1. Introduction

Flexagons were discovered in 1939 by topologist Arthur Stone. A regular flexagon is one that contains 9n equilateral triangular regions on a straight strip of paper. This paper is then rolled into smaller strips of paper and finally into a hexagon with 6 triangular regions called pats, ‘ producing one mathematical face. The pinch flex removes the uppermost triangular regions and replaces them with a new set producing a new face. The flexagon is said to have order 3n because you can color 3n of the faces with 3n different colors. It is well known that when only the pinch flex is used, a flexagon of order 3n is a möbius band with 3(3n?2) half-twists, and has 6n?3 different mathematical faces. Even though a colored face appears more than once, the uppermost triangles might be rotated producing a different mathematical face. When T. Bruce McLean described the V-flex on the flexagon of order 6 in 1979, he showed that it now had 3420 mathematical faces and provided a graph that demonstrated how to reach all of the different faces. This flex scrambles the colors similar to the way the Rubik’s cube does except that a flexagon is flat. It is the purpose of this paper to provide an algorithm that counts the number of mathematical faces for flexagons of order 3n for all n, once the V-flex is included. We will construct a flexagon, with an order of 6. Then we will present the new results that will include an algorithm that counts the mathematical faces of all regular flexagons when the pinch and the V-flex are used.

2. Rationale

3. Construction

Flexagons constructed from straight strips of paper are called regular. It is well known that all regular flexagons have order 3n, n>0, contain 9n equilateral triangles. In this exploration, I will provide an algorithm that counts the number of faces when both the pinch flex and V-flex is included. To demonstrate, we will construct a regular flexagon of order 6, a hexahexaflexagon, by starting with 18 equilateral triangles on a straight strip of paper with a an extra equilateral triangle that we will use to stick the ends together.Hence there will be a total of 19 equilateral triangles. We will number the triangles from 1 to 18 on the front as shown in Fig .1 and the same on the back, but with an apostrophe. This side with apostrophes will be called the prime side

Fig. 1.

By numbering the triangles, it will help construction to be easier as it would be easier to orientate the starting strip. Start by folding the 1′ into 2′, and 3′ into 4′ so on and so forth. Continue to do so until you end up with a straight strip that has 9 triangular regions as shown in Fig. 2 and note that only non-primed side is visible. Complete it as you would for a normal hexaflexagon by folding 7 into 4, and 13 into 10

Fig. 2.

Next, tuck 1 behind 16 and you would find that there will be a free triangle, 18. It would be easy to tape or glue it to 1 which while result in the completion of the hexaflexagon, as shown in Fig 5.

Fig 3.

4. Notation

Putting an apostrophe was to simplfy the construction of the flexagon, but now we will have no use of it and will not treat them as different. We record this beginning face as

F=(Pi)=(P1–P2–P3–P4–P5–P6)

P1=(4 3 , 6 5)

P2=(9 10 7 8 , 13 14 11 12)

P3=(16 15 , 18 17)

P4=(11 12 13 14 , 7 8 9 10)

P5=(2 1 , 4 3)

P6=(15 16 1314 , 1 2 17 18)

This shows the six triangular regions separated by dashes and we will call this a face. Only the 6 left most numbers are visible on the top of the face. Each one of the triangular regions is called a pat. If a pat contains more than one triangle, a comma is placed where the pat splits naturally and this location is called a thumb hole. Two pats that are consecutive in the counter clockwise rotation will be called adjacent and ?p will refer to a pat p turned upside down. If p= 16 15, 18 17 then ?p= 17 18 , 15 16 . We will never have pats with a multiple of three triangles, so we will use odd to mean 3k+1 and even to mean 3k+2. For the definition of a pat, we will do it inductively by starting with pats of degree one.

Definition 1 The Pinch flex

The pinch flex is basically pinching the flexagon into three thirds as seen in Fig. 6 which will which will allow us to flex to a new side as shown in the diagram.

Fig 4.

Definition 2

If p is an ordered set of one natural number, then p is a pat of degree 1 and we will denote that fact by saying D(p)=1.

Definition 3 Continuation of Pat

If s and t are adjacent pats, where both are of odd degree or both are of even degree, then p=?s,?t is a pat of degree

D(p)=D(s)+D(t).

As an example, 2, 3 and 4, 5 are adjacent pats of degree 2, so 32, 54 is a pat of degree 4. Returning to the starting face, we will say that P1= 4 3, 6 5 is in position one and then consecutively number the positions from left to right. This corresponds to placing the uppermost 4 at the top and numbering the other five positions counter clockwise on the flexagon. Thus P4= 11 12 13 14 , 7 8 9 10 is in position 4 at 6 o’clock or straight down. Two pats of the same degree will be pat-equivalent if there is an order preserving one-to-one function from one pat to the other. Pats P1, P3, and P5 are in the same pat-equivalence class, each of degree 4, while P2, P4, and P6 are also in the same pat-class of degree 8.

Pinching down on the 3 o’clock crease between p5 and p6 and pushing in from the external 9 o’clock point that joins P2 and P3 takes you from face 1, shown inside 2 circles to face 2 in Fig . 5. Using the pinch flex, it will allow you to travel from a face to another face. Pinching the flexagon at 30° without rotating the flexagon, will allow one to travel in a loop. For example, 1 > 2 > 3. However in order to obtain other faces, one must rotate the flexagon 60° and apply the pinch flex. This is clearly indicated on the Tuckerman Traverse in Fig. 5. Continuing to follow the traverse allows you to visit the 6 different mathematical faces.

Fig. 5.

Definition 3

If p is a pat of degree D(p)>1, then ?a+b? is a partition of D(p) means that a and bare both odd or both even and a+b=D(p).

For example a pat of degree 5 has 2 partitions. It can be a pat of degree 4 on top of a pat of degree 1 or vice versa and we will say that the pat is wound 4–1 or 1–4. Given a pat p with degree D(p), then h(D(p)) will be the number of different partitions of D(p), and N(D(p)) will be the number of different pat-equivalence classes of degree D(p).

Table 1.

D(p)

All partitions of D(p)

h(D(p))

N(D(p))= Number of pat classes

Examples of p

1

None

0

1

1 2 3 4 5

2

1 + 1

1

1

1, 2 2, 3 3, 4 4, 5

4

2 + 2

1

1

2 1, 4 3 6 5, 8 7 4 3, 6 5

5

4 + 1; 1 + 4

2

2 = 1 + 1

3 4 1 2, 5 4, 7 8 5 6

7

5 + 2; 2 + 5

2

4 = 2 + 2

5 2 1 4 3, 7 6 3 2, 6 5 8 7 4

8

7 + 1; 4 + 4; 1 + 7

3

9 = 4 + 1 + 4

6 7 3 4 1 2 5, 8

10

8 + 2; 5 + 5; 2 + 8

3

22 = 9 + 4 + 9

8 5 2 1 4 3 7 6, 10 9

11

10 + 1; 7 + 4; 4 + 7; 1 + 10

4

52 = 22 + 4 + 4 + 22

9 10 6 7 3 4 1 2 5 8, 11

5. Operations

To present the definitions of the pinch flex and the V-flex, we first define reversing and same functions r1 and s defined on reversed pats and the reversing function r2 defined on pats. If ?p is a pat, define r1(p)=?p, s(p)=p, and r2(?p)=p.

Definition 4

Let f=(pi) be a face where D(pi)?1, and pi=qi,ki for i?{1,3,5}. Then define the pinch flex as P(q1,k1?p2?q3,k3?p4?q5,k5?p6)=r1(q1)?s(k1),r2(p2)?r1(q3)?s(k3),r2(p4)?r1(q5)?s(k5),r2(p6).

As an example, P (4 3, 6 5–9 10 7 8, 13 14 11 12–16 15, 18 17–21 22 19 20, 25 26 23 24–28 27, 30 29–33 34 31 32, 1 2 35 36) = 3, 4–6 5, 12 11 14 13 8 7 10 9–15, 16–18 17, 24 23 26 25 20 19 22 21–27, 28–30 29, 36 35 2 1 32 31 34 33.

The V-flex was defined by B. McLean in 1979. With this twist being added to the operations, the faces become scrambled just as they do in the Rubik’s cube except that a flexagon is flat.

Definition 5

Let f=(pi) be a face where D(pi)?1, and pi=qi,ki for i?{1,4,5}. Then define the V-flex as V(q1,k1?p2?p3?q4,k4?q5,k5?p6)=r1(q1)?s(k1),r2(p2)?r2(p3),s(q4)?r1(k4)?r1(q5)?s(k5),r2(p6).

Note that the definition agrees with the pinch flex for pats in positions 1, 2, 5, and 6. As an example, V(4 3, 6 5–9 10 7 8, 13 14 11 12–16 15, 18 17–21 22 19 20, 25 26 23 24–28 27, 30 29–33 34 31 32, 1 2 35 36) = 3, 4–6 5, 12 11 14 13 8 7 10 9–17 18 15 16, 21 22 19 20–24 23, 26 25–27, 28–30 29, 36 35 2 1 32 31 34 33.

Definition 6

If f=(p1–p2–p3–p4–p5–p6) and g=(q1–q2–q3–q4–q5–q6) are faces, then f is face-equivalent to g means that g can be rotated to g?=(r1–r2–r3–r4–r5–r6) and pi is pat-equivalent to ri for each i, 1?i?6.

Definition 7

If f=(p1–p2–p3–p4–p5–p6) and g=(q1–q2–q3–q4–q5–q6) are faces, then f is math-equivalent to g means that g can be rotated to f.

We will call each face-equivalence class an initial face. Faces that are different except for rotation will be called different mathematical faces. A mathematical face can be distinguished by its upper most triangles or by the way its pats are wound underneath.

6. The count

We begin by reviewing the axioms before the V-flex is included. If n is a natural number and f=(pi)=(p1–p2–p3–p4–p5–p6) is a face of order 3n, then

1.

the sum of the degrees of the 6 pats is 9n,

2.

no pat is a degree of a multiple of 3, and

3.

the sum of the degrees of two adjacent pats is 3n.

Once the V-flex is added, the last axiom is weakened from 3n to a multiple of three. Thus many more faces are possible when both flexes are used. To outline the count, we will count cases, based only upon the degrees of the pats, count initial faces, and finally count mathematical faces.

Theorem 1

Ifh(p)is the number of partitions ofD(p), thenh(p)=?D(p)+13?.

Proof

There are two cases for D(p), 3|D(p)+1 or 3|D(p)+2. For the first case notice that the partitions are 1+(D(p)?1),4+(D(p)?4),7+(D(p)?7),…,(D(p)?1)+1. Notice that we can write them as in the following: 1+(D(p)?1),(1+3)+(D(p)?4),(1+(3)2)+(D(p)?7),…,(1+3(D(p)?23))+1. So as is seen, we get h(p)=D(p)?23+1=D(p)+13. Also notice that since D(p)+13 is an integer we get ?D(p)+13?=D(p)+13. Thus in this case h(p)=?D(p)+13?.

For the second case, 3|D(p)+2, the partitions are 2+(D(p)?2),5+(D(p)?5),…,(D(p)?2)+2. Again in this case we can write them as in the following: 2+(D(p)?2),(2+1(3))+(D(p)?5),…,(2+3(D(p)?43))+1. Thus we get h(p)=(D(p)?43)+1=D(p)?13. Now in this case D(p)?13 is an integer thus: h(p)=D(p)?13=?D(p)?13?=?D(p)?13?+0=?D(p)?13?+?23?. But since D(p)?13 is an integer, ?D(p)?13?+?23?=?D(p)?13+23?=?D(p)+13?. Thus in the second case we get h(p)=?D(p)+13? and the proof is complete. ?

Looking back at Table 1, the pattern for h(p) is straight forward and we offer Theorem 2 as justification.

Theorem 2

If we denote the number of pats forD(p)byN(D(p)), thenN(1)=1and

N(D(p))=?i=0D(p)?23N(1+3i)N(D(p)?(1+3i))where3|D(p)+1and

N(D(p))=?i=0D(p)?43N(2+3i)N(D(p)?(2+3i))where3|D(p)+2.

Proof

We prove it for the first case and the second case is similar. By the previous theorem we have the partitions, 1+(D(p)?1),(1+3)+(D(p)?4),(1+(3)2)+(D(p)?7),…,(1+3(D(p)?23))+1.

If we write each case (1+3i)+(D(p)?(1+3i)), where i=0,1,…,D(p)?23, then if the number of pats for 1+3i is N(1+3i) and the number of pats for D(p)?(1+3i)is N(D(p)?(1+3i)). We get N(1+3i)N(D(p)?(1+3i)) for this case and since i changes from 0 to D(p)?23 we get N(p), the summation of all of the cases, which is N(p)=?i=0D(p)?23N(1+3i)N(D(p)?(1+3i)). The proof for the other case is similar. ?

The number of pat classes for D(p)=29 is N(29)=?i=09N(1+3i)N(29?1+3i)=8,579,560 where we used a recursive formula that matches Theorem 2 and provide a Java applet located at: http://math.georgiasouthern.edu/~bmclean/flex/index.html.

Note that the sum progresses through all of the partitions of 29, 1+28, 4+25,…,28+1. Since the other 5 pats in our representative face can only have 1 pat-class each, there are also 8,579,560 different initial faces that are wound as (1–2–1–2–1–29). For the trihexaflexagon, we only need D(p)? {1, 2}, but for n=4, the dodecahexaflexagon, D(p) can be any non-multiple of 3 up to and including 29. When working with a regular flexagon of order 3n, define

E(n)={(x1,x2,x3,x4,x5,x6)?N6|?i=16xi=9n,3?xi,1?i?6,3|xi+xi+11?i?5, and 3|x6+x1}.

Theorem 3

If(x1,y1,x2,y2,x3,y3)?E(n), thenxi,yi?9n?7.

Proof

To find the maximum of y3 we need to minimize x1,y1,x2,y2,x3. There are two cases for xi, 3|xi+1 or 3|xi+2.

If 3|xi+1, then the minimum of xi is 1 for i=1,2,3. In this case 3|yi+2, so the minimum of y1 and y2 will be 2. Therefore x1+y1+x2+y2+x3?1+2+1+2+1=7. Therefore y3?9n?7.

If 3|xi+2 then the minimum of xi is 2 for i=1,2,3. In this case 3|yi+1, so the minimum of y1 and y2 will be 1. Therefore x1+y1+x2+y2+x3?2+1+2+1+2=8. Therefore y3?9n?8. So xi,yi?9n?7. ?

We summarize these statements in Table 2.

Table 2.

n

Number of triangles

N(k)

k= Largest pat needed

Flexagon name

1

9

1

2

Trihexaflexagon

2

18

52

11

Hexahexaflexagon

3

27

17,710

20

Nonahexaflexagon

4

36

8,579,560

29

Dodecahexaflexagon

?

?

?

?

?

n

9n

Theorem 2

9n?7

Three-n-ahexaflexagon

Insist for a while that 3|x1+2 and define

L(n)={(x1,y1,x2,y2,x3,y3)?E(n)|xi?{1,4,…,9n?8},yi?{2,5,…,9n?7},and(y1+y2+y3)=9n?(x1+x2+x3)}.

Next define

K(n)={(x1,y1,x2,y2,x3,y3)?L(n)|x1=x2=x3andy1=y2=y3}.

It is easy to see that |K(n)|=n and as an example, K(4)= {(1, 11, 1, 11, 1, 11), (4, 8, 4, 8, 4, 8), (7, 5, 7, 5, 7, 5), (10, 2, 10, 2, 10, 2)}.

Here C(n) is the set of cases that must be looked at after defining an equivalence relation on E(n) that ignores the rotation of the elements of E(n), so C(n)=E(n)/?. There are two kinds of case-equivalence classes that make up C(n), those that have 2 elements and those that have 6 elements and will be denoted by K(n)/? and (L(n)?K(n))/? respectively. The total number of cases is |C(n)|=26|L(n)?K(n)|+22|K(n)|=13|L(n)?K(n)|+n where we multiply both terms by 2 because L(n) did not allow for the first number to be even and we divided by the number of elements in each case-equivalence class.

Now we count the number of initial faces. For each of the cases that we have set up, we must now take into account the number of ways that the pats can be wound. Again we will need two cases. The initial faces based upon the cases in L(n)?K(n) will be called F6(n) and those arising from K(n) will be called F2(n). For the cases in L(n)?K(n) there cannot be any kind of symmetry so we just apply the fundamental theorem of counting to get |F6(n)|=13?(x1,x2,x3,x4,x5,x6)?L(n)?K(n)?i=16N(xi).

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