Double Factorial Binomial Coefficients Mitsuki Hanada Submitted in Partial Fulfillment of the Prerequisite for Honors in The

Double Factorial Binomial Coeﬃcients

Mitsuki Hanada

Submitted in Partial Fulﬁllment of the Prerequisite for Honors in the Wellesley College Department of Mathematics under the advisement of Alexander Diesl

May 2021 c 2021 Mitsuki Hanada ii Double Factorial Binomial Coeﬃcients

Abstract Binomial coefficients are a concept familiar to most mathematics students. In particular, n the binomial coefficient k is most often used when counting the number of ways of choosing k out of n distinct objects. These binomial coefficients are well studied in mathematics due to the many interesting properties they have. For example, they make up the entries of Pascal’s Triangle, which has many recursive and combinatorial properties regarding different columns of the triangle. Binomial coefficients are defined using factorials, where n! for n ∈ Z>0 is defined to be the product of all the positive integers up to n. One interesting variation of the factorial is the double factorial (n!!), which is defined to be the product of every other positive integer up to n. We can use double factorials in the place of factorials to define double factorial binomial coefficients (DFBCs). Though factorials and double factorials look very similar, when we use double factorials to define binomial coefficients, we lose many important properties that traditional binomial coefficients have. For example, though binomial coefficients are always defined to be integers, we can easily determine that this is not the case for DFBCs. In this thesis, we will discuss the different forms that these coefficients can take. We will also focus on different properties that binomial coefficients have, such as the Chu-Vandermonde Identity and the recursive relation illustrated by Pascal’s Triangle, and determine whether there exists analogous results for DFBCs. Finally, we will generalize some of our results regarding the form of DFBCs to m-factorial binomial coefficients for arbitrary m ∈ Z>0. Mitsuki Hanada iii

Acknowledgements I would like to thank my advisor, Professor Diesl, for guiding me through this journey of a thesis, as well as all the wonderful professors I have had at Wellesley who have shaped me into the mathematician I am today. In particular, I thank Professor Hirschhorn, who has supported me from Math 120 in my first semester to my first graduate level course, Professor Lange and her Math 309 class which helped me discover new mathematical interests, and Professor McAskill of the Physics Department for helping me explore mathematics through the lense of physics. I would also like to thank Professor Chan, Professor Volić,and Professor Battat for being a part of my thesis committee and reading through this thesis. I thank my parents who have always let me pursue my intellectual interests and my sisters who let me ramble about math and pretend to understand. I am so lucky to have such a supportive and loving family. I also thank my quarantine roommate Woo Young, for helping me work through many of the proofs and mathematical arguments in this thesis, as well as dealing with my general thesis/grad school application stress. Finally, I would like to thank all of the friends I have made throughout my years at Wellesley. I will always cherish our late night problem set sessions in Clapp and Freeman 4th floor kitchen. iv Double Factorial Binomial Coefficients

Contents Abstract ii Acknowledgements iii 1. Background1 1.1. Binomial Coefficients1 1.2. Double Factorials and Double Factorial Binomial Coefficients (DFBCs)2 2. Different forms of DFBCs3 2.1. n even, k odd6 2.2. n odd7 3. Combinatorial Interpretation of DFBCs (Chu-Vandermonde identity)9 4. Pascal’s Triangle 13 4.1. Introducing Pascal’s Triangle and its properties 13 4.2. Finding similar triangles for certain DFBCs 15 5. Generalization to m-Factorial Binomial Coefficients 19 References 26 Mitsuki Hanada 1

1. Background

We will start this thesis by introducing our topic of interest: double factorial binomial coefficients. However, before we do this we must define some basic concepts that help us build the definition of double factorial binomial coefficients.

1.1. Binomial Coefficients. We will first define binomial coefficients and some of their interesting properties. The factorial is a well-known concept in mathematics, where the factorial of a positive integer n denoted (n!) is defined: n! := n(n − 1)(n − 2) ··· (2)(1) and 0! := 1 for n = 0. Recall that these factorials carry some combinatorial meaning: for a nonnegative integer n, it follows that n! denotes the number of permutations of n distinct objects. Though there are many different ways of defining binomial coefficients, the definition we give below involves factorials. n Definition 1.1. For n, k ∈ Z≥0 such that 0 ≤ k ≤ n, the binomial coefficient (denoted k ) is defined n n! := . k k!(n − k)!

Most people who have taken an introductory combinatorics course will be familiar with binomial coefficients. The binomial coefficient has many inherent combinatorial properties, some that are apparent through alternative definitions. We will give one alternative definition that highlights some of these properties.

Definition 1.2. For n, k ∈ Z≥0 such that 0 ≤ k ≤ n, the binomial coefficient is defined to be the coefficient of the term xkyn−k in the polynomial expansion n X n (x + y)n = xkyn−k. k k=0

When we choose k many of the (x + y) terms on the left hand side to take the x value from and take the y value from the remaining n − k terms, the resulting product is xkyn−k. n k n−k Therefore the coefficient k of x y must denote the number of ways we can choose k many of the n total (x + y) terms. This is one of the most well known combinatorial properties of the binomial coefficients: n Lemma 1.3. The binomial coefficient k denotes the number of ways to choose k many of the n total distinct objects.

n Since we have that k counts the number of cases, this implies that these binomial coefficients will always be integer values. This is in fact true and another well known property of binomial coefficients. n Lemma 1.4. For any n, k ∈ Z≥0 such that 0 ≤ k ≤ n, it follows that k ∈ Z>0. 2 Double Factorial Binomial Coefficients

There are many diﬀerent ways of proving this statement using combinatorial, algebraic, or number theory properties. (Many of them are mentioned in [2].) We will mention one method of proving this in Section2.

We can see another well known property of binomial coefficients from both of these definitions. Lemma 1.5. Binomial coefficients are symmetric: for n, k such that 0 ≤ k ≤ n, the equation n n k = n−k always holds.

Proof. This follows from the fact that both deﬁnitions we have given depend symmetrically on both k and n − k. Using Deﬁnition 1.1, we can observe the following: n n! n! n = = = . k k!(n − k)! (n − (n − k))!(n − k)! n − k

1.2. Double Factorials and Double Factorial Binomial Coefficients (DFBCs). Now that we have discussed different properties of binomial coefficients, we will define our topic of interest: double factorial binomial coefficients. The definition of double factorial binomial coefficients utilizes double factorials instead of the standard factorial. Definition 1.6. The double factorial of a non-negative integer n (denoted n!!) is defined  n(n − 2)(n − 4) ··· (4)(2) for n even, n > 0  n!! := n(n − 2)(n − 4) ··· (3)(1) for n odd 1 for n = 0.

These double factorials have many interesting properties. In [3], the authors discuss how certain double factorials have an enumerative combinatorial meaning. For example, the two colored permutation of the set [n] = {1, 2, . . . , n} for positive integer n is deﬁned to be a permutation of [n] where each of the entries are colored by one of two colors. The even double factorial (2n)!! = 2n(n)! denotes the number of possible two colored permutations of [n]. Another example mentioned is for the odd double factorials (2n − 1)!!, which represents the number of matchings in the set [2n] = {1, 2,..., 2n}, where a matching of [2n] is deﬁned to be a partition of [2n] into n unordered pairs. There are also other combinatorial interpretations of double factorials, like those mentioned in [1].

We can also simplify the double factorial n!! using a “standard” factorial for certain choices of n: Lemma 1.7. For an even positive integer 2m, we have (2m)!! = 2m(m)!.

Proof. This comes from simple manipulation of the the terms. Note that we can write (2m)!! = (2m)(2m − 2)(2m − 4) ··· (2). Mitsuki Hanada 3

Note that there are m many terms on the right hand side. If we factor out a factor of 2 from each of them, we get (2m)!! = (2m)(2(m − 1))(2(m − 2)) ··· (2) = 2m(m)(m − 1)(m − 2) ··· 1 = 2m(m!).

Lemma 1.7 will be useful to us later when we are trying to prove results about prime factors of certain double factorials.

Now using these double factorials and the definition of the binomial coefficients, we can finally define double factorial binomial coefficients.

Definition 1.8. For n, k ∈ Z≥0 such that 0 ≤ k ≤ n, the double factorial binomial coefficient n (DFBC) (denoted k ) is defined n n!! := . k k!!(n − k)!!

n n Recall from Lemma 1.5 that binomial coefficients are symmetric: that is, k = n−k . By definition it is evident that this property also holds for DFBCs. Lemma 1.9. Double factorial binomial coefficients are symmetric: for any n, k such that n n 0 ≤ k ≤ n, the equation k = n−k always holds.

This definition and notation of DFBC comes from [3]. Though this paper introduces the concept of DFBCs and discusses different generating functions associated with them, it does not go into depth on the relation between these coefficients and the standard binomial coefficients. In this thesis, we will introduce different combinatorial properties of normal binomial coefficients and see whether there exist analogous properties of DFBCs. We will also discuss what these DFBCs look like, and expand on the idea of DFBCs to m-factorials for any positive integer m.

2. Different forms of DFBCs

One of the most well known properties of the binomial coeﬃcient, which we stated in Lemma 1.4, is that for any pair of integers (n, k) such that 0 ≤ k ≤ n, it follows that n k ∈ Z. This is a crucial property when discussing the combinatorial interpretations of the binomial coeﬃcient.

We will give a proof of this statement using discussion of prime factors. Before giving the proof, we will require a few new deﬁnitions, the ﬁrst being of p-adic valuation.

Definition 2.1. For any n ∈ Z>0 and p prime, we define the p-adic valuation of n (denoted r υp(n)) to be the greatest r ∈ Z≥0 such that p |n. 4 Double Factorial Binomial Coefficients

Note that since these p-adic valuations are equivalent to exponents, we can manipulate them in a similar way to how we manipulate logarithms.

Lemma 2.2. For any a, b ∈ Z>0, it follows that υp(ab) = υp(a) + υ(b).

We will also make use of Legendre’s formula, which gives us the p-adic valuation of factorials. Theorem 2.3. (Legendre’s Formula) For any n ∈ Z>0, the following holds: ∞ X n υ (n!) = . p pi i=1

Before we start our discussion of DFBCs, we will use the following deﬁnition to prove Theorem 2.3. We will also be using this deﬁnition in later lemmas.

Deﬁnition 2.4. For some q ∈ Z>0 and a sequence {a1, a2, . . . , an} where ai ∈ Z≥0, we deﬁne Dq({a1, a2, . . . , an}) to be the number of terms in the sequence {a1, a2, . . . , an} that are divisible by q.

Proof of Theorem 2.3 Let n ∈ Z≥0 and p be some prime. There exists a N ∈ Z≥0 such that for any 0 ≤ i ≤ N i N+1 PN it follows that p ≤ n but p > n. We will ﬁrst show that υp(n!) = i=1 Dpi ([n]) where [n] = {1, 2, . . . , n}. Consider the terms in [n] that are divisible by p. Let us construct a subset [n]1 of [n] of such elements. We can factor out p from each element of [n]1. This gives 0 0 us Dp([n]) factors of p and a new set [n]1. Now consider all terms in [n]1 that are divisible by p. We know that there are Dp2 ([n]) many such elements. Let us construct a subset [n]2 of 0 [n]1 of such elements. We can factor out p from each element of [n]2, which gives us Dp2 ([n]) 0 many factors of p and a new set [n]2. We can continue this N times, until we end up with 0 PN a set [n]N that has no elements divisible by p. Hence we have that there are i=1 Dpi ([n]) many factors of p in n!.

We know that every pi-th elements in [n] is divisible by pi: pi, 2pi, 3pi ... . We can see that j n k i j n k there are pi many block of length p in {1, 2, . . . , n}. Thus, it follows that Dpi ([n]) = pi . i j n k If M < i, then we have that p > n, hence pi = 0. Thus it follows that υp(n!) = N N ∞ P Pj n k Pj n k Dpi (n!) = pi = pi . i=1 i=1 i=1 In our calculations, we will be adding certain ﬂoor functions and will make use of the following fact. Lemma 2.5. For a, b ∈ R, it follows that bac + bbc ≤ ba + bc.

Lemma 2.5 follows from the fact that bac + bbc = ba + bc or ba + bc − 1 depending on if (a − bac) + (b − bbc) ≥ 1 or not. Now, we will use these tools to prove Lemma 1.4. The Mitsuki Hanada 5

following proof is similar to proof 5 in [2], but utilizes notation that we will be using in later proofs.

Proof of Lemma 1.4. n n! We will show that k = k!(n−k)! ∈ Z by showing that for any prime p it follows that υp(k!(n− k)!) ≤ υp(n!). Let p be prime. Note that we can write υp(k!(n − k)!) = υp(k!) + υp((n − k)!). Then, using Theorem 2.3 it follows that

υp(k!(n − k)!) = υp(k!) + υp((n − k)!) ∞ ∞ X k Xn − k = + pi pi i=1 i=1 ∞ X k n − k = + . pi pi i=1

j k k j n−k k We know that for any choice of i, from Lemma 2.5 it follows that pi + pi ≤ j k+(n−k) k pi . Hence, it follows that ∞ ∞ X k n − k X n υ (k!(n − k)!) = + ≤ = υ (n!) p pi pi pi p i=1 i=1

Therefore the claim holds. Lemma 1.4 is a well known result regarding binomial coeﬃcients. However, it is evident that the same statement does not hold for DFBCs. We can see this from the following example. Example 2.6. 4 4!! 4 · 2 8 = = = ∈/ . 1 3!!1!! 3 · 1 · 1 3 Z

The ﬁrst question that comes to mind is whether there exist conditions that guarantee that n k must be an integer. In fact, there is a simple condition that follows from a property of double factorials. 0 0 0 0 Theorem 2.7. For even n, k ∈ Z≥0 where n = 2n , k = 2k for n , k ∈ Z≥0 we have

n n0 = . k k0

Proof. This result follows from Lemma 1.7. Observe that we can rewrite the DFBC using standard factorials, which gives us a traditional binomial coeﬃcient:

n (2n)!! 2n(n)! n = = = . k (2k)!!(2(n − k))!! 2k(k)!2n−k(n − k)! k 6 Double Factorial Binomial Coeﬃcients

Since we know from Lemma 1.4 that binomial coeﬃcients must be integers, the corollary below follows immediately from the lemma: n Corollary 2.8. For even n, k ∈ Z≥0, it follows that k ∈ Z.

However it is more diﬃcult to discuss what happens when these conditions are not met. We will split up the remaining possible DFBCs into two diﬀerent cases: when n is even and k is odd, and when n is odd.

2.1. n even, k odd. In the case where n is even and k is odd, we know from Example 2.6 n n n!! above that there are cases where k is not an integer. Since k = k!!(n−k)!! , where n!! is a product of even numbers and k!!, (n − k)!! are both products of odd numbers, we can see that there are no factors of 2 in the denominator but there are in the numerator. From n a this, we can expect that k is some element b ∈ Q where a is divisible by 2 but b is not. In fact, we can prove an even better result for the forms these coeﬃcients must have:

Theorem 2.9. For n, k ∈ Z≥0 such that n is even and k is odd, it follows that n 2m = k q

where q ≡ 1 (mod 2) and m ∈ Z>0.

In order to prove this, we require the following lemma. Lemma 2.10. For any odd k = 2k0 + 1 , q = 2q0 + 1, it follows that (k0 + 1) + q0 D ({1, 3, 5, . . . , k}) = #(terms in {1, 3, 5, . . . , k} that are divisible by q) = q q

Proof. Let [[k]] = {1, 3, 5, . . . , k} , which is a set of k0 + 1 odd numbers. If q ≤ k, we know that q is a term in [[k]] that is divisible by q. We also know that every q-th term from q in [[k]] (which will have the form 3q, 5q, . . . ) is also divisible by q. Thus, in order to count the number of terms in [[k]] that are divisible by q, we want to count the number of sequences of length q that are contained in the sequence [[k]], when we consider the ﬁrst subsequence {1, 3, . . . , q} of length (q0 + 1) to be a sequence of length q. This is equivalent to counting the number of sequence of length q in a sequence of length (k0 + 1) + (q0), where (k0 + 1) is the length of [[k]] and q0 is the number of additional terms we would need to make the block j (k0+1)+q0 k {1, 3, . . . , q} have length q. This is equivalent to q . If q > k then Dq([[k]]) = 0. Note that if q > k then it follows that q0 > k0. Hence it follows that (k0 +1)+q0 < (q0 +1)+q0 < q, j (k0+1)+q0 k hence q = 0 = Dq([[k]]).

Now that we know that Lemma 2.10 holds, we can prove Theorem 2.9. Mitsuki Hanada 7

Proof of Theorem 2.9 0 0 0 0 Let us denote n = 2n , k = 2k +1 for n , k ∈ Z≥0. By deﬁnition of DFBCs and from Lemma 1.7, we can rewrite: n n!! 2n0 (n0)! = = . k k!!(n − k)!! k!!(n − k)!!

Note that since k, (n − k) are odd, there are no factors of 2 in the denominator. Thus it 0 0 suﬃces to show that for any p 6= 2 such that p ≤ (n ), it follows that υp(n !) ≤ υp(k!!(n−k)!!).

∞ Np j 0 k j 0 k 0 P n P n Np 0 We know that υp(n !) = pi = pi when Np is an integer such that p ≤ n < i=1 i=1 pNp+1. We also know that ∞ X υp(k!!(n − k)!!) = (Dpi ([[k]]) + Dpi ([[n − k]])), i=1 where [[k]] = {1, 3, . . . , k} and [[n − k]] = {1, 3, . . . , n − k}. We will show that for any p 6= 2 j n0 k and 1 ≤ i ≤ Np it follows that pi ≤ Dpi ([[k]]) + Dpi ([[n − k]]).

i pi−1 0 Since p is odd, by Lemma 2.10 (where we denote 2 = pi ∈ Z) 0 0 0 0 0 (k + 1) + pi ((n − k − 1) + 1) + pi D i ([[k]]) + D i ([[n − k]]) = + p p pi pi (k0 + 1) + p0 (n0 − k0) + p0 = i + i pi pi

From Lemma 2.5, it follows that

(k0 + 1) + p0 (n0 − k0) + p0 (k0 + 1) + p0 + (n0 − k0) + p0 i + i ≥ i i − 1 pi pi pi n0 + 2p0 + 1 = i − 1 pi n0 + pi = − 1 pi n0 = . pi

j n0 k 0 Hence it follows that pi ≤ Dpi ([[k]]) + Dpi ([[n − k]]). Thus, it follows that υp(n !) ≤ υp(k!!(n − k!!)).

2.2. n odd. Now let us consider the case where n is odd. Unlike the n even case, there is no need to diﬀerentiate between the k odd and k even cases by Lemma 1.9. After calculating 8 Double Factorial Binomial Coeﬃcients

through some example, like the example below, it is evident that these values are also not necessarily integers. Example 2.11. n 5!! 5 · 3 · 1 5 = = = . k 3!!(5 − 3)!! 3 · 1 · 2 2

Since n!! is a product of odd numbers and either k!! or (n − k)!! is a product of even integers, we can see that there are no factors of 2 in the numerator but there are in the n a denominator. From this, we can expect that k is some element b ∈ Q where b is even but a must be odd. Based oﬀ of this, and our results for the case above, we can make a guess on what these coeﬃcients might look like and try to prove it similarly to how we proved Theorem 2.9.

Theorem 2.12. For n, k ∈ Z≥0 such that n is odd, it follows that n q = k 2m

where q ≡ 1 (mod 2) and m ∈ Z>0.

0 0 Proof. Consider n odd. Let us write n = 2n + 1 for n ∈ Z≥0. Without loss of generality, let us assume k to be even, where k = 2k0. Cancelling out the odd double factorial in the denominator and rewriting terms, we get: n!! n · (n − 2) ··· (n − k + 2) = k!!(n − k)!! k!! n · (n − 2) ··· (n − k + 2) = . 2k0 (k0)! Note that n · (n − 2) ··· (n − k + 2) is a sequence of k0 many odd integers. We will show 0 0 that for any p 6= 2 such that p ≤ k , it follows that υp((k )!) ≤ υp(n · (n − 2) ··· (n − k + 2)). ∞ N 0 Pj k0 k Pj k0 k N 0 We know from Lemma 2.3 that υp((k )!) = pi = pi for N such that p ≤ k but i=1 i=1 ∞ 0 N+1 P k < p . We can calculate υp(n · (n − 2) ··· (n − k + 2)) = Dpi (n · (n − 2) ··· (n − k + 2)). i=1

i Since gcd(p , 2) = 1, we have that 2 is an additive generator of Zpi . Since the sequence {n, (n − 2),..., (n − k + 2)} has k0 many elements and every sequence of pi many elements contains all the equivalence classes of Zpi , the sequence {n, (n − 2),..., (n − k + 2)} contains j k0 k j k0 k at least pi many copies of all of the equivalence classes of Zpi . Hence there are at least pi many elements in {n, (n − 2),..., (n − k + 2)} that are in the equivalence class 0 (mod pi), i which implies that they are divisible by p . Thus for each i it follows that Dpi (n·(n−2) ··· (n− N j k0 k Pj k0 k 0 k + 2)) ≥ pi . Hence it follows that υp(n · (n − 2) ··· (n − k + 2)) ≥ pi = υp(k !). i=1

Many of the properties of binomial coeﬃcients depend on the fact that binomial coeﬃcients are always integers. One of the most essential of such properties is Lemma 1.3, which states Mitsuki Hanada 9

n that k denotes the number of possible ways to choose k many of n distinct objects, which would not make sense if it were not an integer value. However, from these results, it is clear that DFBCs are not necessarily integers, which implies that many of these properties that depend on integers values may not hold. In the following sections, we will provide modiﬁed versions of properties of binomial coeﬃcients that indeed hold for these DFBCs.

3. Combinatorial Interpretation of DFBCs (Chu-Vandermonde identity)

The binomial coefficient is inherently combinatorial: the name “binomial coefficient” , as n k n−k discussed in Definition 1.2, comes from the fact that k is the coefficient of the term x y n n in the polynomial (x + y) . This is another way of stating Lemma 1.3, which says that k represents the number of ways to choose k out of n distinct objects. In this way, it is known that both binomial coefficients and double factorials have combinatorial meaning. However, it is much more difficult to see whether DFBCs also carry some sort of combinatorial meaning. One of our main concerns is that DFBCs are not necessarily integers: since binomial coefficients, as well as double factorials, are inherently defined to be integers, they are able to count objects. In this section, we will be exploring different combinatorial properties and identities that binomial coefficients satisfy, and evaluate whether there exists an analogous property for certain DFBCs.

One known identity that binomial coeﬃcients satisfy is the Chu-Vandermonde identity: Theorem 3.1. Let n ∈ Z, x, y ∈ Q. Then, n x + y X x y = n k n − k k=0

We are interested in this identity because there exists a combinatorial proof in the case where x, y ∈ Z>0, which is the case that we are interested in. We will give this combinatorial proof below.

x+y Proof. Assume that x, y ∈ Z>0. Recall that n denotes the number of ways to choose n P x y n objects from (x + y) total distinct ones. We will show that k n−k also denotes the k=0 number of ways to choose n objects from (x + y) total ones. We can split the (x + y) objects into two diﬀerent types where one has x many objects and the other has the remaining y objects and count the number of choosing n objects from (x + y) total ones in a diﬀerent way. First, we can choose some k where 0 ≤ k ≤ n. Then, we choose k many of the x objects. If we are to choose n total objects we must choose (n − k) many of the remaining x y y options. For any 0 ≤ k ≤ n, it follows that k n−k denotes the number of ways we can choose in this way. Since we can do this for each 0 ≤ k ≤ n, the total number of ways we n P x y can choose n many of the (x + y) objects must be k n−k . Thus the two expressions k=0 are equal combinatorially.

Now, our question becomes whether there exists a similar result for DFBCs, and whether a similar combinatorial proof can be applied to it. This may be a hint to any combinatorial 10 Double Factorial Binomial Coeﬃcients

interpretation we may have for the double binomial coeﬃcients. The author of [4] gives an algebraic proof of 3.1, which proves the statement by induction on n: we will do something similar for certain values of n to prove the following statement.

Lemma 3.2. Let n, x, y ∈ Z>0 such that n is even and n < x, y. Then, n x + y X x y = n k n − k k=0;keven

We will require the following lemma to prove Lemma 3.2.

Lemma 3.3. For any x, n, k ∈ Z>2 and k ≤ x it follows that x x − 2 k x = . n k − 2 n k

Proof. Follows from deﬁnition: x x − 2 kx (x − 2)!! = n k − 2 kn (k − 2)!!(x − k)!! k x(x − 2)!! = n k(k − 2)!!(x − k)!! k x = . n k

Proof of Lemma 3.2 We will prove this statement by induction on n even for arbitrary x, y such that n ≤ x, y. The base case n = 0 follows by deﬁnition.

Now let us assume for any x, y such that n − 2 ≤ x, y that the following holds: n−2 x + y X x y = . n − 2 k (n − 2) − k k=0;keven Fix x, y such that n ≤ x, y holds. This implies that (n − 2) ≤ (x − 2), (y − 2) it follows that the statement above holds for (x − 2) in place of x, and (y − 2) in place of y. That is, we have the following two equations: n−2 (x − 2) + y X x − 2 y = n − 2 k (n − 2) − k k=0;k even n X x − 2 y = , k − 2 n − k k=2;k even

n−2 x + (y − 2) X x y − 2 = . n − 2 k (n − 2) − k k=0;k even Mitsuki Hanada 11

Modifying the two equations using Lemma 3.3, we get the following relation: n x (x − 2) + y X x x − 2 y = n n − 2 n k − 2 n − k k=2;k even n X k x y = n k n − k k=2;k even n X k x y = , n k n − k k=0;k even

n−2 y x + (y − 2) X x y y − 2 = n n − 2 k n (n − 2) − k k=0;k even n−2 X (n − k) x y = n k (n − k) k=0;k even n X (n − k) x y = . n k (n − k) k=0;k even

Note that we also have: x + y x + y (x + y) − 2 = n n n − 2 x y (x + y) − 2 = + n n n − 2 n X k (n − k) x y = + n n k n − k k=0;k even n X x y = . k n − k k=0;k even

Thus the claim holds for n as well. Though we are able to find an analogue for the results of the Chu-Vandermonde identity for k even, this gives little insight into developing new ways of interpreting the DFBCs in a combinatorial way since we do not know how to interpret either side of the equation. In order to combinatorially interpret the expression on the right hand side, we can try to relate the DFBCs with traditional binomial coefficients, which we know how to interpret combinatorially. For the case where x, y, n are all even, we can use Lemma 2.7 to obtain the following result. 12 Double Factorial Binomial Coefficients

Corollary 3.4. Let n, x, y ∈ Z>0 such that n, x, y are all even and n ≤ x, y. If we let n = 2n0, x = 2x0, y = 2y0, it follows that, n0 x + y X x0 y0 = . n k0 n0 − k0 k;=0

Proof. This comes from applying Lemma 2.7 to the results of Lemma 3.2.

x+y Using the results of Corollary 3.4, the coeﬃcient n can be interpreted to be number n x y of possible ways to choosing 2 balls from 2 red ones and 2 white ones. We can also do a similar expansion if x, y are both odd using the following lemma: Lemma 3.5. Let n, x, y ∈ Z such that n, x + y are even but x, y are both odd. (assume that x, y > 1). Then, n n x + y X x − 1 y + 1 X x + 1 y − 1 = = n k n − k k n − k k=0;keven k=0;keven .

x+y x−1 y+1 x+1 y−1 Proof. This is a similar situation. We can split up 2 into either 2 + 2 or 2 + 2 , making both sides even. Without loss of generality let us consider the ﬁrst case. Then x+y x + y 2 = n n 2 n X2 x−1 y+1 = 2 2 k0 n − k0 k0=0 2 n x−1 2 2 x−1 P x−1 y+1 Note that k0 = 2k0 . Hence this sum becomes 2k0 n−2k0 , which is equiva- k0=0 0 lent to the claim when we replace k = 2k .

n The case x, y are both odd could be considered as the number of ways of choosing 2 many x−1 y+1 balls from 2 many red and 2 many white, but that is basically identical to the ﬁrst case minus a shift of variables (shifting odd pair (x, y) to even pair (x + 1, y − 1) or (x − 1, y + 1)).

One open question here is to think about how we would interpret the results in Lemma 3.2 when x is odd and y is even using traditional binomial coeﬃcients. Another open question would be use Lemma 3.2 to think of combinatorial interpretations of DFBCs that did not rely on binomial coeﬃcients.

Note that these results assume that n is even. We encounter an issue when we try to deal with n odd. If we consider the case where x + y is even but n is odd, it seems like there cannot be an expression that works for all possible combinations of (x, y, n). For example, 6 16 1 = 5 has a 5 in the denominator that will never come up unless x or y is larger than or equal to 5, thus the only possible case that there may be a general expression is for Mitsuki Hanada 13

(x, y, n) = (1, 5, 1) or (0, 6, 1). However for (0, 6, 1), the equation would just be the same expression on both sides, so it is not of interest to us. For (1, 5, 1), we can compute this value to see that the claim does not hold. Example 3.6. 1 5 1 5 23 1 + 5 + = 6= 0 1 1 0 8 1

Furthermore, we have more intuition on why this relationship doesn’t hold for n odd. Note that in the proof for Lemma 3.2, we based our induction proof on the fact that the base case was n = 0. We can easily see that the base case where n = 1 does not hold from the example above. Thus it seems like it is diﬃcult to think of Chu-Vandermonde-like relationship, or combinatorial interpretation, for any arbitrary combination (x, y, n). An open question here is to think of a similar relationship that would hold for any choice of (x, y, n), without distinction between the n odd and n even case.

4. Pascal’s Triangle 4.1. Introducing Pascal’s Triangle and its properties. One famous property of the binomial coeﬃcient is that they can be represented using Pascal’s triangle.

0 0 1

1 1 0 1 1 1

2 2 2 0 1 2 1 2 1

3 3 3 3 ⇔ 0 1 2 3 1 3 3 1

4 4 4 4 4 0 1 2 3 4 1 4 6 4 1

Figure 1. Pascal’s Triangle

One defining property of Pascal’s triangle is that any element is the sum of the two elements directly above it. This gives us a recursive relation for binomial coefficients. We can formalize this relation using the equation below. Theorem 4.1. . For 0 ≤ k ≤ n, the following relation holds: n n − 1 n − 1 = + . k k − 1 k n Remark 4.2. Theorem 4.1 assumes that for k > n or k < 0, it follows that k = 0. n n−1 Proof. When n = k , it follows since k = k−1 = 1 and the other term is 0. Similarly, if n n−1 k = 0, then it follows since k = k = 1 and the other term is 0. Assume that 0 < k < n. 14 Double Factorial Binomial Coefficients

Then, by definition of binomial coefficients: n − 1 n − 1 (n − 1)! (n − 1)! + = + k − 1 k (k − 1)!(n − k)! k!(n − k − 1)! (n − 1)! 1 1 = + (k − 1)!(n − k − 1)! (n − k) k (n − 1)! n = (k − 1)!(n − k − 1)! (n − k)k n = . k Hence this relation holds for any 0 ≤ k ≤ n. Pascal’s triangle highlights many interesting properties regarding binomial coefficients. If n we consider the c-th column of Pascal’s Triangle to be elements of the form c for n ≥ c, we can see that there are interesting sequences that arise in certain columns of Pascal’s Triangle. It is evident to see that the 0th column is a constant sequence {1, 1,... } and the 1st column is the sequence of natural numbers {1, 2, 3,... }. For the 2nd sequence, we have {1, 3, 6, 10, 15, 21, 28,... }. We can rewrite this sequence recursively as {an} where a1 = 1 and an = an−1 + n for n > 1. Alternatively, we can Pn write a closed formula for this sequence: an = k=1 k. These numbers are called triangular numbers, where an denotes the number of dots needed to arrange into an equilateral triangle with n dots on one side. It turns out that for any column c, there exists a similar recursive relation that defines the sequence of elements in the column which follows from Theorem 4.1. This gives us another interesting property of Pascal’s Triangle: not only can the elements of the triangle be defined recursively using elements of the row before it, but the sequence of elements in the column are defined recursively as well. Triangular numbers can also be perceived be related to simplices, where the n-th triangular number an is the number of 1-faces of the (n + 1)-simplex. In fact, it follows that for any c ≥ 2, the n-th element in the c-th column corresponds to the number of (c − 1) faces of the (n + c − 1)-simplex, so we are able to give the elements of the columns some geometric meaning. We also have interesting results about the sum of such columns.

Lemma 4.3. For n, c ∈ Z>0 such that 0 ≤ c ≤ n we have n n X k − 1 = . c c − 1 k=c k Proof. Using Theorem 4.1 on terms of the form c for k such that c ≤ k ≤ n and assuming Remark 4.2, we obtain Mitsuki Hanada 15

n n − 1 n − 1 = + c c − 1 c n − 1 n − 2 n − 2 = + + c − 1 c − 1 c n − 1 n − 2 c c = + + ··· + + c − 1 c − 1 c − 1 c n − 1 n − 2 c c − 1 c − 1 = + + ··· + + + c − 1 c − 1 c − 1 c − 1 c n X k − 1 = . c − 1 k=c Lemma 4.3 is stating that we can find entries in Pascal’s Triangle by adding up all the elements above it in the diagonal column. 4.2. Finding similar triangles for certain DFBCs. Now that we have interesting properties about Pascal’s Triangle to motivate our study of it, we will try to find an analogue of Pascal’s Triangle for certain DFBCs. This requires us to divide DFBCs into different cases and analyze the coefficients in the respective cases. First, we can find a similar recursive relation to Theorem 4.1 that holds for certain DFBCs. Lemma 4.4. For n, k ∈ Z such that 0 ≤ k ≤ (n − 2): n n − 2 n − 2 = + k k − 2 k Proof. This follows directly from the definition of DFBCs: n − 2 n − 2 (n − 2)!! (n − 2)!! + = + k − 2 k (k − 2)!!(n − k)!! k!!(n − k − 2)!! (n − 2)!! = (k + (n − k)) k!!(n − k)!! (n)!! n = = . k!!(n − k)!! k Note that this divides our DFBCs into 4 different cases depending on whether n, k are even or odd, similar to when we discussed the different forms of DFBCs. We know that the case where n, k are both even is the same as the normal binomial coefficient from Theorem 2.7, and that by Lemma 1.9 we have that the cases where n odd k even and n odd k odd are similar. Thus, we choose to investigate further the two cases where n is even k is odd, and n is odd k is even. First let us consider the case where n even k odd. We can write out the coefficients in a shape that resembles Pascal’s Triangle to get Figure2. 16 Double Factorial Binomial Coefficients

2 1 2

4 4 23 23 1 3 3 3 6 6 6 ⇔ 24 24 24 1 3 5 5 3 5

8 8 8 8 27 27 27 27 1 3 5 7 5·7 3·5 3·5 5·7

1010101010 28 28 28 28 28 1 3 5 7 9 7·9 3·7 3·5 3·7 7·9

Figure 2. Triangle for Double Factorial Binomial Coeﬃcients with n even k odd

If we factor out certain quantities from each row, we can get a diagram that is closer to Pascal’s Triangle (Figure3), where all of the elements on the outside perimeter are 1. This diagram highlights certain patters along columns of the triangle. Looking at the diﬀerent elements, we can see that the entries seem to be increasing as we increase rows. Analyzing the pattern behind this increase gives us another way of deﬁning these certain DFBCs recursively.

2!! Row 1: × 1!! 1 4!! Row 2: × 3!! 1 1 6!! 5 Row 3: × 5!! 1 3 1 8!! 7 7 Row 4: × 7!! 1 3 3 1 10!! 9 7·9 9 Row 5: × 9!! 1 3 3·5 3 1

Figure 3. Reduced Version of Figure2

Lemma 4.5. For n, k ∈ Z≥2 such that k ≤ (n − 2) it follows that n n n − 2 1 n − 2 = 1 − + . k n − 1 k − 2 k k Proof. This follows by deﬁnition. By manipulation of terms, we can write: n − 2 1 n − 2 (n − 2)!! 1 (n − 2)!! 1 − + = 1 − + k − 2 k k (k − 2)!!(n − k)!! k k!!(n − k − 2)!! (n − 2)!! 1 1 1 = 1 − + (k − 2)!!(n − k − 2)!! n − k k k (n − 2)!! n − 1 = k!!(n − k − 2)!! n − k (n − 1)(n − 2)!! = . k!!(n − k)!! Then, it follows that n n − 2 1 n − 2 n!! n 1 − + = = . n − 1 k − 2 k k k!!(n − k)!! k Mitsuki Hanada 17

Remark 4.6. If we apply this lemma to DFBCs where n is even and k is odd, the relation n highlights the recursive properties of elements in Figure3, where the term n−1 denotes the n−2 1 n−2 values that we factored out from each row and the term k−2 1 − k + k denotes addition within Figure3. Though this formula is more complicated than the one stated in Lemma 4.4, it may be preferable in the cases where we are interested in certain properties that the reduced triangle in Figure3 may have. An open question to consider is whether this identity leads to any interesting properties of the triangle in Figure3. Now, let us consider the case where n is odd and k is even. First let us write out the diagram below in Figure4.

1 0 1

3 3 3 0 2 1 2

5 5 5 ⇔ 5 3·5 0 2 4 1 2 23

7 7 7 7 7 5·7 5·7 0 2 4 6 1 2 23 24

9 9 9 9 9 9 7·9 3·5·7 5·7·9 0 2 4 6 8 1 2 23 24 27

Figure 4. Triangle for Double Factorial Binomial Coeﬃcients with n odd and k even

One evident diﬀerence between this and the other two triangles discussed earlier is the lack of symmetry. In this case, it may make more sense to write the triangle like a right triangle:

1 0 1

3 3 3 0 2 1 2

5 5 5 ⇔ 5 3·5 0 2 4 1 2 23

7 7 7 7 7 5·7 5·7 0 2 4 6 1 2 23 24

9 9 9 9 9 9 7·9 3·5·7 5·7·9 0 2 4 6 8 1 2 23 24 27

Figure 5. Adjusted version of Figure4

Similar to the triangle for n even (Figure2), we have that the elements along the slope of 2n−1 the right triangle are increasing by 2n for each n-th row. The first three columns of this triangle show us a evident pattern of these DFBCs, which alludes to the following lemma that will give us closed formulas for these coefficients: 18 Double Factorial Binomial Coefficients

Lemma 4.7. The following closed formulas hold. n (1) 0 = 1 for any n. n n (2) 2 = 2 for n > 2. n n(n−2) (3) 4 = 23 for n > 4. Proof. All of these closed formulas follow directly from the definition and from the fact that n is odd. We will prove (3). Let n > 4. Then observe: n n!! = 4 4!!(n − 4)!! n(n − 2)(n − 4)!! = 2 · 4 · (n − 4)!! n(n − 2) = . 23 Since Figure5 only contains elements where n is odd, we can see that the entries in each column form an interesting sequence. If we count from the left (where the column of 1 3 entries { 0 , 0 ,... } is column 0), the sequence of elements in the first column will be n of the form { 2 : nodd}, where the sequence of numerators is just the sequence of all the odd numbers greater than 1: {3, 5, 7,... }. Similarly, the sequence of elements in the second n(n−2) column will be of the form { 23 : nodd}, where the sequence of numerators is just the sequence of products of two consecutive odd numbers greater than 1: {3 · 5, 5 · 7, 7 · 9,... }. Though the pattern is more difficult to see in column 3 and beyond using the current notation used in Figure5, if we write the entries in each c-th column so that their denominators are in the form (2c)!!, we can see familiar pattern occur in the numerator. Lemma 4.8. For each column c of Figure5, the sequence of elements in column c is (n)(n − 2) ··· (n − 2(c − 1)) | n odd, n > 2c . (2c)!! Proof. This follows similarly to the proof of Lemma 4.7. Note that we discussed how the sequences of elements in columns of Pascal’s Triangle in Figure1 were related to simplices. An open question would be to determine whether the sequences in Lemma 4.8 have some sort of geometric or combinatorial meaning. Now that we know what the columns look like, let us look at the different sums of columns. Recall that Lemma 4.3 follows from the recursive property of binomial coefficients in Theorem 4.1. From Lemma 4.4, we know that our triangle in Figure5 has a similar recursive n property for all elements k where k ≤ (n − 2).This is equivalent to all the elements in n the triangle except the ones on the right diagonal (of the form n−1 . We can expect to obtain a similar result for these DFBCs if we utilize this recursive relation. However, before we try to state an analogous result for this triangle, we need to modify the triangle so that the recursive relation holds for all elements in the triangle. Since we know what the sequence of elements in each column look like, we can make a guess what element we could add to the beginning of the sequence to make the recursive n!! relation hold: if the first element of the column is of the form (n−1)!! , a natural element to Mitsuki Hanada 19

(n−2)!! add at the head of the column would be (n−1)!! . In general, it turns out that adding this (2c−1)!! (2c)!! term to each column makes the recursive relation holds for all the original elements of Figure5. n (n−2)!! n−2 Lemma 4.9. For n ∈ Z3, we have n−1 = (n−1)!! + n−3 Proof. Observe: (n − 2)!! n − 2 (n − 2)!! (n − 2)!! + = + (n − 1)!! n − 3 (n − 1)!! (n − 3)!! (n − 2)!! 1 = + 1 (n − 3)!! n − 1 (n − 2)!! n = (n − 3)!! n − 1 n!! = . (n − 1)!! Then, we have that the following holds:

Lemma 4.10. For any n, c ∈ Z>0 such that n is odd and 0 ≤ 2c < n, we have n n (2c − 1)!! X k − 2 = + . 2c (2c)!! 2c − 2 k=2c+1,k odd n Proof. We can apply the recursive relation in Lemma 4.4 to 2c until we get n n 2c + 1 X k − 2 = + . 2c 2c 2c − 2 k=2c+3,k odd 2c+1 2c−1 (2c−1)!! Since we have that 2c = 2c−2 + (2c)!! from Lemma 4.9, we get the desired result. Lemma 4.10 is similar to Lemma 4.3: we ﬁnd that any element in the triangle is the sum (2c−1)!! of all of the elements above it in the column to the left, plus an additional term (2c)!! which we added to make the recursive relation holds.

5. Generalization to m-Factorial Binomial Coefficients In this section, we will switch gears from discussing DFBCs to focusing on m-factorial binomial coefficients for m ∈ Z>0 and generalizing results we have about DFBCs (when m = 2) from Section2 and binomial coefficients ( m = 1) to any m. First, we will define m-factorials and m-factorial binomial coefficients. These definitions should not be surprising coming from our definitions in Section 1.2.

Definition 5.1. Let m ∈ Z>0. The m-factorial of a non-negative integer n (denoted (n)!m) is defined (n)!m := n(n − m)(n − 2m) ··· (?), where ? ∈ {1, . . . , m}. 20 Double Factorial Binomial Coefficients

Note that there is an analogous result of Lemma 1.7 that holds for any arbitrary m- factorial. n Lemma 5.2. For m, n ∈ Z>0, we can write (nm)!m = m (n!).

Proof. If we write (nm)!m = nm(nm − m)(nm − 2m) ··· (m) where the RHS is a product of n many terms, and factor out a m from each term in the product, we get

(nm)!m = nm(m(n − 1))(m(n − 2)) ··· (m(1)) = mn(n)(n − 1)(n − 2)) ··· (1) = mn(n!).

Definition 5.3. For m ∈ Z>0 and n, k ∈ Z≥0 such that 0 ≤ k ≤ n, the m-th factorial n binomial coefficient (denoted k m) is defined n n! := m . k m k!m(n − k)!m The definitions above are generalizations of our earlier definitions of double factorials and DFBC. From our discussion on DFBCs, we know that it is premature to assume that these coefficients will be integers: in fact, we know that it will be difficult to know what these coefficients look like at all. However, using our work in Section2, we can conjecture what these m− factorial binomial coefficients may look like. We will first experiment with the m = 3 case, and then try expand to general m. Recall n that in our discussion for DFBC, we divided up all possible coefficients k into four different cases, depending on n, k (mod 2). However, we were also able to combine the cases for n odd k even and n odd k odd, by symmetry of DFBC. In a similar way, we can n divide all possible 3−factorial binomial coefficients k 3 into nine different cases, depending on n, k (mod 3), and then combine cases that behave similarly utilizing the fact that m- factorial binomial coefficients are symmetric: n = n . The four cases we will be k m n−k m considering are: • n ≡ k ≡ 0 (mod 3) • n ≡ 0 (mod 3) but k 6≡ 0 (mod 3) • n 6≡ 0 (mod 3) but k, ≡ 0 (mod 3) • n, k, (n − k) 6≡ 0 (mod 3) (no terms are 0 mod 3). The easiest case is where we have n ≡ k ≡ 0 ( mod 3). This case is analogous to Theorem 2.7 for m = 2. 0 0 0 0 Theorem 5.4. For n, k ∈ Z≥0 such that n = 3n , k = 3k for n , k ∈ Z≥0 we have n n0 = 0 ∈ Z. k 3 k Proof. This follows from Lemma 5.2 similarly to proof of Theorem 2.7 and Corollary 2.8. Since we know that Lemma 5.2 holds for any positive integer m, we can generalize the results for arbitrary m-factorials:

Theorem 5.5. Let m ∈ Z≥0. For any n, k ∈ Z≥0 such that n ≡ k ≡ 0 (mod m), we have n ∈ Z. k m Mitsuki Hanada 21

Proof. The proof is the same as the proof of Theorem 5.4. The next case we can consider is when n ≡ 0 (mod 3) but k, (n − k) 6≡ 0 (mod 3). This case is analogous to Theorem 2.9 for m = 2.

Theorem 5.6. For n, k ∈ Z≥0 such that n ≡ 0 (mod 3) and k 6≡ 0 (mod 3), it follows that n 3t = k 3 q where q 6≡ 0 (mod 3) and t ∈ Z>0. Similar to proof of Theorem 2.9, we will prove this using a lemma: 0 0 0 0 Lemma 5.7. For any k, q ∈ Z>0 such that k = 3k + rk, q = 3q + rq for k , q ∈ Z≥0 and 0 < rk, rq < 3, it follows that j 0 0 k (2q +k )+rq if r = r  q k q 0  Dq({rk, 3 + rk, 3 · 2 + rk,..., 3k + rk}) = j 0 0 k  (q +k )+rq  q if rk 6= rq

0 0 Proof. Let [k]3 = {rk, 3 + rk, 3 · 2 + rk,..., 3k + rk}, which is a set of k + 1 terms. First, assume that rk = rq. This is equivalent to k ≡ q 6≡ 0 (mod 3). If q ≤ k, since q ≡ k (mod 3), it follows that q ∈ [k]3. We also know that every q-th term from q (q + 3q, q + 3 · 2q, . . . ) in [k]3 is also divisible by q. Thus, in order to count the number of terms in [k]3 that are divisible by q, we want to count the number of sequences of length q that are contained in the sequence [k]3, when we consider the ﬁrst subsequence {rk, 3 + rk, . . . , q} to be a sequence of length q. This is equivalent to counting the number of sequence of 0 0 0 0 0 length q in a sequence of length (k + 1) + (2q + (rq − 1)) = 2q + k + rq, where (k + 1) 0 is the length of [k]3 and (2q + (rq − 1) is the number of additional terms we would need 0 to make the sequence {rk, 3 + rk, . . . , q} which is originally of length (q + 1) to be length 0 0 0 j (2q +k )+rq k 0 0 q = 3q + rq. This is equivalent to q . If q > k, it follows that q > k , hence 0 0 0 0 0 j (2q +k )+rq k (2q + k ) + rq < 3q + rq. Thus it follows that q = 0. Note that if q > k then Dq([k]3) = 0, thus the statement holds. Now, assume that rk 6= rq. This implies that q 6≡ k (mod 3). However, we know that 2q ≡ k (mod 3) must hold since k, q 6≡ 0(mod 3). If 2q ≤ k, then it follows that 2q ∈ [k]3. Similarly, since every q−th term from 2q in [k]3 is also divisible by q, we want to count the number of 0 sequences in [k]3 when we consider the subsequence {rk, 3 + rk,..., 2q = 6q + 2rq}, which 0 0 has length 2q + rq, to be a sequence of length q = 3q + rq.This is equivalent to counting the number of sequence of length q in a sequence of length (k0 + 1) + (q0), where (k0 + 1) is the 0 length of [k]3 and q is the number of additional terms we assume exist to make the block 0 0 0 j (q +k )+rq k {rk, 3 + rk,..., 2q} have length q = 3q + rq. This is equivalent to q . If 2q > k, 0 0 0 0 0 0 it follows that 2q > k , hence (q + k ) + rq < (q + 2q ) + rq = q. Thus it follows that 0 0 j (q +k )+rq k q = 0. Note that if q > k then Dq([k]3) = 0, thus the statement holds. Now we will use this lemma to prove Theorem 5.6. Proof of Theorem 5.6 22 Double Factorial Binomial Coeﬃcients

0 0 0 0 Since n ≡ 0 (mod 3) and k 6≡ 0 (mod 3), let us write n = 3n , k = 3k +rk for n , k , rk ∈ Z≥0 0 0 such that 0 < rk < 3. Using this notation, we can write (n − k) = 3(n − k − 1) + (3 − rk) n0 0 From Lemma 5.2 it follows that n!3 = 3 (n )!. We will show that the statement holds by showing that for any prime p 6= 3, it follows 0 j n0 k that υp((n )!) ≤ υp(k!3(n − k)!3), which is equivalent to showing that pi ≤ (Dpi ([k]3) + i 0 Dpi ([n − k]3)) for every i such that p ≤ n . Since p is prime and p 6= 3, it follows that p 6≡ 0 (mod 3). For i such that pi ≤ n0, let us i0 0 write p = 3p + rpi where p , rpi ∈ Z≥0 and 0 < rpi < 3. Note that since k 6≡ 0 (mod 3), it follows that (n − k) 6≡ k (mod 3) and (n − k) 6≡ 0 (mod 3). Thus either k ≡ pi (mod 3) or (n − k) ≡ pi (mod 3). Without loss of generality, assume that k ≡ pi (mod 3). (the (n − k) ≡ pi (mod 3) case follows similarly). Fom Lemma 5.7 and Lemma 2.5, it follows that

i0 0 i0 0 0 (2p + k ) + rpi (p + (n − k − 1) + rpi D i ([k] ) + D i ([n − k] ) = + p 3 p 3 pi pi i0 0 i0 0 0 (2p + k ) + r i + (p + (n − k − 1) + r i ≥ p p − 1 pi i0 0 3p + r i + (n ) + (r i − 1) = p p − 1 pi i 0 p + (n ) + (r i − 1) = p − 1 pi 0 (n ) + (r i − 1) = p . pi

j (n0) k Since (rpi −1) ≥ 0, this implies that Dpi ([k]3)+Dpi ([n−k]3) ≥ pi . Thus the statement holds. Note that in this proof, as well as the proof for Theorem 2.9 we utilize the fact that if p 6= m then pi must be equivalent to either k or (n − k) (mod m) for m = 2 or 3. This follows since there are at most two different equivalence classes that pi can belong to, assuming that pi 6≡ 0 (mod m) and m = 2, 3 only have two and three different equivalence classes respectively. For m > 3, we will have to make modifications to the proof to ensure that this holds. However, though we will not prove it here, we can make a conjecture on what these coefficients will look like for m > 3.

Conjecture 5.8. Let m ∈ Z>3. For n, k ∈ Z≥0 such that n ≡ 0 (mod m) and k 6≡ 0 (mod m), it follows that n mt = k m q where q 6≡ 0 (mod m) and t ∈ Z>0. We may need to divide up the proofs depending on whether m is even, is a prime, or just an odd composite number. The next case we can consider is when n ≡ k (mod 3) but n 6≡ 0 (mod 3).This case is analogous to Theorem 2.12 for m = 2. Mitsuki Hanada 23

Theorem 5.9. For n, k ∈ Z≥0 such that n 6≡ 0 (mod 3) and k ≡ 0 (mod 3), it follows that n q = t k 3 3 where q 6≡ 0 (mod 3) and t ∈ Z>0. Proof. Using the fact that k ≡ 0 (mod 3) and n ≡ (n − k) (mod 3), we can write:

n n! = 3 k 3 k!3(n − k)!3 n!3 0 = k0 0 if k = 3k 3 k !(n − k)!3 n(n − 3) ··· (n − 3(k0 − 1)) = . 3k0 (k0)! Since no factors of 3 appear in n(n − 3) ··· (n − 3(k0 − 1)), we only need to show that for 0 0 0 P∞ j k0 k any p 6= 3 it follows that υp(k !) ≤ υp(n(n−3) ··· (n−3(k −1))). Since υp(k !) = i=1 pi = PN j k0 k N 0 N+1 i=1 pi for N such that p ≤ k < p , we will show that for 1 ≤ i ≤ N it follows that j k0 k 0 0 0 pi ≤ Dpi ({n, (n−3),..., (n−3(k −1))}). Note that {n, (n−3),..., (n−3(k −1))} has k many terms. If p is a prime such that p 6= 3 it follows that gcd(pi, 3) = 1.Thus 3 is a generator 0 j k0 k of Zpi . Then, the sequence {n, (n − 3),..., (n − 3(k − 1))} contains pi many copies of all j k0 k 0 the equivalence classes of Zpi , hence pi ≤ Dpi ({n, (n − 3),..., (n − 3(k − 1))}). Note that this proof only depends on the fact that 3 and p are relatively prime. Hence, it is easy to expand this result for m-factorial such that m is a power of some prime: s Theorem 5.10. Let m = p for some prime p. For n, k ∈ Z≥0 such that n 6≡ 0 (mod p) and k ≡ 0 (mod m), it follows that n q = t k m p where q 6≡ 0 (mod p) and t ∈ Z>0. Proof. Observe: 0 n n(n − m) ··· (n − m(k − 1)) 0 = k0 0 (for k = k m) k m m (k )! n(n − m) ··· (n − m(k0 − 1)) = (for k = k0m). (ps)k0 (k0)! Since mj ≡ 0 (mod p) for any j but n 6≡ 0 (mod p), it follows that (n−m(j)) 6≡ 0 (mod p), hence no factors of p appear in n(n − m) ··· (n − m(k0 − 1)). That is, we only need to show 0 0 0 that for any prime p 6= p, it follows that υp0 (k !) ≤ υp0 (n(n−m) ··· (n−m(k −1))), which is j k0 k 0 0i 0 equivalent to showing ≤ D 0i ({n, (n−m),... (n−m(k −1))}) for all i such that p ≤ k . p0i p 0i s Since p and m = p are relatively prime, we have that m is a generator of Zp0i . Then, the 24 Double Factorial Binomial Coeﬃcients

0 0 j k0 k sequence {n, (n − m),..., (n − m(k − 1)))} of k many terms contains p0i many copies of j k0 k 0 all the equivalence classes of Zp0i , hence p0i ≤ Dp0i ({n, (n − m),... (n − m(k − 1))}). For m which has multiple prime factors, we can work through examples to see how we must modify the statements. For example, we can compute a few examples for m = 6 and n such that n 6≡ 0 (mod p) for all prime factors p of m. Example 5.11. 25 25 · 19 · 13 · 7 · 1 25 · 19 = 2 = 2 12 6 6 (2 · 1)13 · 7 · 2 6 (2 · 1) We have for the numerator we have 25 · 19 ≡ 0 (mod 6) and the denominator is a product of 2 and 3, which are the prime factors of 6.

Q si Conjecture 5.12. Let m = i pi where pi are primes. For n, k ∈ Z≥0 such that n 6≡ 0 (mod pi) for all i and k ≡ 0 (mod m), it follows that n q = Q ti k m i pi where q 6≡ 0 (mod m) and t ∈ Z>0. One open question here would be to determine what happens if we have n 6≡ 0 (mod m) but not necessarily n 6≡ 0 (mod pi) for all prime factors pi of m. The ﬁnal case is when n, k, (n − k) 6≡ 0 (mod 3). For this case, we get the expected result:

Lemma 5.13. For n, k ∈ Z≥0 such that n, k, (n − k) 6≡ 0 (mod 3) it follows that n q = k 3 r where q, r 6≡ 0 (mod 3).

Proof. This follows from the fact that there are no factors of 3 in any of the products. Similar to the previous case, it is easy to expand this result for m− factorial such that m is a power of some prime. s Lemma 5.14. Let m = p for some prime p. For n, k ∈ Z≥0 such that n, k, (n − k) 6≡ 0 (mod m) it follows that n q = k m r where q, r 6≡ 0 (mod p).

Proof. The proof is similar to that of Lemma 5.13 Q si In the case where m = pi for primes pi, it is more diﬃcult to ﬁnd a similar statement. We can see this from the following example. Example 5.15. Let m = 6 and choose n = 7, k = 3. Note that n, k, (n − k) 6≡ 0 (mod m). 7 However if we compute 3 6, Mitsuki Hanada 25

7 7! = 6 3 6 3!64!6 7 = . 12 Note that 7 6≡ 0 (mod 6) but 12 ≡ 0 (mod 6). If we restrict n, k further, we create more cases that we would need to define statements for. However, that would give us more cases we would need to consider. One open question to consider would be to determine whether there exists a form for m-factorial binomial n Q si coefficient k m for m such that m = pi and n, k such that n, k, (n − k) 6≡ 0 (mod m), without adding further restrictions on n or k. 26 Double Factorial Binomial Coefficients

References [1] D. Callan. A Combinatorial Survey of Identities for the Double Factorial, 2009. arXiv: 0906.1317. [2] K. Conrad. Proofs of Integrality of Binomial Coeﬃcients. https://kconrad.math.uconn.edu/blurbs/ proofs/binomcoeffintegral.pdf. (accessed: 05.01.2020). [3] H. Gould and J. Quaintance. Double Fun with Double Factorials. Mathematics Magazine, Vol.85(No.3):177–192, 2012. [4] D. Grinberg. Notes on the combinatorial fundamentals of algebra, 2020. arXiv: 2008.09862.