A Comparison of Integer Partitions Based on Smallest Part

For positive integers $n, L$ and $s$, consider the following two sets that both contain partitions of $n$ with the difference between the largest and smallest parts bounded by $L$: the first set contains partitions with smallest part $s$, while the second set contains partitions with smallest part at least $s+1$. Let $G_{L,s}(q)$ be the generating series whose coefficient of $q^n$ is difference between the sizes of the above two sets of partitions. This generating series was introduced by Berkovich and Uncu in 2019. Previous results concentrated on the nonnegativity of $G_{L,s}(q)$ in the cases $s=1$ and $s=2$. In the present paper, we show the eventual positivity of $G_{L,s}(q)$ for general s and also find a precise nonnegativity result for the case $s=3$.


Introduction
Let n be a nonnegative integer.A partition π = (π 1 , π 2 , . . . ) of n is a weakly decreasing list of positive integers whose sum is n, and we write |π| = n to indicate this.We allow the empty partition as the unique partition of 0. Each π i is known as a part of π.A standard way to visualize π is through its Ferrers diagram, which is a collection of left justified rows of boxes, with the i th row containing π i boxes.
In the present article, it is more convenient to use the notation that expresses the number of parts of each size in a partition.In this notation, we write π = (1 f 1 , 2 f 2 , . ..),where f i is the frequency of i or the number of times a part i occurs in π.Thus, each frequency f i is a nonnegative integer, and if f i = 0, then π has no part of size i.When the frequency of a number is 0, it may or may not be omitted in the expression.In the latter notation, it is clear that |π| = i i • f i .Thus (4, 4, 2, 2, 1), (1 1 , 2 2 , 3 0 , 4 2 , 6 0 ), and (1 1 , 2 2 , 4 2 , 5 0 ) all represent the same partition of 13.We let s(π) and l(π) denote the smallest and largest parts of π, respectively, and ℧ denotes the set of partitions π with |π| > 0.
We emphasize that Theorem 1 is stronger than the conjecture of Berkovich and Uncu about H L,s,k (q) as it explicitly gives the bound for when H L,s,k (q) is positive, and it further states this bound depends only on s.
Berkovich and Uncu [BU19, Theorems 5.1 and 5.2] showed that the series G L,s (q) and H L,s,k (q), for s = 1 and s = 2 and L ≥ 1, satisfy the following simple relationship: A series S(q) = n≥0 a n q n is said to be nonnegative if a n ≥ 0 for all n.The nonnegativity of the series S(q) is denoted by S(q) 0.
In the present article, we explore the nonnegativity properties of G L,s (q) for general positive integers s.We begin the study of this series with the next result.
We prove Theorem 3 in Section 2, but we emphasize that the proof is essentially the one given by Berkovich and Uncu for the cases s = 1 and s = 2, with only minor modifications.We include this proof for completeness.Next we show that for any L ≥ s + 1, the series G L,s (q) is eventually positive, and the bound after which the coefficient of q N is positive can be written explicitly in terms of s only.Define the quantities δ(s) := e 3Γ(s) , and δ ′ (s) := 10s + (s + 2)(s + 3)(δ(s) + 1). (5) Obviously δ ′ (s) > δ(s) for all positive s.
Theorem 4. If s and L ≥ s + 1 are positive integers, then the coefficient of q n in G L,s (q) is positive whenever n ≥ δ ′ (s), so G L,s (q) eventually positive.
We prove Theorem 4 in Section 2. Then we focus on the case s = 3 and obtain an extension of Theorem 2; that is, we show that, with the exception of a few small terms, the series G L,s (q) is nonnegative.The next result states this precisely, and its proof is in Section 3.
We point out that the bound in Theorem 4 is likely far from optimal.Take, for example, the case s = 3.According to Theorem 4, the coefficient of q n in G L,3 (q) is nonnegative whenever n ≥ δ ′ (3), where δ ′ (3) is is extremely large.However, from Theorem 5, the coefficient of q n in G L,3 (q) is nonnegative whenever n ≥ 45.This suggests that the bound in Theorem 4 can be improved greatly.
Theorems 3, 4 and 5 can also be found in the PhD.thesis of the first author [Bin21].

Proofs of Theorems 3 and 4
We begin by proving Theorem 3.
Proof of Theorem 3. The definition of G L,s (q) in (1) is given as the difference of two generating series.We begin by finding a rational expression for the first generating series.All the partitions counted by this generating series have smallest part equal to s and largest part at most L + s.Hence we obtain for the first generating series the expression π∈℧, s(π)=s, l(π)−s(π)≤L For the second generating series in the definition of G L,s (q), we fix the number of parts of the partition to be n and then sum over all n.Suppose π is a partition with n parts, where each part is at least s + 1.Then, in the Ferrers diagram of π, the whole column over the smallest part of π is generated by the q-factor q (s+1)n 1 − q n .Stripping the columns above the smallest part from the far left of the Ferrers diagram of π, we are left with a new partition that has at most n − 1 parts and largest part bounded above by L. It is well known (see for example [Aig07, Proposition 1.1]) that these partitions are generated by the q-binomial coefficient Thus, for the second generating series in the definition of G L,s (q), we have Simplifying the summands on the right hand side, we find where the last step follows from the q-binomial theorem (see [BU19, (2.1)]) ∞ n=0 (a; q) n (q; q) n z n = (az; q) ∞ (z; q) ∞ with a = q L and z = q s+1 .Substituting (6) and (7) into the definition of G L,s (q) gives us Theorem 3 expresses G L,s (q) in terms of H L,s,L (q), while Theorem 1 gives the explicit bound Γ(s) after which coefficients in the series H L,s,L (q) are positive.We use these to show nonnegativity properties of G L,s (q).We first show in Theorem 6 that there is a bound M, which depends only on L and s, such that the coefficient of q n in G L,s (q) is nonnegative whenever n ≥ M.
Theorem 6.Let s and L ≥ s + 1 be positive integers.Then the coefficient of q n in G L,s (q) is nonnegative whenever n ≥ η 3 (L, s).
Proof.Suppose n ≥ η 3 (L, s).We can rewrite (8) as Note that the second sum may be empty.Since n ≥ η 3 (L, s), the first sum on the right hand side of (9) contains at least η 2 (L, s) terms, all of which are positive by Theorem 1. Thus For the second sum in the right hand side of (9), Theorem 1 gives For the third sum in the right hand side of (9), using the triangle inequality, we obtain and thus The result now follows immediately from (9), (10), (11) and (12).
The bound η 3 (L, s) in Theorem 6 guaranteeing when the coefficients of G L,s (q) are nonnegative depends on both L and s.To prove Theorem 4, we need to find a bound that only depends on s, and this is our next goal.We again use the connection between G L,s (q) and H L,s,L (q) in Theorem 3; while Theorem 1 guarantees the series H L,s,L (q) is eventually positive, we also need a lower bound on the size of the coefficients of H L,s,L (q).This is the content of Theorem 8 below, a strengthening of Theorem 1 in the case k = L. Let • D L,s denote the set of nonempty partitions with parts in the set {s + 1, . . ., L + s}, and • I L,s,L be the set of partitions where the smallest part is s, all parts are ≤ L + s, and L does not appear as a part.
From the definition of the series H L,s,L (q) in (2), when L ≥ s+1, elementary partition theory gives the coefficient of q N in H L,s,L (q) as the difference Thus to prove nonnegativity of the coefficient of q N in H L,s,L (q), it suffices to show there exists an injection φ such that Equations ( 13) and ( 14) are central to proving our theorems below in this section and the next.We also need the following result.
Proposition 7.For given positive integers a, b and n with gcd(a, b) = 1, the number of solutions of ax + by = n in nonnegative integer pairs (x, y) is either ⌊ n ab ⌋ or ⌊ n ab ⌋ + 1. See [Tri00].See also [NZM91, Chapter 5] for an elementary proof.
Theorem 8.For positive integers L ≥ 3 and s, with L ≥ s + 1, the coefficient of q N in H L,s,L (q) is greater than or equal to Proof.The proof of Theorem 1, found in [BR21], uses the combinatorial interpretation of the coefficients of H L,s,L (q) in (13); the proof there constructs for N ≥ Γ(s) an injection as in (14) to show nonnegativity of (13).To show positivity of (13), elements of the codomain that are not in the range of φ are given.The details of these injections and elements are dealt with in different cases and theorems of [BR21] depending on the relative sizes of L and s.1 For example, when L ≥ 2s + 3, there is no partition of the form (s 10 , (s + 1) x , (s + 2) y ) in the range of φ, but such partitions are in the codomain.To achieve the present result, we count the number of such partitions; this will then give the desired lower bound for the difference (13) in each case.We then compare the results of all cases.
Table 1 lists the cases, the theorem from [BR21] showing positivity of (13), the partitions in the codomain but not the range of φ showing positivity of (13), and the enumeration of these partitions.To count the number of partitions in Column 3 in Table 1, in Rows 1-4 a straight forward application of Proposition 7 to n = N − ts, where t is the number of parts of s, and a and b set to be the remaining parts, which are coprime, gives the numbers in Column 4.
The remaining rows require a slightly more work.For Row 5, the numbers s + 1 and s + 3 have greatest common factor 2, so we can apply Proposition 7 to n = N −s 2 , a = s+1 2 and b = s+3 2 .The count in Column 4 then follows.The analysis for Rows 6 and 7 are similar.The result is now obtained by observing that all the values in Column 4 exceed Table 1: The first column describes the case, the second the theorem from [BR21] that proves positivity in this case, the third the partitions in the codomain that are not in the range of φ in (14), and the last column contains the number of partitions of the type in Column 3 .

Case of L, s
Theorem Partitions in codomain min.num, of [BR21] not in range of φ of partitions For a positive integer m, let p(m) be the number of partitions of m.We need the following result of de Azevedo Pribitkin [dAP09].
Theorem 9. Let m be a positive integer.Then p(m) ≤ e 3 √ m .
Proof of Theorem 4. As before, let a L,n and b L,n be the coefficient of q n in H L,s,L (q) and G L,s (q), respectively.Further, recall the definitions of Γ(s), δ(s), and δ ′ (s) in (3) and (5).Suppose n ≥ δ ′ (s).Again from (8), we have For m ≥ δ ′ (s), we have m ≥ Γ(s) and ⌊ m−10s (s+2)(s+3) ⌋ ≥ δ(s), so a L,m ≥ δ(s) by Theorem 8.The first sum in the right hand side of (15) contains at least 1 term (the term m = n) and each term in the sum is greater than or equal to δ(s).Thus For the second sum in the right hand side of (15), from Theorem 1 it follows For the third sum on the right hand side of (15), the combinatorial interpretation of H L,s,L (q) in (13) gives a L,m ≥ −p(m) for any m ∈ N. By Theorem 9, we then have where the last inequality follows from the familiar formula for finite geometric sums and the definition of δ(s).The theorem now follows immediately from (15), ( 16), ( 17) and (18).

Proof of Theorem 5
To prove Theorem 5, we use the connection between G L,3 (q) and H L,3,L (q) given in Theorem 3. To use this relationship, we need to understand the coefficients of q N in H L,3,L (q) for small N, so we need a result, in the case s = 3 and L ≥ s + 1, stronger than Theorem 1.Our strategy for proving the coefficient of q N in H L,3,L (q) is nonnegative for small N is to show that the coefficient of q N in H L,3,L (q) is nonnegative when N is larger than a small bound; this is done in Lemmas 15 -19.Then we use the lemmas along with machine computation and Theorem 3 to prove Theorem 5 at the end of the section.
Recall the following fundamental result of Sylvester [Syl82].
Theorem 10 (Sylvester's theorem).Let a and b be positive coprime integers.Then the equation ax + by = n has a nonnegative integer solution (x, y) whenever n ≥ (a − 1)(b − 1).
We additionally need the following lemmas.
Lemma 11.Let n ≥ 4 be a positive integer such that n = 7.Then the equation 4x+5y+6z = n has a solution in nonnegative integer triples (x, y, z).
Lemma 12. Let n ≥ 5 be a positive integer such that n = 8, 9. Then the equation 5x + 6y + 7z = n has a solution in nonnegative integer triples (x, y, z).
The proofs of these lemmas are all simple applications of Sylvester's theorem.For example, the proof of Lemma 11 can be obtained by applying Sylvester's theorem to a = 4 and b = 5, which establishes the conclusion for all n ≥ 12, and then each smaller n can be dealt with individually.The proofs of Lemmas 12 and 13 are similar, so we omit the details.
We also need another lemma.
The next lemma shows that the coefficient of q N in H L,3,L (q) is positive for most values of L and small N. Positivity, as opposed to nonnegativity, is needed in this case to prove Theorem 5 at the end of the section.
Lemma 15.For L ≥ 22 and N ≥ 21, the coefficient of q N in H L,3,L (q) is positive.
Proof.Fix L ≥ 22 and N ≥ 21.Recall the combinatorial interpretation of the coefficients of H L,3,L (q) in (13).We prove nonnegativity of the coefficients of H L,3,L (q) by constructing an injective map φ as in ( 14) for s = 3. Positivity will then be shown at the end of the proof by displaying an element of the codomain of φ that is not in its range.
Let π = 4 f 4 , . . ., L f L , . . ., (L + 3) f L+3 be a partition of N in D L,3 and let f denote f L .Recall that partitions of N in I L,3,L , the codomain of φ, have parts in the set {3, . . ., L + 3}, the number 3 must occur as a part, and L does not occur as a part.Our proof of injectivity of φ is as follows.
• We define φ(π) in cases chiefly determined by f in π, with several subcases that depend on the frequencies of other parts of π.In each case, it will usually be readily apparent that φ is injective, but we will provide some justification for more complicated cases.
• When we analyze why φ is injective overall, we will gather cases by the frequency of 3 in the image of φ.Thus each case is labelled twice: first by its case determined by the frequency f of L in π (see below, for example, Case 2(c)(ii)(α)) and second, in parentheses, by the frequency of 3 in φ(π) (for example, (B2)).Once the cases are gathered by their frequency of 3 in the image of φ, we analyze, for each fixed i, all cases where the frequency of 3 is i in the image of φ, and we argue why φ is injective collectively in these cases.For example, on Page 14 the cases (A * ) are all the cases where the frequency of 3 is 1 in the image of φ.
• We then argue that φ must be injective overall because distinct cases where in the image of φ the frequency of 3 in one case is i and the frequency of 3 in the other case is j, where i = j, cannot contain common elements, so two partitions π and π ′ pertaining to distinct cases i and j cannot have the same image.
We now define φ.
Case 1 (F1, K1) (this case, exceptionally, has many values for the frequency of 3 in a partition in the image of φ, so it has more than one pink label): f ≥ 1.Notice (L − 18)i ≥ 4 for all i ≥ 1, so the equation has a nonnegative integer solution by Lemma 13.For each i ≥ 1, fix such a solution x i , y i , z i and u i .Define φ(π) = 3 6f , 4 f 4 +x f , 5 f 5 +y f , 6 f 6 +z f , 7 f 7 +u f , . . ., L 0 , . . . .
The function φ is injective in this case.Given a partition φ(π) = 3 6f , 4 a , 5 b , 6 c , 7 d , . . ., L 0 , . . . in the range φ, we can infer π comes from this case (no cases below have the same frequency of 3).From the frequency of 3 in φ(π), we can infer f ; then, from (19), we can infer x f , y f , z f and u f ; finally, from f and x f , y f , z f and u f , we can reconstruct π.
Case 2: f = 0. We have the following subcases.Recall that the smallest part of π is denoted by s(π).
Our explanation for why φ is injective in this case is similar to that in Case 1. Suppose that we are given an element in the range of φ of the form φ(π) = 3 1 , 5 1+A , 6 B , 7 C , m fm 1 −1 1 , . . . .Then, using (20), we can find m 1 − 3 from A, B and C, and from there m 1 can be recovered.From m 1 , we can reconstruct the partition π uniquely.
There are cases below where the reasoning that φ is injective is similar to this case and Case 1, so we omit the details there.
To prove the injectivity of the map φ, we organize the cases based on the various frequencies of 3 in φ(π).
First we organize the cases where the frequency of 3 in φ(π) is 1.
, . . ., where m 1 ≥ 8, m 1 = 11, 12, and A, B and C are some nonnegative integers such that at least one of these is positive.
Each case above is individually injective (we can find π from φ(π)).We explain why the map φ is injective overall so far, and we do this by confirming that no two distinct cases contain common partitions.In Case A1, the second smallest and the third smallest parts differ by at least 3, and the frequency of the second smallest part is 1.This distinguishes it from all the other cases.In Case A2, the number 4 is not present as a part, which distinguishes it from Cases A3 and A4, and it contains 5 as a part, which distinguishes it from Cases A5 and A6.In Cases A3 and A4, the number 4 is present as a part, which distinguishes it from Cases A5 and A6.Cases A5 and A6 are distinguished by the frequency of the second smallest part.
We follow the same reasoning below.We collect cases according to the frequency of 3 in partitions in the range of φ.We then explain why all the distinct cases with fixed frequency of 3 have no partitions in common in their range.We leave the verification that each of the different cases for fixed frequency of 3 in the range of φ are individually injective to the reader.
Next we organize the cases where the frequency of 3 in φ(π) is 2.

These cases are distinguished by their frequencies of 4 and 5.
There is only one case where the frequency of 3 in φ(π) is 3: So it is distinguishable from other cases.
Next we organize the cases in which the frequency of 3 in φ(π) is 4.
Thus, when the frequency of 3 in φ(π) is 4, these cases are distinguishable by the frequency of 6 in the image.Next we organize the cases in which the frequency of 3 in φ(π) is 5.
Thus, when the frequency of 3 in φ(π) is 6, these cases are distinguishable.Next we organize the cases in which the frequency of 3 in φ(π) is 7.
Thus, when the frequency of 3 in φ(π) is 7, the next parts after 3 and their frequencies distinguish the various cases.
There is only one case in which the frequency of 3 in φ(π) is 8: , . . ., where m 3 ≥ 12 and f m 3 ≥ 1.
Next we organize the cases in which the frequency of 3 in φ(π) is 9.
Hence all the cases are distinguishable, and the map φ is injective.This shows nonnegativity of the coefficient of q N in H L,3,L (q) when L ≥ 22 and N ≥ 21.To show these coefficients are positive, we find an element of the codomain of φ that is not in its range.In all cases such an element will have frequency of 3 equalling 4, and these can be compared to (D1) -(D3) to ensure they are not in the range of φ.If N = 21 and N = 22, the partitions π 21 = (3 4 , 4 1 , 5 1 ) and π 22 = (3 4 , 5 2 ) are partitions not in the range but in the codomain of φ.
We are left with the cases 4 ≤ L ≤ 21.We deal with 7 ≤ L ≤ 21 in Lemma 16 below.The proof of this lemma is similar in spirit to the proof of [BR21,Theorem 19].
Lemma 16.Let 7 ≤ L ≤ 21 and N L = L 2 + 10L + 7. Then the coefficient of q N in H L,3,L (q) is nonnegative whenever N ≥ N L .
Proof.Again it suffices to show that for fixed 7 ≤ L ≤ 21 and N ≥ N L , there is an injective map φ as in (14).Let π = 4 f 4 , . . ., L f L , . . ., (L + 3) f L+3 be a partition of N in D L,3 and let f = f L .We define φ(π) in cases depending on the value of f .Case 1: f is positive and f ≡ 0 (mod 3).Then define Case 2: f ≡ 1 (mod 3).Then define Case 3: f ≡ 2 (mod 3).Then define Case 4: f = 0. Since N ≥ N L is large enough, either f L+2 ≥ 6 or there exists an i = L+ 2 such that 4 ≤ i ≤ L + 3 and f i ≥ 3. Note that the condition on N is in fact tight for this to happen.We have further subcases.
Proof.Again it suffices to show that for L = 5 and fixed N ≥ 164, there is an injective map φ as in (14).Recall that partitions of N in I 5,3,5 (q), the codomain of φ, have smallest part 3, no part equal to 5, and largest part at most 8. Let π = 4 f 4 , 5 f 5 , . . ., 8 f 8 be a partition of N in D 5,3 and let f denote f 5 .We define φ(π) in cases depending on the value of f .
It is easy to see that φ is injective in each case.To see that φ is injective overall, note that the frequency of 3 in the image distinguishes the cases.In Cases 1, 2 and 3, the frequency of 3 is 0, 4 and 1 modulo 5, respectively.In Cases 4 and 5, it is always 2 or 3 modulo 5 and different for each subcase.Hence the map φ is injective.
Proof.Again it suffices to show that for L = 4 and fixed N ≥ 1042, there is an injective map φ as in (14).Recall that partitions of N in I 4,3,4 (q), the codomain of φ, have smallest part 3, no part equal to 4, and largest part at most 7. Let π = (4 f 4 , 5 f 5 , 6 f 6 , 7 f 7 ) be a partitions of N in D 4,3 and let f = f 4 .We define φ(π) in cases depending on f .
For n ≥ 10, Lemma 12 guarantees that there exists nonnegative integer solutions (x n , y n , z n ) of the equation For each n ≥ 10, fix a nonnegative integer solution (x n , y n , z n ) to the equation.Case 1: 10 ≤ f < 100.Define It is easy to see that φ is injective in this case: given an element φ(π) whose frequency is between 10 and 100, the frequency of 3 gives f , and the values of x f , y f and z f can be found from (21).The partition π can then be reconstructed.Similar arguments can be used to show that φ is injective in the other cases below.Case 2: f ≥ 100.Define Case 3: 0 ≤ f ≤ 9. Since N ≥ 1042 is large enough, at least one of the following conditions is true: (i) f 5 ≥ 62; (ii) f 6 ≥ 57; or (iii) f 7 ≥ 53.We deal with each case below.Note that the condition on N is in fact tight here.
Case 3(i): f 5 ≥ 62. Define Case 3(ii): f 5 ≤ 61 and f 6 ≥ 57.Define It is easy to see that φ is injective in each case.To see that φ is injective overall, note that the frequency of 3 in the image distinguishes the cases.Hence the map φ is injective.
Lemmas 15, 16, 17, 18 and 19 show that for N larger than a small number, the coefficient of q N in H L,s,L (q) is nonnegative.With these results, the use of computer searches, and Theorem 3 (with s = 3), we can now prove Theorem 5.

Proof of Theorem
First we focus on the case L ≥ 22.By Lemma 15, the coefficients satisfy a L,N ≥ 1 whenever N ≥ 21.For N ≤ 20, we observe that a L,N is independent of L. To see why, the combinatorial interpretation of a L,N in (13) gives that, when L ≥ N + 1, the number a L,N is the difference between the number of partitions of N with smallest part 3 and the number of partitions of N with smallest part at least 4; that is, the condition on the largest part of the partitions becomes superfluous.Thus, when 1 ≤ N ≤ 20, since L ≥ N + 1, we have a L,N = a N +1,N , and these 20 values can all be found by a computer search.The search finds that a N +1,N is negative only when N is one of 4, 5, 8, 10, 12, 14 or 16, and in each case a N +1,N is exactly −1.Thus, for any N ≥ 1, the right hand side of (22) contains at most one term equal to -1, while the rest of the terms are positive.It follows from (22) that b L,N is negative only when N is one of 4, 5, 8, 10, 12, 14 or 16, and in each case b L,N is exactly −1.This gives Theorem 5 for L ≥ 22.
The remaining cases are easier.For 7 ≤ L ≤ 21, Lemma 16 renders the unknown values of a L,N to the cases N ≤ N L , a finite set of values that can be searched using a computer.These computations along with (22) give Theorem 5. Similarly, for 4 ≤ L ≤ 6, the Lemmas 17, 18 and 19 leave only a finite number of unknown values for a L,N for small N, which can all be found using a computer.These values and an application of (22) complete the proof of Theorem 5.
Remark.The programming for 7 ≤ L ≤ 21 and N ≤ N L turned out to be a difficult task in Magma.For example, it is hard to calculate the number of partitions of 250 with all parts in the set {4,5, . . ., 17} using Magma.The command P artitions(250, min part = 4, max part = 17).cardinality() in Sage also does not work (it takes too long and ultimately stops working).We overcame this problem through another related command and some mathematics.In Sage, we noticed that the command P artitions(n, max part = 17).cardinality() is very fast even for large n (even until n = 1000000, it is fast!)Thus, we calculate the number of partitions with all parts in the set {4, 5, . . ., 17} in terms of the number of partitions p 17 (n) of n with maximum part at most 17.We do this by viewing partitions with all parts in the set {4, 5, . . ., 17} as partitions with maximum part 17 and no part 1, 2 and 3. Let A, B and C denote the set of partitions of n with maximum part 17 and also having 1, 2 and 3 as a part, respectively.Then we need to find the cardinality of the set A ∁ ∩ B ∁ ∩ C ∁ .Using inclusion and exclusion principle, we get that the number of partitions of n with all parts in the set {4, 5, . . ., 17} is given by p 17 (n) − p 17 (n − 1) − p 17 (n − 2) + p 17 (n − 4) + p 17 (n − 5) − p 17 (n − 6), and thus can be easily computed.
4 The series G L,s (q) for s ≥ 4 and other generalizations For fixed L and s, let p L,s (q) be the smallest degree polynomial in q with smallest coefficients such that G L,s (q) + p L,s (q) 0. This paper finds p L,s (q) for s = 3 and all L, while the cases s = 2 (found in Theorem 2) and s = 1 (found in [BU19]) were found earlier.One goal is to find p L,s (q) for general s and L. Our numerical computations for s = 4 and s = 5 do not suggest a nice form for p L,s (q).A simpler problem, though also interesting, is to determine the degree of p L,s (q) for general L and s.Furthermore, we suspect that for fixed s ≥ 4 that p L,s (q) stabilizes; that is, there is an L 0 such that p L,s (q) = p L 0 ,s (q) for all L ≥ L 0 , as in the case for s = 3 (Theorem 5 gives L 0 = 10 when s = 3).Finding L 0 as a function of s is also another open problem.
Moreover, series analogous to G L,s (q) for partitions with further restrictions on parts, such as for partitions with only odd parts or for self conjugate partitions, may prove interesting.