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Annals of Combinatorics

, Volume 23, Issue 3–4, pp 589–611 | Cite as

Quasipolynomials and Maximal Coefficients of Gaussian Polynomials

  • Angelica Castillo
  • Stephanie Flores
  • Anabel Hernandez
  • Brandt KronholmEmail author
  • Acadia Larsen
  • Arturo Martinez
Article
  • 11 Downloads

Abstract

We establish an algorithm for producing formulas for p(nmN), the function enumerating partitions of n into at most m parts with no part larger than N. Recent combinatorial results of H. Hahn et al. on a collection of partition identities for p(n, 3, N) are considered. We offer direct proofs of these identities and then place them in a larger context of the unimodality of Gaussian polynomials \(N+m\brack m\) whose coefficients are precisely p(nmN). We give complete characterizations of the maximal coefficients of \({M\brack 3}\) and \({M\brack 4}\). Furthermore, we prove a general theorem on the period of quasipolynomials for central/maximal coefficients of Gaussian polynomials. We place some of Hahn’s identities into the context of some known results on differences of partitions into at most m parts, p(nm), which we then extend to p(nmN).

Keywords

Integer partition Gaussian polynomial Quasipolynomial q-Series 

Mathematics Subject Classification

Primary 11P81 Secondary 05A15 

1 Prologue

There is no question that George Andrews knows his mathematics. He also knows his mathematicians, his students in particular. The origins of this paper begin with the following message sent by George Andrews to the fourth author.

It occurred to me that you might think about applying your methods to the coefficients in the Gaussian polynomial (i.e. p(j,k,n)). Given the success of Hahn et al., this seems like a likely venture. Best wishes, George email, 6/7/2017

As happy as we are that Professor Andrews knows us, we are ever more thankful that we know him.

2 Introduction

Along with the celebrated Hardy–Ramanujan [19] asymptotic formula for the unrestricted partition function, given by
$$\begin{aligned} p(n)\sim \frac{\text {e}^{\pi \sqrt{\frac{2n}{3}}}}{4n\sqrt{3}} \end{aligned}$$
as \(n\rightarrow \infty \), formulas for partition functions remain at the leading edge of research. In this paper, we introduce a very natural way of producing formulas for p(nmN), the function that enumerates partitions of n into at most m parts with no part larger than N. For example, following the methods in this paper, the formula for \(p(12k+2,4,12j+5)\) obtained after a few straightforward computations:
$$\begin{aligned} p(12k+2,4,12j+5)&= 2\genfrac(){0.0pt}1{k+3 }{ 3}+39\genfrac(){0.0pt}1{k+2}{ 3}+30\genfrac(){0.0pt}1{k+1}{3}+\genfrac(){0.0pt}1{k}{3} -41\genfrac(){0.0pt}1{k+2-j}{3} \\&\quad -194\genfrac(){0.0pt}1{k+1-j}{3}-53\genfrac(){0.0pt}1{k-j}{3} +2\genfrac(){0.0pt}1{k+2-2j}{3}+160\genfrac(){0.0pt}1{k+1-2j}{3} \\&\quad +250\genfrac(){0.0pt}1{k-2j}{3}+20\genfrac(){0.0pt}1{k-1-2j}{3} -16\genfrac(){0.0pt}1{k+1-3j}{3}-173\genfrac(){0.0pt}1{k-3j}{3} \\&\quad -98\genfrac(){0.0pt}1{k-1-3j}{3} -\genfrac(){0.0pt}1{k-2-3j}{3} +15\genfrac(){0.0pt}1{k-4j}{3}\\&\quad +48\genfrac(){0.0pt}1{k-1-4j}{3}+9\genfrac(){0.0pt}1{k-2-4j}{3}. \end{aligned}$$
With a computer algebra program, one might compute \(p(420k+297, 7,420j)\), which begins with \(248893190\genfrac(){0.0pt}1{k+6 }{ 6}+40291579602\genfrac(){0.0pt}1{k+5 }{ 6} +\cdots \) and concludes approximately 50 terms later with \(\cdots -8212122234\genfrac(){0.0pt}1{k+1- 7j}{ 6} -1805085\genfrac(){0.0pt}1{k- 7j}{ 6}.\)

The techniques employed in this paper lead us to a complete characterization of the maximal coefficients of all Gaussian polynomials of the form \({N+3\brack 3}\) in Sect. 4. This characterization is complete in terms of quantity and quality. For example, we prove that for \(\ell \ge 0\), there are exactly three largest coefficients of \({4\ell +1 \brack 3}\), and they are \(p(6\ell -4,3,4\ell -2),~p(6\ell -3,3,4\ell -2)\) and \(p(6\ell -2,3,4\ell -2)\) and that they are equal to \(2\ell ^2\). We go on to characterize the maximal coefficients of \({N+4\brack 4}\) and then establish a general theorem on the period of quasipolynomials for maximal coefficients of Gaussian polynomials in Sect. 5. Section 6 details a general result on the difference \(p(n,m,N)-p(n-1,m,N-1)\).

Despite the fact that the generating function for p(nmN) is the well-known Gaussian polynomial denoted by \(N+m\brack m\), the authors could find very little information on formulas for p(nmN). However, a Hardy–Ramanujan formula can be found in [1] and formulas based on Sylvester’s “waves” can be found in [24]. Further asymptotic behavior of the coefficients of Gaussian polynomials can be found in [26]. Although Sylvester [27] was the first to prove the unimodality of Gaussian polynomials, we make special mention of Kathleen O’Hara’s celebrated constructive proof of unimodality [22]. Recent results on strict unimodality may be found in [8] and [23].

In this paper, we will also encounter the function p(nm) which enumerates partitions of n into at most m parts. Unlike p(nmN), formulas for p(nm) are plentiful. Many formulas for specific small values of m are catalogued in [12, 13, 15, 21]. Andrews [3] notes several varieties of formulas for p(n, 4). The simplest among these may be the 1896 result of Glösel [14]:
$$\begin{aligned} p(n,4)=\Big \lfloor \Big \lfloor \frac{n+4}{2} \Big \rfloor ^2 \left( 3\Big \lfloor \frac{n+9}{2} \Big \rfloor - \Big \lfloor \frac{n+10}{2} \Big \rfloor \right) \frac{1}{36}\Big \rceil , \end{aligned}$$
(2.1)
where \(\lfloor \cdot \rceil \) is the nearest integer function. In [25], one will find a computer algebra (Maple) package that “completely automatically discovers, and then proves, explicit expressions \(\ldots \) for p(nm) for any desired m” in similar fashion to (2.1).

The methods in this paper can also be applied to compute formulas for p(nm), although the resulting formulas are of a different nature than (2.1) and those in the previously cited references. Our methods come from a novel manipulation of generating functions and q-series initially informed by the arithmetic of E. Ehrhart’s polyhedral combinatorics [9, 10, 11]. In short, we create a piece-wise polynomial, otherwise known as a quasipolynomial, and our results follow after a bit of arithmetic.

Definition 2.1

A function f(n) is a quasipolynomial if there exist d polynomials \(f_0{(n)},\ldots , f_{d-1}{(n)}\), such that
$$\begin{aligned} f(n) = {\left\{ \begin{array}{ll} f_{0}(n), &{} \text {if } n\equiv 0 \pmod {d},\\ f_{1}(n), &{} \text {if } n\equiv 1 \pmod {d},\\ ~~~\vdots &{} ~~~~~~\vdots \\ f_{d-1}(n), &{} \text {if } n\equiv {d-1} \pmod {d}, \end{array}\right. } \end{aligned}$$
for all \(n\in {\mathbb {Z}}\). The polynomials \(f_i\) are called the constituents of f and the number of them, d, is the period of f.

The formulas which we present for p(n, 3, N) and p(n, 3) will be displayed following the format of Definition 2.1.

We will express these formulas in a binomial basis \(\genfrac(){0.0pt}1{a}{b}\) and/or a monomial basis \(\alpha k^n+\beta k^{n-1}+\cdots +\omega k+z\), as appropriate, because there is geometric and combinatorial meaning inferred from each format. The accompanying Ehrhart Theory and polyhedral geometry will not be considered beyond just a few scattered comments in this paper. However, the geometric implications could be of significant interest and may be considered in future research.

Breuer’s paper [6] is a very nice and comprehensive introduction to Ehrhart’s ideas and [7] features an exploration of the deeper geometric, algebraic, and combinatorial mathematics related to partitions.

3 Background Information

The generating functions for the partition functions p(nm) and p(nmN) are well known:
$$\begin{aligned} \sum _{n=0}^{\infty }p(n,m)q^n= & {} \frac{1}{(1-q)(1-q^2)\cdots (1-q^m)}=\frac{1}{(q;q)_m}, \end{aligned}$$
(3.1)
$$\begin{aligned} \sum _{n=0}^{mN}p(n,m,N)q^n= & {} \frac{(q;q)_{N+m}}{(q;q)_m(q;q)_N}=\frac{(q^{N+1};q)_{m}}{(q;q)_m}={N+m \brack m}. \end{aligned}$$
(3.2)
For \(n<0\), we agree that \(p(n,m)=p(n,m,N)=0\) and whenever \(n>Nm\), \(p(n,m,N)=0\).

Gaussian polynomials \({N+m \brack m}\) are q-analogues of binomial coefficients. A proof that \({N+m \brack m}\) is a polynomial can be found in Chapter 3 of [2]. Also, in Chapter 14 of [2], one can find a small table of coefficients for a selection of Gaussian polynomials.

We require the following definition.

Definition 3.1

Let \(\text {lcm}(m)\) denote the least common multiple of the numbers \(1,2,3,\ldots ,m\).

For example, \(\text {lcm}(3)=6\), \(\text {lcm}(4)=12\), \(\text {lcm}(5)=60\), etc.

3.1 Quasipolynomials for p(nm)

Because it is somewhat simpler, we begin with p(nm). Our procedure is to recast the generating function for p(nm) as the product of a polynomial and a generating function for binomial coefficients:
$$\begin{aligned} \sum _{n= 0}^{\infty } p(n,m)q^{n} = \frac{1}{(q;q)_m}&= \frac{1}{(q;q)_m}\times \frac{\left( \frac{(1-q^{\text {lcm}(m)})^{m}}{(q;q)_{m}}\right) }{\left( \frac{(1-q^{\text {lcm}(m)})^{m}}{(q;q)_{m}}\right) } \nonumber \\&=\frac{\prod _{j=1}^{m}\sum _{i=0}^{\frac{\text {lcm}(m)-j}{j}}q^{ij}}{(1-q^{\text {lcm}(m)})^m}\nonumber \\&=\prod _{j=1}^{m}\sum _{i=0}^{\frac{\text {lcm}(m)-j}{j}}q^{ij}\times \sum _{k=0}^{\infty }{k+m-1\atopwithdelims ()m-1}q^{\text {lcm}(m)k} \nonumber \\&= E_m(q)\times \sum _{k=0}^{\infty }{k+m-1\atopwithdelims ()m-1}q^{\text {lcm}(m)k}. \end{aligned}$$
(3.3)
The penultimate equality in (3.3) comes from the generating function for binomial coefficients:
$$\begin{aligned} \frac{1}{(1-q)^b}=\sum _{a\ge 0}{a+b-1\atopwithdelims ()b-1}q^{a}, \end{aligned}$$
and the final equality is simply introducing some notation:
$$\begin{aligned} E_m(q) = \prod _{j=1}^{m}\sum _{i=0}^{\frac{\text {lcm}(m)-j}{j}}q^{ij}. \end{aligned}$$
(3.4)
To establish a quasipolynomial and obtain formulas for p(nm), one fixes m and then proceeds to multiply and collect like terms in the far right side of (3.3). With the exponent on q being \(\text {lcm}(m)k\) in the far right side of (3.3), it is natural to set \(n=\text {lcm}(m)k+r\) for \(0\le r <\text {lcm}(m)\) and expect a quasipolynomial of period \(\text {lcm}(m)\).
As a worked example and because we will use this information later in Sect. 6.2, we compute the quasipolynomial for p(n, 3):
$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }p(n,3)q^{n}=\frac{1}{(q;q)_{3}} =E_3(q)\times \sum _{k=0}{k+2 \atopwithdelims ()2}q^{6k}. \end{aligned} \end{aligned}$$
Noting that
$$\begin{aligned} E_3(q){=}1 {+} q {+} 2 q^2 {+} 3 q^3 {+} 4 q^4 {+} 5 q^5 {+} 4 q^6 {+} 5 q^7 + 4 q^8 {+} 3 q^9 {+} 2 q^{10} {+} q^{11} {+} q^{12}, \end{aligned}$$
we compute for instance, the formula for \(p(6k+1,3)\):
$$\begin{aligned} \sum _{k=0}^{\infty }p(6k+1,3)q^{6k+1}&=\left( q+5q^7\right) \times \sum _{k= 0}^{\infty }{k+2\atopwithdelims ()2}q^{6k} \nonumber \\&=\sum _{k=0}^{\infty }\left( {k+2\atopwithdelims ()2}+5{k+1\atopwithdelims ()2}\right) q^{6k+1}. \end{aligned}$$
(3.5)
Hence
$$\begin{aligned} p(6k+1,3)={k+2 \atopwithdelims ()2} + 5{k+1 \atopwithdelims ()2}= 3k^2+4k+1. \end{aligned}$$
Multiplying and collecting like terms five more times give us the complete quasipolynomial for p(n, 3) displayed below:
$$\begin{aligned} p(n,3)= {\left\{ \begin{array}{ll} p(6k,3) &{} = 1{k+2 \atopwithdelims ()2} + 4{k+1 \atopwithdelims ()2} + 1{k \atopwithdelims ()2} = 3k^2+3k+1,\\ p(6k+1,3) &{} = 1{k+2 \atopwithdelims ()2} + 5{k+1 \atopwithdelims ()2} = 3k^2+4k+1,\\ p(6k+2,3) &{} = 2{k+2 \atopwithdelims ()2} + 4{k+1 \atopwithdelims ()2} = 3k^2+5k+2,\\ p(6k+3,3) &{} = 3{k+2 \atopwithdelims ()2} + 3{k+1 \atopwithdelims ()2} = 3k^2+6k+3,\\ p(6k+4,3) &{} = 4{k+2 \atopwithdelims ()2} + 2{k+1 \atopwithdelims ()2} = 3k^2+7k+4,\\ p(6k+5,3) &{} = 5{k+2 \atopwithdelims ()2} + 1{k+1 \atopwithdelims ()2} = 3k^2+8k+5. \end{array}\right. }\nonumber \\ \end{aligned}$$
(3.6)

3.2 Quasipolynomials for p(nmN)

Following the notation \(E_m(q)\) in (3.4), for Gaussian polynomials, we set
$$\begin{aligned} (q^{N+1};q)_{m}=G_{m,N}(q) \end{aligned}$$
(3.7)
and write
$$\begin{aligned} {N+m \brack m}= & {} \sum _{n= 0}^{mN} p(n,m,N)q^{n} \nonumber \\= & {} G_{m,N}(q)\times E_m(q)\times \sum _{k=0}^{\infty }{k+m-1\atopwithdelims ()m-1}q^{\text {lcm}(m)k}. \end{aligned}$$
(3.8)
Despite the fact that we are expressing the Gaussian polynomial as the product of a polynomial and a power series, the procedure for computing a quasipolynomial for p(nmN) is the same as it was for p(nm). After setting
$$\begin{aligned} n=\text {lcm}(m)k+r \end{aligned}$$
for \(0\le r <\text {lcm}(m)\) and
$$\begin{aligned} N=\text {lcm}(m)j+t \end{aligned}$$
for \(0\le t <\text {lcm}(m)\), we multiply and collect like terms in the right side of (3.8). Because there are \(\text {lcm}(m)\) choices for N in \(G_{m,N}\) and there are \(\text {lcm}(m)\) constituents part and parcel with \(E_m(q)\), the natural quasipolynomial in two variables for p(nmN) consists of \(\text {lcm}(m)^2\) constituents.
We compute the 36 constituents of the quasipolynomial for p(n, 3, N):
$$\begin{aligned} {N+3 \brack 3}&=\sum _{n= 0}^{3N} p(n,3,N)q^{n} = \frac{(q;q)_{N+3}}{(q;q)_{3}(q;q)_N} \nonumber \\&= G_{3,N}(q)\times E_3(q)\times \sum _{k=0}^{\infty }{k+2 \atopwithdelims ()2}q^{6k}. \end{aligned}$$
(3.9)
We expand \( G_{3,N}(q)\times E_3(q)\) below:
$$\begin{aligned}&G_{3,N}(q)\times E_3(q)\nonumber \\&\quad = 1+ q + 2 q^2 + 3 q^3 + 4 q^4 + 5 q^5 + 4 q^6 + 5 q^7 + 4 q^8 + 3 q^9 + 2 q^{10} + q^{11} + q^{12}\nonumber \\&\qquad - q^{1 + N} - 2 q^{2 + N} - 4 q^{3 + N} - 6 q^{4 + N} - 9 q^{5 + N} - 12 q^{6 + N} - 13 q^{7 + N}\nonumber \\&\qquad - 14 q^{8 + N} - 13 q^{9 + N} - 12 q^{10 + N} - 9 q^{11 + N} - 6 q^{12 + N} - 4 q^{13 + N}- 2 q^{14 + N}\nonumber \\&\qquad - q^{15 + N} + q^{3 + 2 N} + 2 q^{4 + 2 N} + 4 q^{5 + 2 N} + 6 q^{6 + 2 N} + 9 q^{7 + 2 N}+ 12 q^{8 + 2 N}\nonumber \\&\qquad + 13 q^{9 + 2 N} + 14 q^{10 + 2 N} + 13 q^{11 + 2 N} + 12 q^{12 + 2 N}+ 9 q^{13 + 2 N}+ 6 q^{14 + 2 N}\nonumber \\&\qquad + 4 q^{15 + 2 N} + 2 q^{16 + 2 N} + q^{17 + 2 N} - q^{6 + 3 N}- q^{7 + 3 N} - 2 q^{8 + 3 N} - 3 q^{9 + 3 N}\nonumber \\&\qquad - 4 q^{10 + 3 N} - 5 q^{11 + 3 N} - 4 q^{12 + 3 N} - 5 q^{13 + 3 N} - 4 q^{14 + 3 N} - 3 q^{15 + 3 N}\nonumber \\&\qquad - 2 q^{16 + 3 N} - q^{17 + 3 N} - q^{18 + 3 N}. \end{aligned}$$
(3.10)
Depending on N modulo 6 in (3.10), we establish the quasipolynomial for p(n, 3, N). The 36 constituents for this quasipolynomial appear in Appendix A and are referenced from there throughout the remainder of the paper.

We display one constituent followed by a definition as fuel for discussion.

Example 3.2

Replacing N with \(6j+4\) in (3.10), expanding \(G_{3,6j+4}(q)\times E_3(q)\), and selecting exponents of the form \(n=6k+3\) as one multiplies and collects like terms, we obtain the following constituent:
$$\begin{aligned} p(6k+3,3,6j+4)&= 3\genfrac(){0.0pt}1{k+2 }{ 2} + 3\genfrac(){0.0pt}1{k+1 }{ 2}- 9\genfrac(){0.0pt}1{k+1-j }{ 2} -9\genfrac(){0.0pt}1{k-j }{ 2} + 9\genfrac(){0.0pt}1{k-2j }{ 2} \nonumber \\&\quad + 9\genfrac(){0.0pt}1{k-1-2j }{ 2} -3\genfrac(){0.0pt}1{k-1-3j }{ 2} -3\genfrac(){0.0pt}1{k-2-3j }{ 2}. \end{aligned}$$
(3.11)

Regarding the constituents of p(nmN), we adhere to a strict interpretation of binomial terms.

Definition 3.3

Let a and b be natural numbers.
  • Whenever \(a<b\), then \({a \atopwithdelims ()b}=0\).

  • Whenever \(a\ge b\), we allow the usual translation to the monomial basis: \({a \atopwithdelims ()b}=\frac{a!}{b!(a-b)!}\).

The immediate reason for this comes from our q-series arithmetic where we recast the standard generating function for p(nmN) from an index of n to a new index of k for \(k\ge 0\) and the fact that N is non-negative. Another reason for this interpretation comes from the polyhedral geometry in which a term like \(9{k-2j\atopwithdelims ()2}\) in (3.11) indicates a certain “tiling” of a slice of a certain partition cone. See [6, 7] for more information. There are partition theoretic interpretations for some negative variables when it comes to theorems of combinatorial reciprocity [5].

4 A Characterization of the Maximal Coefficients of \({N+3\brack 3}\)

4.1 Motivation from Recent Results

We are motivated by the following results of Hahn et al. in [17] and [18].

Theorem 4.1

([17]). For any integer \(\ell \ge 1\), one has:
$$\begin{aligned} p(6\ell -3,3,4\ell -2)-p(6\ell -4,3,4\ell -2)=0,\end{aligned}$$
(4.1)
$$\begin{aligned} p(6\ell ,3,4\ell )-p(6\ell -1,3,4\ell )=1,\end{aligned}$$
(4.2)
$$\begin{aligned} p(6\ell -3,3,4\ell -1)-p(6\ell -4,3,4\ell -1)=1,\end{aligned}$$
(4.3)
$$\begin{aligned} p(6\ell ,3,4\ell +1)-p(6\ell -1,3,4\ell +1)=1. \end{aligned}$$
(4.4)

Theorem 4.2

([18]). For any integer \(\ell \ge 1\), one has:
$$\begin{aligned} p(6\ell ,3, 4\ell )-p(6\ell -3,3,4\ell -1)=\ell +1,\end{aligned}$$
(4.5)
$$\begin{aligned} p(6\ell -1,3,4\ell )-p(6\ell -4,3,4\ell -1)=\ell +1,\end{aligned}$$
(4.6)
$$\begin{aligned} p(6\ell +3,3,4\ell +2)-p(6\ell ,3,4\ell +1)=\ell +1,\end{aligned}$$
(4.7)
$$\begin{aligned} p(6\ell +2,3,4\ell +2)-p(6\ell -1,3,4\ell +1)=\ell +2. \end{aligned}$$
(4.8)

Theorem 4.1 was first established in [17] while working on detection of subgroups of \(\mathrm{GL}_n\) by representations and are part of Langlands’ beyond endoscopy proposal [16]. Combinatorial proofs of both theorems appear in [18].

Notice that both Theorem 4.1 and Theorem 4.2 deal with coefficients near or at the middle of Gaussian polynomials \({N+3\brack 3}\). It is possible to directly verify each line of Theorem 4.1 simply by referring to the relevant constituents from the quasipolynomial for p(n, 3, N) in Appendix A. Instead, we prove a result that places Theorem 4.1 in the context of the unimodality of all Gaussian polynomials of the form \({N+3\brack 3}\). The first part of the characterization is established by Theorem 4.6 which, depending on N, describes how many maximal coefficients for \({N+3\brack 3}\) which one should expect. The second part of this characterization is found in Corollary 4.8 in which we create a quasipolynomial of period 4 for the maximal coefficients of \({N+3\brack 3}\). We state analogous results for \(N+4\brack 4\) in Sect. 4.4.

4.2 Enumerating the Maximal Coefficients of \({N+3\brack 3}\)

We require the following definitions before we can prove our results.

Definition 4.3

A polynomial
$$\begin{aligned} P(q)=a_0+a_1q+a_2q^2+\cdots +a_dq^d \end{aligned}$$
of degree d is said to be reciprocal if for each i, one has \(a_i=a_{d-i}\).

Definition 4.4

A polynomial
$$\begin{aligned} P(q)=a_0+a_1q+a_2q^2+\cdots +a_dq^d \end{aligned}$$
of degree d is called unimodal if there exists m, such that
$$\begin{aligned} a_0\le a_1 \le a_2 \le \cdots \le a_m \ge a_{m+1} \ge a_{m+2} \ge \cdots \ge a_d. \end{aligned}$$

Because the coefficients of Gaussian polynomials are positive, unimodal and reciprocal, it follows that the value of the central coefficient(s) of a Gaussian polynomial will be greater than or equal to all the other coefficients.

Remark 4.5

Whenever \(\deg {N+m \brack m}\) is odd, there will be a pair of central coefficients. Otherwise, there will be a single central coefficient. As such, \(p\left( \lfloor \frac{mN}{2}\rceil ,m,N\right) \) is one of the central and hence one of the maximal coefficient(s) of \(N+m \brack m\), where \(\lfloor \cdot \rceil \) is the nearest integer function.

Theorem 4.6

There are at most four but never exactly two maximal coefficients for any Gaussian polynomial of the form \({N+3\brack 3}\).

Proof

We will prove Theorem 4.6 in four cases. We will show that for \(k\ge 0\),

Case 1. There is exactly one largest coefficient for Gaussian polynomials of the form \({{4k+3}\brack 3}\).

Case 2. There are exactly four largest coefficients for Gaussian polynomials of the form \({{4k+4}\brack 3}\).

Case 3. There are exactly three largest coefficients for Gaussian polynomials of the form \({{4k+5}\brack 3}\).

Case 4. There are exactly four largest coefficients for Gaussian polynomials of the form \({{4k+6}\brack 3}\).

Case 1 is proved entirely from (4.2) of Theorem 4.1. We will make use of (4.1), from Theorem 4.1, and the quasipolynomial for p(n, 3, N) in Appendix A to prove Case 3. The remaining Cases 2 and 4 are verified in the same way as Case 3 and are left to the reader.

We now treat Case 1. With \(\deg {{4k+3}\brack 3}=12k\), there are exactly \(12k+1\) terms in these Gaussian polynomials, the central of which is \(p(6k,3,4k)q^{6k}\). For \(k\ge 0\), we will show:
$$\begin{aligned} p(6k-1,3,4k)<p(6k, 3, 4k)>p(6k+1,3,4k). \end{aligned}$$
We note that Case 1 is satisfied in the instance of \(k=0\) where we have \({{3}\brack 3}=1\). Now, by (4.2) of Theorem 4.1 and the reciprocity of Gaussian polynomials, we immediately obtain:
$$\begin{aligned} \begin{aligned} p(6\ell ,3,4\ell )-p(6\ell -1, 3, 4\ell )=1,\\ p(6\ell ,3,4\ell )-p(6\ell +1,3,4\ell )=1. \end{aligned} \end{aligned}$$
(4.9)
Replacing \(\ell \) with k in (4.9), we observe that
$$\begin{aligned} p(6k-1,3,4k)<p(6k, 3, 4k)>p(6k+1,3,4k). \end{aligned}$$
Hence, Case 1 is proved.
We prove Case 3 by showing
$$\begin{aligned} p(6k+1,3,4k+2) < p(6k+2,3,4k+2) = p(6k+3,3,4k+2)\\ = p(6k+4,3,4k+2) > p(6k+5,3,4k+2). \end{aligned}$$
Since \(\deg {{4k+5}\brack 3}=12k+6\), we find that the central coefficient of \({{4k+5}\brack 3}\) is:
$$\begin{aligned} p(6k+3,3,4k+2). \end{aligned}$$
(4.10)
Because Gaussian polynomials are reciprocal, we have:
$$\begin{aligned} p(6k+2,3,4k+2)=p(6k+4,3,4k+2). \end{aligned}$$
(4.11)
Setting \(k=\ell -1\) in both (4.10) and (4.11) and taking the result of (4.1) from Theorem 4.1, we obtain:
$$\begin{aligned} p(6\ell -4,3,4\ell -2)=p(6\ell -3,3,4\ell -2)=p(6\ell -2,3,4\ell -2), \end{aligned}$$
and hence
$$\begin{aligned} p(6k+2,3,4k+2)=p(6k+3,3,4k+2)=p(6k+4,3,4k+2). \end{aligned}$$
(4.12)
We now show that these three coefficients are maximal by proving:
$$\begin{aligned} p(6k+4,3,4k+2)-p(6k+5,3,4k+2) = 1, \end{aligned}$$
(4.13)
which, because of the reciprocity of the coefficients of Gaussian polynomials, will simultaneously show:
$$\begin{aligned} p(6k+2,3,4k+2)-p(6k+1,3,4k+2) = 1. \end{aligned}$$
(4.14)
We prove (4.13), hence (4.14), by making use of the quasipolynomial for p(n, 3, N) in Appendix A. We have three cases to consider in (4.13): \(4k+2 = 6j, 6j+2\) and \(6j+4\). For the case \(4k+2=6j \), we set \(k=3f+1\), so that we may rewrite (4.13) for \(f\ge 0\) as:
$$\begin{aligned} p(6(3f+1)+4,3,6(2f+1))-p(6(3f+1)+5,3,6(2f+1)). \end{aligned}$$
(4.15)
Turning to (A.5) and (A.6), we compute (4.15):
$$\begin{aligned}&p(6(3f+1)+4,3,6(2f+1))-p(6(3f+1)+5,3,6(2f+1))\nonumber \\&\quad =4\genfrac(){0.0pt}1{3f+3}{ 2} + 2\genfrac(){0.0pt}1{3f+2}{ 2}- 6\genfrac(){0.0pt}1{f+2}{ 2} - 12\genfrac(){0.0pt}1{f+1}{ 2} +2\genfrac(){0.0pt}1{-f+1}{ 2} +14\genfrac(){0.0pt}1{-f}{ 2} \nonumber \\&\qquad +\genfrac(){0.0pt}1{-f-1}{ 2} -4\genfrac(){0.0pt}1{-3f-1}{ 2} -2\genfrac(){0.0pt}1{-3f-2}{ 2} -5\genfrac(){0.0pt}1{3f+3}{ 2} - \genfrac(){0.0pt}1{3f+2}{ 2} +9\genfrac(){0.0pt}1{f+2}{ 2} \nonumber \\&\qquad + 9\genfrac(){0.0pt}1{f+1}{ 2} -4\genfrac(){0.0pt}1{-f+1}{ 2} -13\genfrac(){0.0pt}1{-f}{ 2} -\genfrac(){0.0pt}1{-f-1}{ 2} +5\genfrac(){0.0pt}1{-3f-1}{ 2} -\genfrac(){0.0pt}1{-3f-2}{ 2} \nonumber \\&\quad =-\genfrac(){0.0pt}1{3f+3}{ 2} + \genfrac(){0.0pt}1{3f+2}{ 2}+3\genfrac(){0.0pt}1{f+2}{ 2} -3\genfrac(){0.0pt}1{f+1}{ 2} -0 \nonumber \\&\quad =1. \end{aligned}$$
(4.16)
For the case \(4k+2=6j +2\), we set \(k=3f\) and rewrite (4.13) with the relevant constituents (A.17) and (A.18), so that for \(f\ge 0\), we compute
$$\begin{aligned} p(6(3f)+4,3,6(2f)+2)-p(6(3f)+5,3,6(2f)+2)=1. \end{aligned}$$
(4.17)
For the case \(4k+2=6j +4\), we set \(k=3f+2\) in (A.29) and (A.30) and following (4.17), we compute
$$\begin{aligned} p(6(3f+2)+4,3,6(2f+1)+4)-p(6(3f+2)+5,3,6(2f+1)+4)=1.\nonumber \\ \end{aligned}$$
(4.18)
Hence, (4.16), (4.17) and (4.18) establish (4.13) and, simultaneously, by reciprocity of coefficients, (4.14). Hence, combining (4.12), (4.13) and (4.14), we obtain:
$$\begin{aligned} p(6\ell -5,3,4\ell -2) < p(6\ell -4,3,4\ell -2) = p(6\ell -3,3,4\ell -2) \\ = p(6\ell -2,3,4\ell -2) > p(6\ell -1,3,4\ell -2). \end{aligned}$$
Thus, Case 3 is settled.

Case 2 and Case 4 are similar to Case 3 and can be verified by the reader to complete the proof of Theorem 4.6. \(\square \)

4.3 Formulas for the Maximal Coefficients of \({N+3\brack 3}\)

In this section, we establish a quasipolynomial for the maximal coefficients of Gaussian polynomials \(N+3\brack 3\).

Definition 4.7

For non-negative integers m and N, let \(M_m(N)\) denote the function whose value is equal to the maximal coefficient(s) of the Gaussian polynomial \(N+m\brack m\). Note that for all N, \(M_0(N)=1\).

This is a corollary to Theorem 4.6.

Corollary 4.8

For \(\ell \ge 0\), the quasipolynomial for \(M_3(N)\) has period 4 and is given by:
$$\begin{aligned} M_3(4\ell -2)= & {} \left\{ \begin{array}{lr} p(6\ell -4,3,4\ell -2)\\ p(6\ell -3,3,4\ell -2)\\ p(6\ell -2,3,4\ell -2) \end{array} \right\} =2{\ell +1 \atopwithdelims ()2}+2{\ell \atopwithdelims ()2} =2\ell ^2, \end{aligned}$$
(4.19)
$$\begin{aligned} M_3(4\ell -1)= & {} \left\{ \begin{array}{lr} p(6\ell -3,3,4\ell -1)\\ p(6\ell -2,3,4\ell -1)\\ p(6\ell -1,3,4\ell -1) \\ p(6\ell ,3,4\ell -1) \end{array} \right\} =3{\ell +1 \atopwithdelims ()2}+{\ell \atopwithdelims ()2} =2\ell ^2+\ell , \end{aligned}$$
(4.20)
$$\begin{aligned} M_3(4\ell )= & {} \begin{array}{lr} p(6\ell ,3,4\ell ) \end{array} ={\ell +2 \atopwithdelims ()2}+2{\ell +1 \atopwithdelims ()2}+{\ell \atopwithdelims ()2} =2\ell ^2+2\ell +1, \end{aligned}$$
(4.21)
$$\begin{aligned} M_3(4\ell +1)= & {} \left\{ \begin{array}{lr} p(6\ell ,3,4\ell +1)\\ p(6\ell +1,3,4\ell +1)\\ p(6\ell +2,3,4\ell +1)\\ p(6\ell +3,3,4\ell +1) \end{array} \right\} ={\ell +2 \atopwithdelims ()2}+3{\ell +1 \atopwithdelims ()2} =2\ell ^2+3\ell +1.\nonumber \\ \end{aligned}$$
(4.22)

Corollary 4.8 allows for a straightforward proof of Theorem 4.2 which is completed later in Sect. 6.1.

We prove only (4.21) from Corollary 4.8. With the multiplicity of the maximal coefficients established by Theorem 4.6, the remainder of Corollary 4.8 is proved simply by referring to the quasipolynomial and is left to the reader.

Proof

For (4.21), we have the following three cases: \(4\ell = 6j, 6j+2\) and \( 6j+4\). For the case \(4\ell =6j\), we set \(\ell =3f\), so that we may rewrite \(p(6\ell ,3,4\ell )\) as p(6(3f), 3, 6(2f)). We now refer to (A.1) in the quasipolynomial for p(n, 3, N) and compute:
$$\begin{aligned} p(6(3f),3,6(2f))&=\genfrac(){0.0pt}1{{3f+2}}{ 2}+4\genfrac(){0.0pt}1{{3f+1}}{ 2}+\genfrac(){0.0pt}1{{3f}}{ 2}-12\genfrac(){0.0pt}1{{f+1}}{ 2} -6\genfrac(){0.0pt}1{{f}}{ 2} +6\genfrac(){0.0pt}1{-f+1}{ 2} \\&\quad +12\genfrac(){0.0pt}1{-f}{ 2}-\genfrac(){0.0pt}1{-3f+1}{ 2}-4\genfrac(){0.0pt}1{-3f}{ 2}-\genfrac(){0.0pt}1{-3f-1}{ 2} \\&=18f^2+6f+1. \end{aligned}$$
Since \(\ell =3f\), we replace f with \(\ell /3\) to arrive at:
$$\begin{aligned} \displaystyle p(6\ell ,3,4\ell )= 2\ell ^2+2\ell +1. \end{aligned}$$
(4.23)
For the case \(4\ell =6j+2\), we set \(\ell =3f+2\), so that we may rewrite \(p(6\ell ,3,4\ell )\) as \(p(6(3f+2),3,6(2f+1)+2)\) and working from the constituent for \(p(6k,3,6j+2)\), we compute:
$$\begin{aligned} p(6(&3f+2),3,6(2f+1)+2) \\&=\genfrac(){0.0pt}1{{3f+4}}{ 2}+4\genfrac(){0.0pt}1{{3f+3}}{ 2}+\genfrac(){0.0pt}1{{3f+2}}{ 2} -6\genfrac(){0.0pt}1{{f+2}}{ 2} -12\genfrac(){0.0pt}1{{f+1}}{ 2} \\&\quad +12\genfrac(){0.0pt}1{-f}{ 2}+6\genfrac(){0.0pt}1{-f-1}{ 2} -\genfrac(){0.0pt}1{-3f-2}{ 2} -4\genfrac(){0.0pt}1{-3f-3}{ 2}-\genfrac(){0.0pt}1{-3f-4}{ 2}\\&=18f^2+30f+13. \end{aligned}$$
Since \(\ell =3f+2\), we replace f with \((\ell -2)/3\) to arrive at:
$$\begin{aligned} p(6\ell ,3,4\ell )= 2\ell ^2+2\ell +1. \end{aligned}$$
(4.24)
We finish the proof of (4.21) by considering the case \(4\ell =6j+4\). We set \(\ell =3f+1\), so that we may rewrite \(p(6\ell ,3,4\ell )\) as \(p(6(3f+1),3,6(2f)+4)\) and working from the constituent for \(p(6k,3,6j+4)\), we compute:
$$\begin{aligned}&p(6(3f+1),3,6(2f)+4) \\&\quad =\genfrac(){0.0pt}1{{3f+3}}{ 2}+4\genfrac(){0.0pt}1{{3f+2}}{ 2}+\genfrac(){0.0pt}1{{3f+1}}{ 2}-2\genfrac(){0.0pt}1{{f+2}}{ 2}-14\genfrac(){0.0pt}1{{f+1}}{ 2}-2\genfrac(){0.0pt}1{{f}}{ 2} \\&\qquad +2\genfrac(){0.0pt}1{-f+1}{ 2}+14\genfrac(){0.0pt}1{-f}{ 2}+2\genfrac(){0.0pt}1{-f-1}{ 2}-\genfrac(){0.0pt}1{-3f }{ 2}-4\genfrac(){0.0pt}1{-3f-1}{ 2}-\genfrac(){0.0pt}1{-3f-2}{ 2}\\&\quad =18f^2+18f+5. \end{aligned}$$
Since \(\ell =3f+1\), we replace f with \((\ell -1)/3\) to arrive at:
$$\begin{aligned} p(6\ell ,3,4\ell )= 2\ell ^2+2\ell +1. \end{aligned}$$
(4.25)
Since (4.23), (4.24), and (4.25) are equal, it follows that for \(\ell \ge 0\), the maximal coefficient of the Gaussian polynomial \({{4\ell +3}\brack 3}\) is \(p(6\ell ,3,4\ell )\) and is equal to \(2\ell ^2+2\ell +1\) which proves (4.21) of Corollary 4.8.\(\square \)
As an after-the-fact observation, we note that the quasipolynomial for \(M_3(N)\) goes hand-in-hand with the following generating function
$$\begin{aligned} \sum _{N=0}^{\infty }M_3(N-2)q^N = \frac{q^2(1+q^3)}{(1-q)(1-q^2)(1-q^4)}. \end{aligned}$$
(4.26)
After applying the arithmetic from Sect. 3.1, (4.26) yields the same constituents as Corollary 4.8 but without the full context of Theorem 4.6. Nevertheless, this would appear to be a very fruitful area of further inquiry.

4.4 A Characterization of the Maximal Coefficients of \({N+4\brack 4}\)

We offer the following results on maximal coefficients of \({N+4 \brack 4}\). Proofs are omitted as they are done similarly to Theorem 4.6 and Corollary 4.8.

Theorem 4.9

The maximal coefficient of Gaussian polynomials of the form \({N+4\brack 4}\) is unique except when \(N=1\) in which case there are exactly 5 coefficients, each of which is 1.

Corollary 4.10

For \(\ell \ge 0\), the quasipolynomial for \(M_4(N)\) has period 6 and is given by:
$$\begin{aligned} M_4(6\ell )=p(12\ell ,4,6\ell )&= \genfrac(){0.0pt}1{\ell +3}{ 3} +14\genfrac(){0.0pt}1{\ell +2}{ 3}+20\genfrac(){0.0pt}1{\ell +1}{ 3}+\genfrac(){0.0pt}1{\ell }{ 3} \\&= 6\ell ^3+\frac{15\ell ^2}{2}+\frac{7\ell }{2}+1 , \\ M_4(6\ell +1)= p(12\ell +2,4,6\ell +1)&= \genfrac(){0.0pt}1{\ell +3}{ 3} +20\genfrac(){0.0pt}1{\ell +2}{ 3}+14\genfrac(){0.0pt}1{\ell +1}{ 3}+\genfrac(){0.0pt}1{\ell }{ 3} \\&= 6\ell ^3+\frac{21\ell ^2}{2}+\frac{13\ell }{2}+1, \\ M_4(6\ell +2)= p(12\ell +4,4,6\ell +2)&= 3\genfrac(){0.0pt}1{\ell +3}{ 3}+21\genfrac(){0.0pt}1{\ell +2}{ 3}+12\genfrac(){0.0pt}1{\ell +1}{ 3} \\&= 6\ell ^3+\frac{27\ell ^2}{2}+\frac{21\ell }{2}+3, \\ M_4(6\ell +3)= p(12\ell +6,4,6\ell +3)&= 5\genfrac(){0.0pt}1{\ell +3}{ 3}+23\genfrac(){0.0pt}1{\ell +2}{ 3}+8\genfrac(){0.0pt}1{\ell +1}{ 3} \\&= 6\ell ^3+\frac{33\ell ^2}{2}+\frac{31\ell }{2}+5, \\ M_4(6\ell +4)= p(12\ell +8,4,6\ell +4)&= 8\genfrac(){0.0pt}1{\ell +3}{ 3}+23\genfrac(){0.0pt}1{\ell +2}{ 3}+5\genfrac(){0.0pt}1{\ell +1}{ 3} \\&= 6\ell ^3+\frac{39\ell ^2}{2}+\frac{43\ell }{2}+8, \\ M_4(6\ell +5)= p(12\ell +10,4,6\ell +5)&= 12\genfrac(){0.0pt}1{\ell +3}{ 3}+21\genfrac(){0.0pt}1{\ell +2}{ 3}+3\genfrac(){0.0pt}1{\ell +1}{ 3}, \\&= 6\ell ^3+\frac{45\ell ^2}{2}+\frac{57\ell }{2}+12. \end{aligned}$$
One cannot help, but notice that
$$\begin{aligned} p(12\ell +6,4,6\ell +2)\equiv p(12\ell +10,4,6\ell +5)\equiv 0\pmod {3}. \end{aligned}$$
And, similar to (4.26), we have:
$$\begin{aligned} \sum _{N=0}^{\infty }M_4(N)q^n = \frac{(1+q^3)}{(1-q)(1-q^2)^2(1-q^3)}. \end{aligned}$$
(4.27)

5 On the Period of Quasipolynomials for Maximal Coefficients of \({N+m\brack m}\)

One might expect that the period of the corresponding quasipolynomial for \(M_m(N)\) would naturally be \(\text {lcm}(m)\). However, Corollary 4.8 and Corollary 4.10 lead us to the following result.

Theorem 5.1

The quasipolynomial for \(M_m(N)\) has period: \(\frac{2\cdot \text {lcm}(m)}{m}\).

We note that \(\frac{2\cdot \text {lcm}(m)}{m}\le \text {lcm}(m)\) for \(m\ge 2\), with equality for \(m=2\).

Proof

Beginning with Remark 4.5, we set \(N=\text {lcm}(m)j+b\) for \(0\le b\le \text {lcm}(m)-1\) and write:
$$\begin{aligned} p\left( \Big \lfloor \frac{mN}{2} \Big \rceil ,m,N \right) = p\left( \frac{m\text {lcm}(m)j}{2}+\Big \lfloor \frac{mN}{2} \Big \rceil ,m,\text {lcm}(m)j+b \right) . \end{aligned}$$
(5.1)
Set
$$\begin{aligned} \frac{m\text {lcm}(m)j}{2}+\Big \lfloor \frac{mN}{2} \Big \rceil = \text {lcm}(m)k+a \end{aligned}$$
for \(0\le a\le \text {lcm}(m)-1\) and solve for j which depends on k to obtain:
$$\begin{aligned} j=\frac{2\left( \text {lcm}(m)k+a-\Big \lfloor \frac{mN}{2} \Big \rceil \right) }{m\text {lcm}(m)}. \end{aligned}$$
(5.2)
Now, set \(k=\frac{\text {lcm}(m)f}{2}+i\), for those i with \(|i|<\frac{\text {lcm}(m)}{4}\), such that
$$\begin{aligned} \text {lcm}(m)i+a-\Big \lfloor \frac{mN}{2} \Big \rceil =\text {lcm}(m)t \end{aligned}$$
for some integer \(t\ge 0\). We are interested in only these k.
Setting
$$\begin{aligned} j=\frac{2\left( \frac{\text {lcm}(m)f}{2}+t\right) }{m}, \end{aligned}$$
we rewrite the left side of (5.1) as:
$$\begin{aligned} p\left( \frac{m\text {lcm}(m)\left( \frac{2\left( \frac{\text {lcm}(m)f}{2}+t\right) }{m}\right) }{2}+\Big \lfloor \frac{mN}{2} \Big \rceil ,m,\text {lcm}(m)\left( \frac{2\left( \frac{\text {lcm}(m)f}{2}+t\right) }{m}\right) +b \right) .\nonumber \\ \end{aligned}$$
(5.3)
Writing
$$\begin{aligned} \Big \lfloor \frac{mN}{2} \Big \rceil =x\text {lcm}(m)+r_1 \end{aligned}$$
for \(0\le r_1\le \text {lcm}(m)-1\) and
$$\begin{aligned} b=x\frac{2\left( \frac{\text {lcm}(m)f}{2}+t\right) }{m}+r_2 \end{aligned}$$
for \(0\le r_2\le \frac{2\cdot \text {lcm}(m)}{m}-1\), we rewrite (5.3) as:
$$\begin{aligned} p\left( \text {lcm}(m) \left( \frac{\text {lcm}(m)f}{2}+t+x \right) +r_1, m, \frac{2\cdot \text {lcm}(m)}{m}\left( \frac{\text {lcm}(m)f}{2}+t+x \right) +r_2 \right) .\nonumber \\ \end{aligned}$$
(5.4)
Finally, allowing
$$\begin{aligned} \frac{\text {lcm}(m)f}{2}+t+x =\ell , \end{aligned}$$
we have
$$\begin{aligned} p\left( \Big \lfloor \frac{mN}{2} \Big \rceil ,m,N \right) = p\left( \text {lcm}(m) \ell +r_1, m, \frac{2\cdot \text {lcm}(m)}{m}\ell +r_2 \right) . \end{aligned}$$
(5.5)
Hence, quasipolynomials for \(M_m(N)\) have a period of length \(\frac{2\cdot \text {lcm}(m)}{m}\).\(\square \)

It can be shown that the quasipolynomial for \(M_5(N)\) has period 24, and curiously, for \(M_6(N)\), the period is even shorter with length 20.

The \(\frac{2\cdot \text {lcm}(m)}{m}\) period of \(M_m(N)\) appears to extend well beyond the collection of maximal/central coefficients of Gaussian polynomials. For example, just as the period of \(M_4(N)\) in Corollary 4.10 is 6, it can be shown that the quasipolynomials for the coefficients that are one and two terms away from the center of \(N+4 \brack 4\) also have a period of 6.

6 On the Difference \(p(n,m,N)-p(n-1,m,N-1)\)

In this section, we provide a direct proof of Theorem 4.2, which, having already completed the relevant computations from the quasipolynomial in Appendix A, is straightforward.

We are further motivated to extend known theorems on first differences of p(n, 3) and p(n, 4) to p(n, 3, N) and p(n, 4, N) respectively after establishing a generalization on first differences of p(nmN).

6.1 Proof of Theorem 4.2

Proof

We first prove (4.5) of Theorem 4.2. Since both \(p(6\ell ,3, 4\ell )\) and \(p(6\ell -3,3,4\ell -1)\) from (4.5) are maximal coefficients of their respective Gaussian polynomials, we use (4.21) and (4.20) from Corollary 4.8 to compute:
$$\begin{aligned} p(6\ell ,3, 4\ell )-p(6\ell -3,3,4\ell -1)=2\ell ^2+2\ell +1-(2\ell ^2+\ell ) =\ell +1, \end{aligned}$$
(6.1)
which verifies (4.5).

The proof of (4.6) is slightly different, because \(p(6\ell -1,3,4\ell )\) and \(p(6\ell -4,3,4\ell -1)\) are not central coefficients. However, (4.2) and (4.3) tell us respectively, that \(p(6\ell ,3, 4\ell )-1=p(6\ell -1,3,4\ell )\) and \(p(6\ell -3,3, 4\ell -1)-1=p(6\ell -4,3,4\ell -1)\), which together with (6.1) yields (4.6).

To prove (4.7), we replace \(\ell \) with \(\ell +1\) in \(p(6\ell -3,3,4\ell -2)\) in (4.19) to obtain:
$$\begin{aligned} p(6\ell +3,3,4\ell +2)=2\ell ^2+4\ell +2. \end{aligned}$$
(6.2)
Computing the difference between (6.2) and (4.22) proves (4.7).

The proof of (4.8) is done similarly to (4.7) and completes the proof of Theorem 4.2. \(\square \)

Corollary 4.8 lets us expand Theorem 4.2 slightly. For example, with (4.5) in mind, one has:
$$\begin{aligned} p(6\ell ,3,4\ell )-\left\{ \begin{array}{lr} p(6\ell -3,3,4\ell -1)\\ p(6\ell -2,3,4\ell -1)\\ p(6\ell -1,3,4\ell -1) \\ p(6\ell ,3,4\ell -1) \end{array} \right\} =\ell +1. \end{aligned}$$

6.2 Known Results on First Differences of p(n, 3) and p(n, 4) are Extended to p(n, 3, N) and p(n, 4, N)

Theorem 4.2 describes the differences of Gaussian polynomial coefficients which are near the middle of their respective generating polynomials. What can be said of related differences that are away from the middle of their generating polynomials? To answer this question, we consider Theorem 6.1 and Theorem 6.2 on differences of partitions into three and four parts, respectively. These theorems are proved combinatorially in [4]. Quasipolynomials allow for direct proofs.

Theorem 6.1

([4]). For \(k\ge 0\), one has:
$$\begin{aligned} \begin{aligned} \left. \begin{array}{lr} p(6k+1,3)-p(6k,3) \\ p(6k-1,3)-p(6k-2,3)\\ p(6k-2,3)-p(6k-3,3) \end{array} \right\}&=p(2k-1,2),\\ \left. \begin{array}{lr} p(6k+3,3)-p(6k+2,3)\\ p(6k+2,3)-p(6k+1,3)\\ p(6k,3)-p(6k-1,3) \end{array} \right\}&=p(2k,2). \end{aligned} \end{aligned}$$

Theorem 6.2

([4]). For \( k \ge 0\), one has:
$$\begin{aligned} \begin{array}{lr} p(2k-3,4)-p(2k-4,4)=p(k-3,3),\\ p(2k-4,4)-p(2k-5,4)=p(k-2,3). \end{array} \end{aligned}$$

We provide a new and very direct proof of Theorem 6.1 below.

Proof

Referring to the six constituents of the quasipolynomial for p(n, 3) in (3.6), one computes:
$$\begin{aligned} \begin{aligned} \left. \begin{array}{lr} p(6k+1,3)-p(6k,3)=3k^2+4k+1-(3k^2+3k+1)\\ p(6k-1,3)-p(6k-2,3)=3k^2+2k-(3k^2+k)\\ p(6k-2,3)-p(6k-3,3)=3k^2+k-(3k^2) \end{array} \right\} =k=p(2k-1,2) \end{aligned} \end{aligned}$$
and
$$\begin{aligned}\begin{aligned} \left. \begin{array}{lr} p(6k+3,3)-p(6k+2,3)=3k^2+6k+3-(3k^2+5k+2)\\ p(6k+2,3)-p(6k+1,3)=3k^2+5k+2-(3k^2+4k+1)\\ p(6k,3)-p(6k-1,3)=3k^2+4k+1-(3k^2+2k) \end{array} \right\} =k+1=p(2k,2). \end{aligned} \end{aligned}$$
\(\square \)

The direct proof of Theorem 6.2 follows from the 12 constituents of the quasipolynomial for p(n, 4) and is left to the reader. A forthcoming paper by the fifth author [20] will generalize the difference \(p(n,m)-p(n-1,m)\) in terms of \(p(n,m-1)\).

Motivated by Theorem 4.2, we extend Theorem 6.1 and Theorem 6.2 to the coefficients of Gaussian polynomials. The extensions are corollaries of a general theorem, Theorem 6.3 below, that we state and prove and which requires the q-binomial theorem:
$$\begin{aligned} (z;q)_m = \sum _{h=0}^m \begin{bmatrix}m\\h \end{bmatrix}(-1)^hq^{\frac{h(h-1)}{2}}z^h. \end{aligned}$$
(6.3)

Theorem 6.3

For \(n < 2N+2\):
$$\begin{aligned} p(n,m,N)-p(n-1,m,N-1)=p(n,m)-p(n-1,m). \end{aligned}$$

Proof

We manipulate a difference of generating functions to show
$$\begin{aligned} p(n,m,N)-p(n-1,m,N-1)=p(n,m)-p(n-1,m) \end{aligned}$$
for all \(n <2N+2\):
$$\begin{aligned} \sum _{n=0}^\infty&\left( p(n,m,N)-p(n-1,m,N-1)\right) q^n\nonumber \\&= \frac{(q^{N+1};q)_m}{(q;q)_m}- q\frac{(q^{N};q)_m}{(q;q)_m} \nonumber \\&= \frac{(q^{N+1};q)_{m-1}((1-q^{m+ N}) - q(1-q^{N}))}{(q;q)_m} \nonumber \\&= \frac{(q^{N+1};q)_{m-1}(1-q)\left( \sum \limits _{i=0}^{m+N-1} q^i - \sum \limits _{j=1}^N q^j\right) }{(q;q)_m}\nonumber \\&= \left( \sum \limits _{n=0}^\infty \left( p(n,m)-p(n-1,m)\right) q^n \right) (q^{N+1};q)_{m-1}\left( 1+ \sum \limits _{i=1}^{m-1} q^{N+i}\right) . \end{aligned}$$
(6.4)
We apply the q-binomial theorem (6.3) to re-express \((q^{N+1};q)_{m-1}\) in (6.4):
$$\begin{aligned} (&q^{N+1};q)_{m-1}\nonumber \\&= \sum _{h=0}^{m-1} \begin{bmatrix}m-1\nonumber \\ h \end{bmatrix}q^{\frac{h(h-1)}{2}+h( N+1)} \nonumber \\&= \sum _{h=0}^{m-1} \sum _{i=0}^{h(m-1 -h)}(-1)^{h} p(i,h,m-1-h) q^{\frac{h(h-1)}{2}+h( N+1)+i} \nonumber \\&=1- \sum _{j=0}^{m-2}q^{N+1+j}+ \sum _{h=2}^{m-1} \sum _{i=0}^{h(m-1 -h)}(-1)^{h} p(i,h,m-1-h) q^{\frac{h(h-1)}{2}+h( N+1)+i}. \end{aligned}$$
(6.5)
Note that the following polynomial from (6.5), denoted by A(q), has degree strictly greater than \(2N+2\):
$$\begin{aligned} A(q) = \sum _{h=2}^{m-1} \sum _{i=0}^{h(m-1 -h)}(-1)^{h} p(i,h,m-1-h) q^{\frac{h(h-1)}{2}+h( N+1)+i}. \end{aligned}$$
Now, we have:
$$\begin{aligned} (&q^{N+1};q)_{m-1}\left( 1+\sum \limits _{i=1}^{m-1} q^{N+i}\right) \nonumber \\&\quad = \left( 1-q^{N+1}\sum _{j=0}^{m-2}q^{j}+ A(q) \right) \left( 1+ q^{N+1}\sum \limits _{i=0}^{m-2} q^i\right) \nonumber \\&\quad =1 - q^{2N+2}\left( \sum \limits _{j=0}^{m-2} q^{j}\right) ^2 +A(q)+ q^{N+1}\sum \limits _{i=0}^{m-2} q^{i}A(q). \end{aligned}$$
(6.6)
Therefore, the smallest non-zero power of q in (6.6) is \(2N+2\), and Theorem 6.3 is proved. \(\square \)

The cases for \(m=3,4\) echo Theorem 6.1 and Theorem 6.2 and are left to the reader to verify.

Corollary 6.4

For \(k<\frac{2N-3}{6}\), one has:
$$\begin{aligned} \begin{aligned} \left. \begin{array}{lr} ~~~~~~p(6k+1,3,N)-p(6k,3,N-1)~~~~~\\ ~~~~~~p(6k-1,3,N)-p(6k-1,3,N-1)\\ ~~~~~~p(6k-2,3,N)-p(6k-3,3,N-1) \end{array} \right\} =k=p(2k-1,2),~~~~~\\ \left. \begin{array}{lr} p(6k+3,3,N)-p(6k+2,3,N-1)\\ p(6k+2,3,N)-p(6k+1,3,N-1)\\ p(6k,3,N)-p(6k-1,3,N-1) \end{array} \right\} =k+1=p(2k,2).~~~~ \end{aligned} \end{aligned}$$

Corollary 6.5

For \(k<\frac{2N-9}{12}\), one has:
$$\begin{aligned} \begin{array}{lr} p(2k-3,4,N)-p(2k-4,4,N-1)=p(k-3,3),\\ p(2k-4,4,N)-p(2k-5,4,N-1)=p(k-2,3). \end{array} \end{aligned}$$

Notes

Acknowledgements

This material is based upon work supported by, or in part by, the NSF-Louis Stokes Alliance for Minority Participation program under grant number HRD-1202008. The authors would like to thank the referee for providing helpful comments, especially with some insight regarding (4.26) and (4.27). The authors are especially grateful for the input and advice of both Drew Sills and Dennis Eichhorn regarding the revisions of this paper. The authors offer hearty thanks to George E. Andrews for alerting us to the recent results of Hahn in [18]. In forthcoming work, we will set

\(\displaystyle \sum _{k=0}^{\infty }{k+m-1 \atopwithdelims ()m-1}q^{\text {lcm}(m) k}=A_m(q)\)

and consider generalizations to p(nmN) and \({N+m\brack m}\) by further inquiry into

\(\displaystyle {N+m \brack m}= G_m(q)E_{m}(q)A_m(q). \;\qquad \qquad \qquad \qquad (6.7)\)

It is clear that GEA in (6.7) is a great mathematical object rich with wisdom and generosity and worthy of our attention and admiration.

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Copyright information

© Springer Nature Switzerland AG 2019

Authors and Affiliations

  • Angelica Castillo
    • 1
  • Stephanie Flores
    • 1
  • Anabel Hernandez
    • 2
  • Brandt Kronholm
    • 1
    Email author
  • Acadia Larsen
    • 1
  • Arturo Martinez
    • 1
  1. 1.School of Mathematical and Statistical SciencesUniversity of Texas Rio Grande ValleyEdinburgUSA
  2. 2.College of Engineering and Computer ScienceUniversity of Texas Rio Grande ValleyEdinburgUSA

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