Decompositions of Unit Hypercubes and the Reversion of a Generalized M\"obius Series

Let $s_d(n)$ be the number of distinct decompositions of the $d$-dimensional hypercube with $n$ rectangular regions that can be obtained via a sequence of splitting operations. We prove that the generating series $y = \sum_{n \geq 1} s_d(n)x^n$ satisfies the functional equation $x = \sum_{n\geq 1} \mu_d(n)y^n$, where $\mu_d(n)$ is the $d$-fold Dirichlet convolution of the M\"obius function. This generalizes a recent result by Goulden et al., and shows that $s_1(n)$ also gives the number of natural exact covering systems of $\mZ$ with $n$ residual classes. We also prove an asymptotic formula for $s_d(n)$ and describe a bijection between $1$-dimensional decompositions and natural exact covering systems.


Introduction 1.Decomposing the d-dimensional unit hypercube
Suppose we start with (0, 1) d , the d-dimensional unit hypercube, and iteratively perform the following operation: • choose a region in the current decomposition, a coordinate i ∈ {1, . . ., d}, and an arity p ≥ 2; • partition the selected region into p equal smaller regions with cuts orthogonal to the i th axis.
For example, Figure 1 illustrates one way to decompose the unit square (0, 1) 2 into 8 regions using a sequence of 4 partitions.
n n-region decompositions of (0, 1) 2 We are interested in the following question: Given integers d, n ≥ 1, how many distinct compositions of (0, 1) d are there with n regions?In Figure 2, we list all the possible decompositions of (0, 1) 2 with n ≤ 4 regions.
Let's introduce some notation so we can discuss these decompositions more precisely.Let N = {1, 2, . . ., } denote the set of natural numbers, and [n] = {1, 2, . . ., n} for every n ∈ N. Given a region (all regions mentioned in this manuscript will be rectangular), a coordinate i ∈ [d], and an arity p ≥ 2, define for every j ∈ {0, 1, . . ., p}, and Thus, H i,p (R) is a set consisting of the p regions obtained from splitting R along the i th coordinate.Then we define the set of hypercube decompositions S d recursively as follows: • (0, 1) d ∈ S d -this is the trivial decomposition with one region, the entire unit hypercube.
In other words, S d consists of the decompositions of (0, 1) d that can be achieved by a sequence of splitting operations.In particular, the coordinate and arity used in each splitting operation are arbitrary and can vary over the splitting sequence.Furthermore, given S ∈ S d we let |S| denote the number of regions in S, and define S d,n = {S ∈ S d : |S| = n} for every integer n ≥ 1.Note that distinct splitting sequences can result in the same decomposition.For an example, the decomposition S = H 1,6 ((0, 1)) ∈ S 1,6 can be obtained from simply 6-splitting the unit interval (0, 1), or 2-splitting (0, 1) followed by 3-splitting each of (0, 1/2) and (1/2, 1).
We are interested in enumerating the number of decompositions s d (n) = |S d,n |.For small values of d, we obtain the following sequences: Note that the hyperlink at s 1 (n) in the table above directs to the sequence's entry in the On-Line Encyclopedia of Integer Sequences (OEIS) [7].Similar hyperlinks are placed throughout this manuscript for all integer sequences that are already in the OEIS at the time of this writing.

A generalized Möbius function
To state our main result, we will need to describe the following generalization of the well-studied Möbius function.Given n ∈ N, the Möbius function is defined to be µ(n) = (−1) k if n is a product of k distinct primes; 0 if p 2 |n for some p > 1.
(The convention is that µ(1) = 1.)A well-known property of µ(n) is that i|n More generally, given two arithmetic functions α, β : N → R, their Dirichlet convolution α * β : N → R is the function For an example, let ½ : N → R be the function where ½(n) = 1 for all n ≥ 1, and let δ : N → R denote the function where δ(1) = 1 and δ(n) = 0 for all n ≥ 2. Then (1) can be equivalently stated as Next, given integer d ≥ 1, we define the d-fold Möbius function where .
The following table gives the first few terms of µ d (n) for d ≤ 3. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 The generalized Möbius function µ d was independently discovered several times, first by Fleck in 1915 [10,Section 2.2].The reader may refer to [10] for the historical developments of this (and other) generalizations of µ.
Next, we state two formulas for µ d (n) that will be useful subsequently.
Lemma 1.For all integers d, n ≥ 1 • if α, β are multiplicative, so is their Dirichlet convolution α * β.Thus, µ d is multiplicative as well, and so it follows that Our main result of the manuscript is the following.

Natural exact covering systems
Theorem 2 generalizes a recent result on natural exact covering systems, which we describe here.Given a ∈ Z, n ∈ N, we let a, n denote the residue class {x ∈ Z : x ≡ a mod n}.
• For all C ∈ C, residue class a, n ∈ C, and integer r ≥ 2, Notice that 0, 1 = Z, and each NECS C ∈ C consists of a collection of residue classes that partition Z. Also, given n ∈ N, let C n ⊆ C denote the set of NECS with exactly n residue classes, and let c(n) = |C n |.For example, { 0, 4 , 2, 8 , 6, 8 , 1, 6 , 3, 6 , 5, 6 } is an element in C 6 as it can be obtained from 2-splitting 0, 1 , then 2-splitting 0, 2 , then 2-splitting 2, 4 , then 3-splitting 1, 2 .Recently, Goulden, Granville, Richmond, and Shallit [6] showed the following: Not only does Theorem 3 provide a formula for the number of NECS with a given residual classes, it also shows that the generating series with coefficients c(n) is exactly the compositional inverse of the familiar Möbius power series.In that light, Theorem 2 is a generalization of Theorem 3, as we provide a formula that gives s d (n) while also showing that they are the coefficients of the compositional inverse of the generalized Möbius series.Moreover, Theorem 2 implies that s 1 (n) = c(n) for every n ≥ 1 (i.e., the number of ways to split (0, 1) into n subintervals using a sequence of splitting operations is equal to the number of NECS of Z with n classes).
To the best of our knowledge, NECS first appeared in the mathematical literature in Porubskỳ's work in 1974 [8].More broadly, a set of residual classes , the residual classes need not be disjoint or result from a sequence of splitting operations).An ample amount of literature has been dedicated to studying covering systems since they were introduced by Erdős [4].The reader may refer to [9,11,12,13] and others for broader expositions on the topic.

Roadmap of the manuscript
This manuscript is organized as follows.In Section 2, we introduce some helpful notions (such as the gcd of a decomposition) and prove Theorem 2. Then in Section 3, we take a small detour to study a d (n), the coefficients of the multiplicative inverse of the generalized Möbius series.We establish a combinatorial interpretation for a d (n), and prove some results that we'll rely on in Section 4, where we prove an asymptotic formula for s d (n) and study its growth rate.Finally, we establish a bijection between S 1,n and C n in Section 5, and give an example of how studying hypercube decompositions can lead to results on NECS.

Proof of Theorem 2
In this section, we prove Theorem 2 using a d-fold variant of Möbius inversion.We remark that the structure of our proof is somewhat similar to Goulden et al.'s corresponding argument for NECS [6,Theorem 3].
Given integers r 1 , . . ., r d ∈ N, let D (r 1 ,...,r d ) ∈ S d denote the decomposition obtained by • starting with (0, 1) d ; • for i = 1, . . ., d, r i -split all regions in the current decomposition in coordinate i. Figure 3   Also, it will be convenient to define the notion of scaling.Given regions R, R ′ ⊆ (0, 1) d , notice there is a unique affine function L R→R ′ : R d → R d that preserves the lexicographical order of vectors, while satisfying {L R→R ′ (x) : for every i ∈ [d] is the function that satisfies the aforementioned properties.Next, given regions R, R ′ , R ′′ ⊆ (0, 1) d , we define We will always apply this scaling function in cases where R ′′ ⊆ R. gives a decomposition of (0, 1) d that refines D (r 1 ,...,r d ) .On the other hand, given D ∈ S d that refines D r 1 ,...,rn , define D j ⊆ S such that By assumption that S D (r 1 ,...,r d ) , we know that D j can be obtained from applying a sequence of splitting operations to B j for every j.Hence, each of S j = scale B j →(0,1) n (R) : R ∈ D j is an element of S d .Doing so for each of D 1 , . . ., D n results in an n-tuple of decompositions that corresponds to S. Thus, it follows that Next, given S ∈ S d , we say that gcd(S) = (r 1 , . . ., r d ) if • S D (r 1 ,...,r d ) ; • there doesn't exist (r ′ 1 , . . ., r ′ d ) = (r 1 , . . ., r d ) where (r ′ 1 , . . ., r ′ d ) ≥ (r 1 , . . ., r d ) and S D (r ′ For example, in Figure 4, gcd(S 1 ) = gcd(S 3 ) = (3, 2), while gcd(S 2 ) = (3, 1).As we shall see later, this notion of the gcd of a decomposition corresponds well to the existing notion of the gcd of an NECS.Next, define In particular, consider the case when for some n ≥ 2, then the splitting sequence for S is non-empty, and so the gcd(S) = (1, . . ., 1).Thus, we see that the only decomposition S where gcd(S) = (1, . . ., 1) is the trivial decomposition (0, 1) d .This implies that Hence, it follows that S d,(1,...,1) (x) = x.For general r 1 , . . ., r d , we have the following.
Lemma 5.For all d, n, r 1 , . . ., r d ∈ N, Thus, if we let ℓ i be the least common multiple of r i and b i , it follows that S H i,ℓ i ((0, 1) d ) .Since b i was chosen maximally (by the definition of gcd), it must be that b i = ℓ i for every i.This means that r i |b i , and so there exists a i ∈ N where b i = a i r i .This finishes our proof.
With Lemmas 4 and 5, we are now ready to prove a result that is slightly more general than Theorem 2.
Proof.From Lemmas 4 and 5, we obtain that for every fixed n ≥ 1, In fact, we see that y ( d i=1 r i) n = a 1 ,...,a d ≥1 S d,(a 1 q 1 ,...,a d q d ) (x) for all q 1 , . . ., q d ∈ N where Now we multiply both sides of ( 6) by µ d (n) and sum them over n ≥ 1.On the left hand side, we obtain For the right hand side, we have = This finishes the proof. When and thus Theorem 2 follows as a consequence.
then the functional equation in Theorem 2 can be restated simply as M d (y) = x.We have shown earlier that s d (n) gives the coefficient of x n in y, hence finding a combinatorial interpretation of the coefficients of the compositional inverse of the series M d (z).In this section, we'll mainly focus on the coefficients of what's essentially the multiplicative inverse of M d (z).For every d ≥ 1 and n ≥ 0, define Then it follows that for all n ≥ 1, with the initial condition a d (0) = 1.The following table gives some values of a d (n) for small d and n: The main goal of this section is to establish two results (Proposition 10 and Lemma 12) that we will need when we study the asymptotic behavior and growth rate of s d (n).While we do so, we will take somewhat of a scenic route and uncover some properties of a d (n) along the way.

A lower bound for a d (n + 1)/a d (n)
When we prove an asymptotic formula for s d (n) in Section 4, one of the requirements will be to show that a d (n) ≥ 0 for every d ≥ 1 and n ≥ 0. Likewise, Goulden et al. [6,Theorem 2] showed that a 1 (n) ≥ 0 with a complex analysis argument as a part of their work establishing an asymptotic formula for c(n).
In this section, we describe the set Ãd,n and show that | Ãd,n | = a d (n) for all d ≥ 1 and n ≥ 0. Having a combinatorial interpretation for a d (n) implies (among other things) that these coefficients are indeed nonnegative.Before we are able to define Ãd,n , we first need to introduce some intermediate objects B d,n , A d,n and mention some of their relevant properties along the way.Given d ∈ N, we call a multiset S = {s 1 , . . ., s ℓ } a d-coloured prime set if the following holds: • The elements s 1 , . . .s ℓ are (not necessarily distinct) prime numbers.
• Each of s 1 , . . ., s ℓ is assigned a colour from [d], such that each instance of the same prime number must be assigned distinct colours.
We also define the weight of S to be w(S) = ℓ i=1 s i − 1, and the sign of S to be v(S) = (−1) ℓ+1 .For every n ≥ 0, let B d,i denote the set of d-coloured prime sets with weight i.For instance, (where the assigned colour of a number is indicated in its subscript), and B 2,26 = ∅.More generally, notice that if n + 1 has prime factorization n . Thus, we see that Next, for every d ≥ 1 and n ≥ 0, let A d,n to be the set of sequences (A 1 , . . ., A k ) where In other words, v(A) = 1 if there is an odd number of even-sized sets in the sequence A, and v(A) = −1 otherwise.For example, the following table lists all elements of A 1,n for 0 ≤ n ≤ 7, as well as their signs.
To reduce cluttering, we used | to indicate separation of sets, and suppressed the colour assignment of the numbers since there is only d = 1 available colour in this case.
Lemma 7.For every d ≥ 1 and n ≥ 0, Proof.We prove the claim by induction on n.First, a d (0) = 1, and A d,0 consists of just the empty sequence A = (), and v(A) = 1.Thus, the base case holds.Now suppose n ≥ 1, and let = a d (n).
Next, we show that there are always more sequences in A d,n with positive signs than those with negative signs.Given a sequence A = (A 1 , . . ., A k ) ∈ A d,n , we say that a subsequence (A j , A j+1 , . . ., A j+ℓ ) of A is odd, ascending, and repetitive (OAR) if 1. (Odd) A j = {ℓ}, a singleton set, and |A j+1 |, . . ., |A j+ℓ | are all odd.
2. (Ascending) Let c 0 be the colour assigned to the element ℓ ∈ A j .Then every element of A j+1 is either greater than ℓ, or is equal to ℓ and assigned a colour greater than c 0 .
. Define indices i 1 (A), i 2 (A) as follows: • i 1 (A) is the smallest index where |A i | is even; • i 2 (A) is the smallest index where an OAR subsequence begins.
Notice that, when both defined, i 1 (A) and i 2 (A) must be distinct for any sequence A, since an OAR subsequence must begin with a singleton set, which has odd size.Next, given sequence A, we define the sequence f (A) as follows: • Case 1: i 1 (A) is defined, and i 2 (A) > i 1 (A) or is undefined.Let i = i 1 (A), and let p 0 ∈ A i be the smallest prime number that is assigned the minimum colour.Let • Case 2: i 2 (A) is defined, and i 1 (A) > i 2 (A) or is undefined.Let i = i 2 (A) and let p 0 be the unique element in A i .Define f (A) by taking A and replacing the OAR subsequence A j , A j+1 , . . ., A j+p 0 by the set A j ∪ A j+1 , with the colour assignment to elements unchanged.
For example, given A = ({2} , {3} , {11} , {11} , {11} , {5, 7}), then i 1 (A) = 6 and i 2 (A) = 2, and we have Notice that if v(A) = −1, then i 1 (A) must be defined, and so f (A) is defined.Since f either replaces one even set by a number of odd sets (Case 1) or vice versa (Case 2), we see that v(f (A)) = −v(A) whenever f (A) is defined.It is also not hard to check that w(f (A)) = w(A) in both cases.Furthermore, notice that if f (A) is defined, then f (f (A)) = A.This shows that f is a bijection between the sets {A ∈ A d,n : v(A) = −1} and {f (A) : Thus, f is injective when we consider it as a mapping from {A ∈ A d,n : v(A) = −1} to | {A ∈ A d,n : v(A) = 1}, and our claim follows.
Thus, the number of elements A ∈ A d,n with v(A) = −1 never outnumber those with v(A) = 1.In fact, the function f defined in the proof above can be seem as "pairing up" the sequences in A d,n that contains an even set or an OAR subsequence (or both).Thus, we have obtained the following.
(Again, we have used subscripts to denote assigned colours of elements.)The function f can be seen as appending the set {2 c } at the end of the given sequence (which increases the total weight by 1 and does not create an even set).If this does create an OAR subsequence, it must be that the last three sets of the new sequence are {2 c ′ } , {2 c } , {2 c } where c ′ < c.In this case, we further replace the two instances of {2 c } by one instance of {3 c }, which does not change the weight of the whole sequence, and now guarantees that it does not have an OAR subsequence.
Since f (A, c) must have weight n + 1 and does not contain any even sets nor OAR subsequences, it is an element of Ãd,n+1 .It is also easy to see that A and c are uniquely recoverable from f (A, c), and so f (A, c) is injective.Thus, we conclude that We illustrate in Figure 5 the values of a d (n+1) a d (n) for some small values of d and n.While we have shown that the ratio is bounded below by d, the figure suggests that d + 1 is a tight upper bound.We provide an algebraic proof that this is indeed true for all d ≥ 3.
Proposition 11.For every d ≥ 3 and n ≥ 0, To do so, we'll need a lemma that will also be useful in Section 4.
Proof.We first prove (i).Notice that for all n ∈ 2 k , . . ., 2 k+1 − 1 , n has at most k prime factors (counting multiplicities), and so For the last inequality, notice that 2 i−1 ≥ i for all i ≥ 1. Substituting i = ℓ − k + 1 and then multiplying both sides by 2 k gives 2 Next, when d ≥ 3, k ≥ 3, and , and so For the case d = 2, notice that µ 2 (2 k ) = 0 for all k ≥ 3. Thus, using essentially the same chain of inequalities above, we obtain that Next, we prove (ii) using similar observations.Given d ≥ 2, k ≥ 3, and x ≤ 1 2d , We are now ready to prove Proposition 11.
Proof of Proposition 11.We prove our claim by induction on n.First, using (9), we obtain that From the above, one can check that a d (n+1) a d (n) ≤ d + 1 for all 0 ≤ n ≤ 5. Next, assume n ≥ 6. Observe that Using the inductive hypothesis as well as Proposition 10, we see that Also, from Proposition 10 again, a d (n − 6 − i) ≤ d −i a d (n − 6) for every i ≥ 1, and so using Lemma 12(i).Thus, where the last inequality relies on the assumption that d ≥ 3.
As seen in Figure 5, it appears that Proposition 11 is also true for d = 1 and d = 2.A combinatorial proof for that would be interesting.

Asymptotic Formula and Growth Rate of s d (n)
In this section, we consider the asymptotic behavior and growth rate of s d (n).We will also end the section by describing a mapping from labelled plane rooted trees to hypercube decompositions that will help provide additional context to the growth rate of s d (n).

An asymptotic formula
The main tool we will rely on is the following result from Flajolet and Sedgewick [5, Theorem VI.6, p. 404]: Theorem 13.Let y be a power series in x.Let φ : C → C be a function with the following properties: (i) φ is analytic at z = 0 and φ(0) > 0; (ii) y = xφ(y); (iii) [z n ]φ(z) ≥ 0 for all n ≥ 0, and [z n ]φ(z) = 0 for some n ≥ 2.
(iv) There exists a (then necessarily unique) real number s ∈ (0, r) such that φ(s) = sφ ′ (s), where r is the radius of convergence of φ. Then, Using Theorem 13, we obtain the following: Theorem 14.Let d ≥ 1, and let s > 0 be the smallest real number such that M ′ d (s) = 0.Then, Proof.We first verify the analytic conditions listed in Theorem 13.Since M d (y) = x, to satisfy (ii) we have y = xφ(y) where where the coefficients a d (n) were defined in (8) and studied in Section 3. It is easy to see that φ(0) = 1 (for all d) and that Φ is analytic at z = 0, and so (i) holds.Condition (iii) follows readily from Proposition 10.For (iv), notice that 3  .
(11) Let r be the radius of convergence of φ at z = 0. Since φ(z) = z M d (z) , r is the smallest positive solution to M d (r) = 0. Given M d (0) = M d (r) = 0 and that M d (z) is differentiable over (0, r), there must exist s ∈ (0, r) where M ′ d (s) = 0. Now observe that Thus, condition (iv) holds.Now that the analytic assumptions on φ(z) have been verified, we may establish the asymptotic formula.When M ′ d (s) = 0, the expressions in (11) simplifies to Therefore, we have This completes the proof.

Growth rate
We define the growth rate of s d (n) to be Goulden et al. [6,Theorem 2] showed that K 1 ≈ 5.487452 in their work on NECS.Here, we show that d = 1 turns out the only case where K d < 4d +  s) where s is the smallest positive real number where M ′ d (s) = 0.For convenience, we define the polynomials Then from Lemma 12 we know that , which is negative over [0, 1  2d ].Thus, M + d (x) is concave down over this interval.and ).The second inequality holds because we showed above that s + maximizes M + d (x) over [0, 1  2d ].The third inequality holds since M + d (x) is concave down, and so is an increasing function, and we obtain that L(s + ) ≤ L(k 2 ).

Relating trees and hypercube decompositions
We end this section by describing a mapping from plane rooted trees to hypercube decompositions, which will also add some perspective to the bounds we found for K d in Proposition 15.
Given an integer d ≥ 1, let T d be the set of plane rooted trees where each internal node has a label from [d] and at least 2 children, while the leaves of the tree are unlabelled.Furthermore, let T d,n ⊆ T d denote the set of trees with exactly n leaves.Then we can define a tree-to-decomposition mapping Ψ : T d,n → S d,n recursively as follows: • Ψ maps the tree with a single leaf node to the trivial decomposition (0, 1) d .
• Now suppose T ∈ T d,n has root node labelled i with r children.Let T 1 , . . ., T r be the subtrees of the root node ordered from left to right.Also, for each j ∈ [r], let   The mapping Ψ is a natural extension of another tree-to-decomposition mapping that Bagherzadeh, Bremner, and the author used to study a generalization of Catalan numbers [2].It is not hard to see that the mapping Ψ is onto -given S ∈ S d,n , we can use the sequence of splitting operations that resulted in S to generate a tree T ∈ T d,n where Ψ(T ) = S. On the other hand, Ψ is not one-to-one.Figure 8 gives two types of situations where two distinct trees are mapped to the same decomposition.
In general, if we let t d (n) = |T d,n |, then we see that t d (n) ≥ s d (n) for every d, n ≥ 1, with the inequality being strict for all n ≥ 4. We remark that when d = 1, we can consider the nodes of the trees in T 1,n as being unlabelled, and so t 1 (n) gives the well-studied small Schröder numbers [7, A001003].Thus, the sequences t d (n) can be seen as a generalization of small Schröder numbers, which the author recently studied in another manuscript [1].
In particular, we have the following for the growth rate of t d (n):

Relating Decompositions and NECS
An immediate consequence of Theorem 2 is that s 1 (n) = c(n) for all n ≥ 1.In this section, we highlight some combinatorial connections between hypercube decompositions and NECS.

A bijection between S 1,n and C n
Given an NECS C = { a i , n i : i ∈ [k]}, define gcd(C) to be the greatest common divisor of the moduli n 1 , . . ., n k .Furthermore, when we discuss the gcd of a decomposition in S ∈ S 1,n , we'll slightly abuse notation and write gcd(S) = r (instead of gcd(S) = (r), for ease of comparing gcd's of decompositions and NECS.Next, we define the mapping Φ : S 1,n → C n recursively as follows: • (n = 1) Φ maps the trivial decomposition {(0, 1)} to { 0, 1 }, the NECS with a single residual class.
(Recall that E j,r ( a, n ) = jn + a, rn , as defined in Section 1.3.) Figure 9 illustrates the mapping Φ applied to elements in S 1,n for n ≤ 4. Next, we prove a result that will help us show that Φ is in fact a bijection.
Since Φ is gcd-preserving from Lemma 17, we are able to use a single column in Figure 9 to indicate the gcd of both the input decomposition and the output NECS.Next, we prove that Φ is indeed a bijection.Proposition 18. Φ : S 1,n → C n is a bijection.
Proof.From Theorem 2, we know that |S 1,n | = |C n |, and so every function from S 1,n to C n is either both one-to-one and onto, or neither.Thus, it suffices to show that Φ is one-to-one, which we will prove by induction on n.The base case n = 1 obviously holds.

Counting decompositions/NECS with a given LCM
We have established a map Φ between 1-dimensional decompositions and NECS that is not only bijective, but also preserves some key combinatorial properties, such as the gcd as shown in Lemma 17.Thus, one can see hypercube decompositions as a generalization of NECS, and any result we prove for hypercube decompositions may also specialize to a corresponding implication for NECS.In this section, we provide one such example.Similar to how we defined the gcd of a given hypercube decomposition, we can also define its lcm.Given a S ∈ S d , we say that lcm(S) = (r 1 , . . ., r d ) if • S D (r 1 ,...,r d ) ; • there doesn't exist (r ′ For the general case, we have the following recursive formulas: , where the sum is over q 1 , . . ., q d ∈ N where q i |r i for every i ∈ [d], and d i=1 q i = 1. (ii) where the sum is over q 1 , . . ., q d ∈ N where q i |r i for every i ∈ Proof.We first prove (i).Given n ∈ N, let P(n) be the set of prime divisors of n (e.g., P(40) = {2, 5}).Now suppose S D (r 1 ,...,r d ) .Then either S = (0, 1) d , or S refines H i,p ((0, 1) d ) for some i ∈ [d] and p ∈ P(r i ).Thus, if we define then we see that We apply the principle of inclusion-exclusion to the above to compute the size of G (r 1 ,...,r d ) .
For convenience, given a finite set S ⊆ N, we let π(S) denote the product of all elements in S. Then notice that each intersection of a collection of k sets from {W i,p : i ∈ [d], p ∈ P(r i )} can be uniquely written as for some choice of P 1 , . . ., P d where P i ⊆ P(r i ) for every i ∈ [d], and d i=1 |P i | = k (note that some of the P i 's could be empty, in which case π(P i ) = 1).Now, the decomposition D (π(P 1 ),...,π(P d )) consists of m = d i=1 π(P i ) regions -let's denote them by B 1 , . . ., B m .Now, if S is a decomposition that belongs to (12), then S D (π(P 1 ),...,π(P d )) , and so is a decomposition in its own right for every j ∈ [m].Furthermore, S D (r 1 ,...,r d ) implies that S j D (r 1 /π(P 1 ),...,r d /π(P d )) .Thus, we see that there is a bijection between ( 12) and G m (r 1 /π(P 1 ),...,r d /π(P d )) , and so (12) has size g r 1 π(P 1 ) , . . ., . This gives Next, notice that if we let q i = π(P i ) for every i ∈ [d], then Thus, the sum above can be rewritten as , where the sum is over q 1 , . . ., q d where q i is a product of prime divisors of r i for every i, and the condition d i=1 |P i | = k ≥ 1 translates to d i=1 q i = 1.Moreover, since µ(q i ) = 0 when q i is divisible by a non-trivial square, the sum above would remain the same if we expanded it to include all q i 's that are divisors of r i , which results in the claimed formula.
Next, we prove (ii) using a similar inclusion-exclusion argument as above.Given S ∈ G (r 1 ,...,r d ) , then either lcm(S) = (r 1 , . . ., r d ) and S ∈ H (r 1 ,...,r d ) , or there exists i ∈ [d] and p ∈ P(r i ) such that S ∈ G (r 1 ,...,r i /p,...,r d ) .Thus, µ(q i ) g r 1 q 1 , . . ., r d q d , where we made the substitution q i = π(P i ) in the last equality and applied (13).For the same rationale as in the proof of (i), the above summation can be taken over all q 1 , . . ., q d where q i |r i , and our claim follows.
With Proposition 19, we obtain a recursive formula to compute the number of hypercube decompositions with a given lcm.In the case of d = 1, we obtain the following sequences: Moreover, given an NECS C = { a i , n i : i ∈ [k]}, if we define lcm(C) to be the least common multiple of n 1 , . . ., n k , then it is not hard to adapt Lemma 17 to show that Φ preserves lcm as well.For example, the last column in Figure 9 gives the lcm of the corresponding decomposition S, as well as that of the NECS Φ(S).Thus, we see that g(n) gives the number of NECS whose lcm divides n, and h(n) gives the number of NECS whose lcm is exactly n. (Goulden et al. [6,Proposition 6] also obtained a recursive formula for finding the number of NECS with a given lcm, gcd, and number of residual classes.)Also, recall that µ(ab) = µ(a)µ(b) whenever a, b are coprime.Thus, if r 1 , . . ., r d ∈ N are pairwise coprime, then it follows from Proposition 19 that g(r 1 , . . ., r d ) = g ).This geometric interpretation of NECS is similar to the "low-dimensional mapping" used by Berger, Felzenbaum, and Fraenkel, who obtained a series of results on covering systems by mapping them to lattice parallelotopes and employing geometric and combinatorial arguments.The reader can refer to [3] and the references therein for their results.It would be interesting to investigate further what more can we learn about covering systems by relating them to hypercube decompositions.

)
Proof.(2) follows immediately from the definition of the Dirichlet convolution and a simple induction on d.For (3), one can use induction on d to show that µ d (p m 1 1 ) = (−1) m 1 d m i for all prime powers p m 1 1 , and then use the facts that • µ is multiplicative (i.e., µ(ab) = µ(a)µ(b) if a, b are coprime);

Figure 4 :
Figure 4: Illustrating the refinement relation on S d .

Corollary 9 .
Define Ãd,n ⊆ A d,n to be the set of sequences that neither contains an even set nor an OAR subsequence.Then a d (n) = | Ãd,n | for every d ≥ 1 and n ≥ 0. Corollary 9 gives us a set whose elements is counted by a d (n).Using this combinatorial description, we prove the following result.Proposition 10.For every d ≥ 1, a d (0) = 1 and a d (n + 1) a d (n) ≥ d for all n ≥ 0. Proof.Given A = (A 1 , . . ., A k ) ∈ Ãd,n , we define f : Ãd,n × [d] → Ãd,n+1 as follows:

3. 2
An upper bound for a d (n + 1)/a d (n)
the values of K d − 4d − 3 2 for 2 ≤ d ≤ 30, as well as the upper bound given in Proposition 15 for comparison.

Figure 7
Figure 7  gives an example of this mapping.

Figure 7 :
Figure 7: Illustrating the mapping Ψ from trees to hypercube decompositions.