x�), then

∏�

i=1 xi is the number of ways to dothis. Each composition gives rise to distinct choices, and so distinct nets. The totalnumber of nets obtained is therefore given by equality (5):

sn(h, k) =m∑

�=1

∑(x1,x2,...,x�)∈Cm

�∏i=1

xi. (5)

Finally, we use the following fact:

m∑�=1

∑(x1,x2,...,x�)∈Cm

�∏i=1

xi = F2m. (6)

(Formula (6) seems to have first been given by Moser and Whitney in 1961 [11].See [3, 7, 14] for other proofs of this and many other Fibonacci-related results aboutcompositions.) By (6) and (5) we obtain sn(h, k) = F2m = F2(n−h−k), as claimed, andour lemma is proved.

The online supplement to this article contains a completely different, elementary,inductive proof of Lemma 1. That proof exploits a pretty self-similarity between thecounts of certain natural subfamilies (based on neck and lapel sizes) of Sn. As aresult we generate the members of Sn+1 directly, one-at-a-time, from those of Sn. Withthat, this article becomes completely self-contained, not relying on the external resultsinvolving compositions of integers. It should be stressed that with the second proofwe truly enumerate, rather than merely “count” the objects in question—we actuallycrank them out. The proof above does allow us to construct the members of Sn, giventhat we can produce all the compositions of the integers involved. This is a routinematter using available software and known algorithms (and can certainly be done byhand), but please see the supplement for a more visual, direct, and revealing way.

Also note that if we were not interested in “fine structure” counts such as sn(h, k),we could avoid Lemma 1 and use the compositions of n to directly obtain sn = F2n.Figure 17 illustrates (using (5, 3, 1, 2) ∈ C11) how this works—just replace m with n

in (6), letting x1 = h + k. In this way, each of our symmetric nets can be coded using

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a composition (x1, x2, . . . x�) of n and a sequence (y1, y2, . . . , y�), where 1 ≤ yi ≤ xi

for each i = 1, 2, . . . , �.

Figure 17. One of the 5 · 3 · 1 · 2 members of the subset of S11 generated by the composition (5, 3, 1, 2) of11.

4. THE COMPLETE SET. Our final task is to prove the following, confirming oursecond main goal (equation (2)).

Theorem 2. There are precisely sn(sn + 1)/2 nets of the n-antiprism. That is, tn =F2n(F2n + 1)/2.

The above formula entices us to look for a way to take each pair† of members of Sn

and somehow create from the two symmetric nets a unique member of Tn. We will doexactly that. When the pair consists of two identical members of Sn, our scheme willreturn that very same net; when the pair consists of distinct elements, our constructionwill return a member of Tn \ Sn.

Proof of Theorem 2. Let N1 and N2 be any two nets, not necessarily distinct, chosenfrom Sn. We will first construct from these a unique net N3 ∈ Tn, thereby determiningan injective function V : S (2)

n → Tn, that is, an injective function V from the set of allpairs from Sn into Tn. After that we will show that V is invertible, which will finishthe proof.

Case 1: N1 and N2 have equal neck sizes. When the two selected nets have the sameneck size, we simply take the right side of N1 and the left side of N2 and join them toform N3. Refer to Figure 18, where these right and left sides are shaded. By “right”and “left” sides we simply mean the partial arrow-grams on those respective sides ofvertical lines drawn through the centers of the necks on their respective arrow-grams.

Figure 18. Two to one—same neck sizes.

Clearly, the resulting net N3 will be symmetric if and only if N1 = N2.†By a pair of elements from a set X we mean a 2-multiset of X, as in [14]. Loosely speaking, for a non-

negative integer k, a k-multiset of a set X is a collection of k elements from X with repetitions (multiplicities)allowed.

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It remains to be verified that the resulting arrow-gram N3 is truly that of a net ofsome n-antiprism. In Case 1 it is fairly easy to see, since we have not added a wrap-around edge. The neck is preserved, and every face retains some connection to a head,because it had one already on its respective side of the original arrow-gram. All thearrow-gram criteria for a net are met.

Case 2: N1 and N2 have different neck sizes. We may assume that the respectiveneck sizes are h1 and h2 with h2 > h1. Let �h = h2 − h1. A twist (literally) occurswhen �h is odd. See Figures 19 (even �h) and 20 (odd �h) for examples, after whichthe details are given.

Figure 19. Two to one—differing neck sizes, even �h; h2 = 3 > 1 = h1.

Figure 20. Two to one—differing neck sizes, odd �h; h2 = 2 > 1 = h1.

Notice that the arrow-grams of the nets denoted by N3 in Figures 19 and 20 eachhave a wrap-around edge. (It is important here to recall that a neck-centered arrow-gram of a symmetric net can never have a wrap-around edge.)

(Step 1 of 3.) Start by connecting the two sides as in Case 1, using the right sideof N1, the left side of N2. (When �h is odd, an illegal arrow-gram is the result of thisstep. When �h is even, a legal arrow-gram is obtained for the same reasons as givenin Case 1, but it turns out that the results are then not unique.) See Figure 21, whichbegins with arrow-grams for N1 and N2 from Figure 19.

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Figure 21. Two to one, even �h, step 1 of 3.

(Step 2 of 3.) Shift and compress the left side of this new arrow-gram (the side withthe long neck parts) toward the center by removing �h vertices (and edges), starting atthe center. These vertices and edges come only from the neck and their removal resultsin a neck size of h1. The remainder of the left side will be shifted right (and flipped if�h is odd) by �h places. It also means there are now �h too few edges and vertices.See Figure 22.

Figure 22. Two to one, even �h, step 2 of 3.

(Step 3 of 3.) Add �h isolated band vertices at the far left ({3, 8} in the example) torefill the band. Add a wraparound edge and starting there, add �h edges from leftto right. See Figure 23, which is the final result. In each case, the symmetric netsN1, N2 ∈ Sn combine to form a net N3 with neck size of min(h1, h2) and lapel sizesk1 and k2. (The lapels were not altered in the construction.)

Figure 23. Two to one, even �h, step 3 of 3.

There is no danger of making a second connection between the two heads with thesenew edges, since the lack of the wrap-around edge in the original N2 results in a breakat the position h2 − h1 places from the left. But the new set of edges, which includes a

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new wrap-around edge, will connect the replaced vertices to a head on the right. Theresulting N3 satisfies the arrow-gram criteria and is clearly asymmetric.

Not only that, but the result is unique. To see this we will briefly describe an injec-tive function � : Tn → S (2)

n . To that end, start with any net N ∈ Tn. Construct thearrow-gram of N , centering its neck.

If there is no wrap-around edge, then split the arrow-gram into two pieces at theneck’s center. Convert each of these pieces to a member of Sn via symmetry. Thenecks will clearly be of the same size. The only way both of the symmetric results willbe identical is if N ∈ Sn.

If there is a wrap-around edge, then it connects to exactly one head. Orient thearrow-gram so that the vertex (band face) that connects to that head is to the right ofthe center. Then there is now a chain of edge-vertex pairs, starting from the left at thewrap-around edge and moving right, that terminates at some vertex. Take the numberof edges in this chain to be h2 − h1. Setting h1 to be the necksize of N , both h1 and h2

are thereby determined, with h2 > h1. Again chop the arrow-gram in two by cuttingthe neck at the center, but keep the wrap-around edge with the left piece. The rightand left pieces will become symmetric nets N1 and N2 with neck sizes h1 and h2,respectively. Remove h2 − h1 edges/vertex pairs from the left side of the left piece andjoin them to the right end of the left piece (to become part of a neck of size h2). Fromthese right and left pieces create two members of Sn via symmetry, as before.

This process described in the previous two paragraphs describes an injective func-tion � mapping Tn into the set S (2)

n of pairs from Sn. We had already described aninjective function V that goes the other way, and this implies that the cardinalities ofTn and S (2)

n are equal. (In fact, V −1 = �.) Our theorem follows.

5. MISCELLANEOUS. There is plenty more counting to be done, and we will justmention a few directions.

5.1. Other proofs. The generation of our antiprism nets can be accomplished in manyother ways. Even in the spirit of this article, there are surely other ways to combinepairs of symmetric nets of Sn to form the entire family Tn. It would be nice to see onethat is a bit simpler than ours.

As mentioned earlier, the online supplement to this article contains a completelydifferent, self-contained proof of Lemma 1, using mathematical induction. It employsgrafting and transmogrifying tricks like those in Section 4, and generates the membersof Sn+1 directly from those of Sn.

A different idea, based on counting certain strings, has been applied to collect allthe nets of antiprisms and other “banded” polyhedra, such as prisms, cupolas, andmore [9], so the story is still unfolding.

Another approach may have already been used. After the author made the discov-eries of the main formulas and constructions presented here, an internet search turnedup mention of a “refereed but unpublished” paper [10] on the enumeration of nets ofprisms, antiprisms, and pyramids, but attempts to retrieve that paper were unsuccess-ful.

Recently, Horiyama and Shoji [8] have made great strides, giving the numbersof nets of Platonic solids (which were known), all the Archimedean solids andJohnson–Zalgaller solids, along with formulas for the numbers of uniform prismsand antiprisms. It should be emphasized that unlike the methods in the present articleand [9], no enumerations are given in [8], just the counts. (The difficulties with enu-meration are mentioned.) See [9] for a bit more about that and for references to muchearlier results regarding nets of Platonic solids.

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5.2. Path nets. How many path nets of the n-antiprism are there? Figure 24 showsthe 27 nets for n = 5 whose spanning trees in G5 are simply paths. (Such nets arecalled “hamiltonian” in [13], “single-chain” in [5], and “serpentine” in [2].) It is easyto show there are 2n − 2 symmetric path nets in Tn when n ≥ 2. It turns out that thetotal number Pn of path nets in Tn is (3n2 − 5n + 4)/2. One way to prove this is byfirst showing that the number Pn(h) of such path nets with neck-size h is given byPn(n) = 1 (obvious), Pn(n − 1) = 2 (fairly obvious), and Pn(n − j) = 3j − 1 when1 < j < n (not as obvious).

Figure 24. The 27 path nets for n = 5, the first 8 of which are symmetric.

See the subsection “Path nets” in the supplement for proofs and details.

5.3. Pyramids and prisms. The title of the paper [10] refers to enumerations ofthe nets of antiprisms, but also to those of prisms and pyramids. Figures 25 and 26give a small sampling. In [9], Fibonacci numbers appear in connection with pyramids,namely that the number of nets of pyramids with regular n-gonal bases and congruentisosceles triangles comprising the other sides is given by

−1 + Fn

2+ 1

2n

n∑j=1

(F2 gcd(n,j)+1 + F2 gcd(n,j)−1).

Furthermore, the number of such nets with bilateral symmetry is precisely Fn.

Figure 25. Some of the 320 nets of the 9-pyramid.

For the nets of regular prisms the author has also discovered recursive constructions,along with proofs, similar to those presented throughout the present article (namely, viavarious dismemberings and rememberings). If the regular prisms seem to be simplerobjects than the antiprisms while the following formulas for the counts seem morecomplicated for the regular prisms, notice that with the regular prisms there are netswith bilateral symmetries as well as those with point symmetries. When n is odd thereis a net having both symmetries. Here we simply record the formulas.

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Let an denote the number of distinct nets of the n-prism, and let pn and bn denotethe number of those having, respectively, point symmetry and bilateral symmetry. Thenp2 = 1, b2 = 2, p3 = b3 = 3, and for n ≥ 4,

bn ={

bn−1 + pn−1 if n is evenbn−1 + bn−2 − (n − 3)/2 if n is odd

and

pn ={

bn − n/2 if n is evenbn + bn−1 − (n − 1) if n is odd.

Furthermore,

an = 1

4

(�n

n+ 2pn + 2bn − (n mod 2)

)

for n ≥ 2, where �n = nTn(2) is the number of distinct labeled nets of the prism andTn(x) is the Chebyshev polynomial of the first kind. (See [9] for another approachleading to equivalent, but different-looking formulas.)

Figure 26. Some nets of the 10-prism.

ACKNOWLEDGMENTS. The author thanks the Sklar Foundation for generous support during the years2010–2015. This paper developed during the last few years of the author’s tenure (1989–2015) as a full-timemember of the Department of Mathematics at LSU Shreveport. The straightening out of creases in some of theideas presented in the paper is thanks to the great numbers (not to be confused with large numbers) of studentand faculty participants in the department’s long-running “Friday Seminars” (especially the ever-present, ever-watchful Professor Zsolt Lengvarszky). Visitors to the resulting talks during “Jim Roberts Week at LSUS”in 2013 and 2015 are thanked and applauded for their endurance. On that same note, the author thanks thereferees and editors for their guidance and patience over the course of several revisions.

REFERENCES

[1] Alexandrov, A. D. (2005). Convex Polyhedra. Springer Monographs in Mathematics (Dairbekov, N. S.,Kutateladze, S. S., Sossinsky, A. B., trans.) Berlin: Springer.

[2] Demaine, E., Demaine, M., Hart, V., Iacono, J., Langerman, S., O’Rourke, J. (2011). Continuous bloom-ing of convex polyhedra. Graphs Combin. 27(3): 363–376.

[3] Gessel, I. M., Li, J. (2013). Compositions and Fibonacci identities. J. Integer Seq. 16(4): Art. 13.4.5, 16pp.

[4] Ghomi, M. (2018). Durer’s unfolding problem for convex polyhedra. Notices Amer. Math. Soc. 65(1):25–27.

[5] Grunbaum, B. (1991). Nets of polyhedra II. Geombinatorics. 1(3): 5–10.[6] Grunbaum, B. (2002). No-net polyhedra. Geombinatorics. 11(4): 111–114.[7] Hoggatt, Jr., V. E., Lind, D. A. (1969). Compositions and Fibonacci numbers. Fibonacci Quart. 7: 253–

266.[8] Horiyama, T., Shoji, W. (2013). The number of different unfoldings of polyhedra. In: Cai, L., Cheng,

S.-W., Lam, T.-W., eds. Algorithms and Computation. Lecture Notes in Computer Science, Vol. 8283.Heidelberg: Springer, pp. 623–633.

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[9] Lengvarszky, Z., Mabry, R. (2017). Enumerating nets of prism-like polyhedra. Acta Sci. Math. (Szeged).83(3–4): 377–392.

[10] Lloyd, E. K., Meredith, G. H. F. Enumeration of face nets for pyramids, prisms and antiprisms, unpub-lished manuscript.

[11] Moser, L., Whitney, E. L. (1961). Weighted compositions. Canad. Math. Bull. 4: 39–43.[12] O’Rourke, J. (2011). How to Fold It: The Mathematics of Linkages, Origami, and Polyhedra. Cam-

bridge, MA: Cambridge Univ. Press.[13] Shephard, G. C. (1975). Convex polytopes with convex nets. Math. Proc. Cambridge Philos. Soc. 78(3):

389–403.[14] Stanley, R. P. (2012). Enumerative Combinatorics, Vol. 1. Cambridge Studies in Advanced Mathematics,

Vol. 49, 2nd ed. Cambridge, MA: Cambridge Univ. Press.[15] Weisstein, E. (2018). Net—From MathWorld, a Wolfram Web Resource. mathworld.wolfram.com/

Net.html

RICK MABRY is professor emeritus at LSU Shreveport, now living in the city of Kempten (im schonenAllgau) in Bavaria. His main goal is to catch up on mathematical paper writing and musical interests, weatherpermitting. (Nice weather often curtails these activities in favor of bike riding and picture taking in the hillyfarmlands and nearby towns.) He hopes to make many trips to Shreveport during spring semesters, timed tovisit the LA/MS MAA Section Meeting. (Some of the results in this article were presented as a contributedpaper during the 2018 meeting on the beautiful campus of the University of Louisiana at Lafayette.)Department of Mathematics, Louisiana State University Shreveport, Shreveport, LA 71115-2399, USARichard.Mabry@LSUS.edu

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