n-Color Partitions into Distinct Parts as Sums over Partitions

Merca, Mircea; Simion, Emil

doi:10.3390/sym15112067

Open AccessArticle

n-Color Partitions into Distinct Parts as Sums over Partitions

by

Mircea Merca

^1,2,3,*

and

Emil Simion

^1,4

¹

Department of Mathematical Methods and Models, National University of Science and Technology Politehnica Bucharest, 060042 Bucharest, Romania

²

Fundamental Sciences Applied in Engineering Research Center, National University of Science and Technology Politehnica Bucharest, 060042 Bucharest, Romania

³

Academy of Romanian Scientists, 050044 Bucharest, Romania

⁴

Center for Research and Training in Innovative Techniques of Applied Mathematics in Engineering “Traian Lalescu”, National University of Science and Technology Politehnica Bucharest, 060042 Bucharest, Romania

^*

Author to whom correspondence should be addressed.

Symmetry 2023, 15(11), 2067; https://doi.org/10.3390/sym15112067

Submission received: 21 October 2023 / Revised: 10 November 2023 / Accepted: 12 November 2023 / Published: 15 November 2023

(This article belongs to the Special Issue Symmetry and Graph Theory)

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Abstract

:

The partitions in which the parts of size n can come in n different colors are known as n-color partitions. For

r \in {0, 1}

, let

Q L_{r} (n)

be the number of n-color partitions of n into distinct parts which have a number of parts congruent to r modulo 2. In this paper, we consider specializations of complete and elementary symmetric functions in order to establish two kinds of formulas for

Q L_{0} (n) \pm Q L_{1} (n)

as sums over partitions of n in terms of binomial coefficients. The first kind of formulas only involve partitions in which the parts of size n appear at most n times, while the second kind of formulas involve unrestricted partitions. Similar results are obtained for the first differences of

Q L_{0} (n) \pm Q L_{1} (n)

and the partial sums of

Q L_{0} (n) \pm Q L_{1} (n)

.

Keywords:

partitions; symmetric functions; binomial coefficients

MSC:

11P81; 11P82; 05A19; 05A20

1. Introduction

Compositions and partitions are fascinating topics in number theory, and they have many applications in combinatorics and other fields. For example, compositions and partitions can be used to count the number of ways to arrange objects, encrypt messages, solve equations, and more. They are also related to other concepts such as Fibonacci numbers, modular arithmetic, and symmetric groups. Compositions and partitions are examples of how simple ideas can lead to rich and beautiful mathematics.

Compositions and partitions are two ways of writing an integer as a sum of positive integers, but they differ in how they treat the order of the terms. For example,

4 = 2 + 1 + 1

and

4 = 1 + 2 + 1

are two different compositions of 4, but they are the same partition of 4. In general, a composition of n is an ordered list of positive integers whose sum is n, and a partition of n is an unordered list of positive integers whose sum is n.

In other words, a composition of a positive integer n is a sequence

λ = (λ_{1}, λ_{2}, \dots, λ_{k})

of positive integers such that

n = λ_{1} + λ_{2} + \dots + λ_{k} .

(1)

The terms of the sequence are called parts [1]. If the order of positive integers

λ_{i}

is irrelevant, then the representation (1) is known as an integer partition and can be rewritten as

n = t_{1} + 2 t_{2} + \dots + n t_{n},

where each positive integer i appears

t_{i}

times in the partition (the

t_{i}

are non-negative integers). For consistency, a partition of n will be written with the parts in nonincreasing order. As usual, the number of compositions and partitions of n are denoted by

c (n)

and

p (n)

, respectively. For convenience, we define

c (0) = p (0) = 1

.

For example, the sixteen compositions of five are:

\begin{matrix} (5), (4, 1), (3, 2), (3, 1, 1), (2, 3), (2, 2, 1), (2, 1, 2), (2, 1, 1, 1), (1, 4), \\ (1, 3, 1), (1, 1, 3), (1, 2, 2), (1, 2, 1, 1), (1, 1, 2, 1), (1, 1, 1, 2), (1, 1, 1, 1, 1), \end{matrix}

while the seven partitions of five are:

\begin{matrix} (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1) . \end{matrix}

(2)

One way to count the number of compositions of n is to use the generating function. It is easily verified that

\begin{matrix} \sum_{n = 1}^{\infty} c (n) q^{n} = \sum_{n = 1}^{\infty} {(q + q^{2} + q^{3} + \dots)}^{n} = \sum_{n = 1}^{\infty} {(\frac{q}{1 - q})}^{n} = \frac{q}{1 - 2 q} = \sum_{n = 1}^{\infty} 2^{n - 1} q^{n} . \end{matrix}

So, we have

c (n) = 2^{n - 1}

. This formula was discovered by Euler in the 18th century. However, the number of partitions of n is much harder to find. There is no simple formula for

p (n)

. Dealing with series and products, Euler showed that the generating function of

p (n)

can be expressed as an elegant infinite product

\begin{matrix} \sum_{n = 0}^{\infty} p (n) q^{n} = \frac{1}{{(q; q)}_{\infty}} . \end{matrix}

(3)

Here and throughout the paper, we use the following customary q-series notation:

\begin{matrix} {(a; q)}_{n} = \{\begin{matrix} 1, & for n = 0, \\ (1 - a) (1 - a q) \dots (1 - a q^{n - 1}), & for n \in N; \end{matrix} \\ {(a; q)}_{\infty} = lim_{n \to \infty} {(a; q)}_{n} . \end{matrix}

Here, q is a complex number with

| q | < 1

. Whenever

{(a; q)}_{\infty}

appears in a formula, we shall assume

| q | < 1

. All definitions and identities may be understood in the sense of formal power series in q.

The concept of n-color partitions is a natural generalization of the concept of ordinary partitions. An n-color partition of a positive integer is a partition in which a part of size n can come in n different colors denoted by subscripts

n_{1}, n_{2}, \dots, n_{n}

. The parts satisfy the following order:

\begin{matrix} 1_{1} < 2_{1} < 2_{2} < 3_{1} < 3_{2} < 3_{3} < 4_{1} < 4_{2} < 4_{3} < 4_{4} < \dots \end{matrix}

In this paper, we denote by

Q L (m)

the number of n-color partitions of m into distinct parts. For convenience, we define

Q L (0) = 1

. For example, there are sixteen n-color partitions into distinct parts of five:

\begin{matrix} (5_{5}), (5_{4}), (5_{3}), (5_{2}), (5_{1}), (4_{4}, 1_{1}), (4_{3}, 1_{1}), (4_{2}, 1_{1}), (4_{1}, 1_{1}), \\ (3_{3}, 2_{2}), (3_{3}, 2_{1}), (3_{2}, 2_{2}), (3_{2}, 2_{1}), (3_{1}, 2_{2}), (3_{1}, 2_{1}), (2_{2}, 2_{1}, 1_{1}) . \end{matrix}

(4)

We remark that n-color partitions were introduced to mathematics in 1987 by A. K. Agarwal and G. E. Andrews [2] for giving combinatorial interpretations of several q-series identities. We mention that n-color partitions were used indirectly in many studies of planar partitions before Andrews and Agarwal started studying n-color partitions [3,4,5]. For further reading on n-color partitions, we refer to [6,7,8,9,10,11,12,13,14].

For

r \in {0, 1}

, we denote by

Q L_{r} (m)

the number of n-color partitions of m into distinct parts which have a number of parts congruent to r modulo 2. For example, there are ten n-color partitions of five into distinct parts with an even number of parts:

\begin{matrix} (4_{4}, 1_{1}), (4_{3}, 1_{1}), (4_{2}, 1_{1}), (4_{1}, 1_{1}), (3_{3}, 2_{2}), (3_{3}, 2_{1}), (3_{2}, 2_{2}), (3_{2}, 2_{1}), (3_{1}, 2_{2}), (3_{1}, 2_{1}) \end{matrix}

and six n-color partitions of five into distinct parts with an odd number of parts:

\begin{matrix} (5_{5}), (5_{4}), (5_{3}), (5_{2}), (5_{1}), (2_{2}, 2_{1}, 1_{1}) . \end{matrix}

It is clear that

Q L (n) = Q L_{0} (n) + Q L_{1} (n) .

Elementary techniques in the theory of partitions [1] give the following generating function:

\begin{matrix} \sum_{n = 0}^{\infty} (Q L_{0} (n) \pm Q L_{1} (n)) q^{n} = \prod_{n = 1}^{\infty} {(1 \pm q^{n})}^{n} . \end{matrix}

(5)

The expansions start as

\begin{matrix} \prod_{n = 1}^{\infty} {(1 + q^{n})}^{n} = 1 + q + 2 q^{2} + 5 q^{3} + 8 q^{4} + 16 q^{5} + 28 q^{6} + 49 q^{7} + 83 q^{8} + \dots . \end{matrix}

(6)

and

\begin{matrix} \prod_{n = 1}^{\infty} {(1 - q^{n})}^{n} = 1 - q - 2 q^{2} - q^{3} + 4 q^{5} + 4 q^{6} + 7 q^{7} + 3 q^{8} - 2 q^{9} - \dots . \end{matrix}

(7)

If in a partition of n we have

t_{k}

parts of size k, then we can color these parts in distinct colors in

(\binom{k}{t_{k}})

ways. This remark allows us to immediately derive the following formula.

Theorem 1.

Let

r \in {0, 1}

. For

n ⩾ 0

,

Q L_{r} (n) = \sum_{\begin{matrix} t_{1} + 2 t_{2} + \dots + n t_{n} = n \\ t_{1} + t_{2} + \dots + t_{n} \equiv r (mod 2) \end{matrix}} (\binom{1}{t_{1}}) (\binom{2}{t_{2}}) \dots (\binom{n}{t_{n}}) .

By (4), we see that

Q L_{0} (5) = 10

and

Q L_{1} (5) = 6

. According to Theorem 1, we can write

\begin{matrix} Q L_{0} (5) & = (\binom{1}{1}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{1}) (\binom{5}{0}) + (\binom{1}{0}) (\binom{2}{1}) (\binom{3}{1}) (\binom{4}{0}) (\binom{5}{0}) + (\binom{1}{3}) (\binom{2}{1}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) \\ = 4 + 6 + 0 = 10 \end{matrix}

and

\begin{matrix} Q L_{1} (5) & = (\binom{1}{0}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{1}) + (\binom{1}{2}) (\binom{2}{0}) (\binom{3}{1}) (\binom{4}{0}) (\binom{5}{0}) + (\binom{1}{1}) (\binom{2}{2}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) \\ + (\binom{1}{5}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) \\ = 5 + 0 + 1 + 0 = 6 \end{matrix}

We remark the following equivalent form of Theorem 1.

Theorem 2.

For

n ⩾ 0

,

Q L_{0} (n) \pm Q L_{1} (n) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} (\binom{1}{t_{1}}) (\binom{2}{t_{2}}) \dots (\binom{n}{t_{n}}) .

The sum in the right-hand side of this equation runs over all the partitions of n, but not all terms are non-zero. Because for

t_{k} > k

we have

(\binom{k}{t_{k}}) = 0

, in this sum we can consider only the partitions of n into at most k copies of parts of size k, for each

k \in {1, 2, \dots, n}

. On the other hand, it is known that the number of these partitions of n is equal to the number of partitions of n into non-pronic numbers (cannot be written as

i (i + 1)

) ([15] A002378, A052335). This combinatorial interpretation follows easily if we take into account the following relation

\begin{matrix} \prod_{k = 1}^{\infty} (1 + q^{k} + q^{2 k} + \dots + q^{k^{2}}) = \prod_{i = 1}^{\infty} \frac{1 - q^{i (i + 1)}}{1 - q^{i}} . \end{matrix}

The left-hand side of this equation is the generating function for the number of partitions of n into at most k copies of parts of size k, while the right-hand side is the generating function for the number of partitions of n into parts which cannot be written as

i (i + 1)

. Recall that the numbers that can be arranged to form a rectangle are called rectangular numbers (also known as pronic numbers). For example, the number 12 is a rectangular number because it is three rows by four columns.

There is a more general result which combines compositions and partitions, where our Theorem 2 is the first entry.

Theorem 3.

Let m be a positive integer. For

n ⩾ 0

,

\sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) \pm Q L_{1} (t_{k})) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m \cdot k}{t_{k}}) .

We note that the sum on the left-hand side of this identity runs over all the compositions of n into exactly m parts, while the sum on the right-hand side runs over all the partitions of n into at most

m \cdot k

copies of parts of size k, for each

k \in {1, 2, \dots, n}

.

Symmetry is an important concept in mathematics, and it plays an important role in the study of integer partitions. There are many interesting results and theorems related to integer partitions and symmetry (see, for example, [16]), and they have applications in many areas of mathematics and beyond. In this paper, we take into account specializations of the elementary symmetric functions in order to provide an analytic proof of Theorem 3. Our approach allows us to obtain other results involving n-color partitions into distinct parts.

In order to introduce the following result, we consider the sequence

{(a_{n})}_{n ⩾ 1}

, defined by

a_{n} = \{\begin{matrix} n - 1, & for n = 2^{k}, k = 0, 1, 2, \dots, \\ n, & otherwise . \end{matrix}

Theorem 4.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L (t_{k}) - Q L (t_{k} - 1)) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m \cdot a_{k}}{t_{k}}) . \end{matrix}

In this context, we remark that the first differences of

Q L (n)

can be expressed as a sum over the partitions of n into parts greater than one, in terms of binomial coefficients.

Corollary 1.

For

n ⩾ 0

,

\begin{matrix} Q L (n) - Q L (n - 1) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} (\binom{a_{2}}{t_{2}}) (\binom{a_{3}}{t_{3}}) \dots (\binom{a_{n}}{t_{n}}) . \end{matrix}

By (6), we see that

Q L (5) - Q L (4) = 16 - 8 = 8

. According to Corollary 1, we can write

\begin{matrix} Q L (5) - Q L (4) & = (\binom{2 - 1}{0}) (\binom{3}{0}) (\binom{4 - 1}{0}) (\binom{5}{1}) + (\binom{2 - 1}{1}) (\binom{3}{1}) (\binom{4 - 1}{0}) (\binom{5}{0}) = 5 + 3 = 8 . \end{matrix}

We have a similar result for the first differences of

Q L_{0} (n) - Q L_{1} (n)

.

Theorem 5.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} ((Q L_{0} (t_{k}) - Q L_{1} (t_{k})) - (Q L_{0} (t_{k} - 1) - Q L_{1} (t_{k} - 1))) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + t_{3} + \dots + t_{n}} (\binom{2 m}{t_{1}}) \prod_{k = 2}^{n} (\binom{m \cdot k}{t_{k}}) . \end{matrix}

Corollary 2.

For

n ⩾ 0

,

\begin{matrix} (Q L_{0} (n) - Q L_{1} (n)) - (Q L_{0} (n - 1) - Q L_{1} (n - 1)) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + t_{3} + \dots + t_{n}} (\binom{2}{t_{1}}) (\binom{2}{t_{2}}) (\binom{3}{t_{3}}) \dots (\binom{n}{t_{n}}) . \end{matrix}

By (7), we see that

(Q L_{0} (5) - Q L_{1} (5)) - (Q L_{0} (4) - Q L_{1} (4)) = 4 - 0 = 4 .

According to Corollary 2, we can write

\begin{matrix} (Q L_{0} (5) - Q L_{1} (5)) - (Q L_{0} (4) - Q L_{1} (4)) \\ = - (\binom{2}{0}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{1}) + (\binom{2}{1}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{1}) (\binom{5}{0}) + (\binom{2}{0}) (\binom{2}{1}) (\binom{3}{1}) (\binom{4}{0}) (\binom{5}{0}) \\ - (\binom{2}{2}) (\binom{2}{0}) (\binom{3}{1}) (\binom{4}{0}) (\binom{5}{0}) - (\binom{2}{1}) (\binom{2}{2}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) + (\binom{2}{3}) (\binom{2}{1}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) \\ - (\binom{2}{5}) (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{0}) \\ = - 5 + 8 + 6 - 3 - 2 + 0 - 0 = 4 . \end{matrix}

The following results show that the partial sums of

Q L_{0} (n) \pm Q L_{1} (n)

can be be expressed as a sum over all the partitions of n in terms of binomial coefficients. The sequence

{(b_{n})}_{n ⩾ 1}

is given by

b_{n} = \{\begin{matrix} n + 1, & for n = 2^{k}, k = 0, 1, 2, \dots, \\ n, & otherwise . \end{matrix}

Theorem 6.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} \sum_{j = 0}^{t_{k}} Q L (j) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m \cdot b_{k}}{t_{k}}) . \end{matrix}

By Theorem 6, with m replaced by one, we obtain the following identity.

Corollary 3.

For

n ⩾ 0

,

\begin{matrix} \sum_{k = 0}^{n} Q L (k) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{b_{1}}{t_{1}}) (\binom{b_{2}}{t_{2}}) \dots (\binom{b_{n}}{t_{n}}) . \end{matrix}

By (6), we see that

Q L (0) + Q L (1) + Q L (2) + Q L (3) + Q L (4) + Q L (5) = 1 + 1 + 2 + 5 + 8 + 16 = 33 .

According to Corollary 3, we can write

\begin{matrix} Q L (0) + Q L (1) + Q L (2) + Q L (3) + Q L (4) + Q L (5) \\ = (\binom{2}{0}) (\binom{3}{0}) (\binom{3}{0}) (\binom{5}{0}) (\binom{5}{1}) + (\binom{2}{1}) (\binom{3}{0}) (\binom{3}{0}) (\binom{5}{1}) (\binom{5}{0}) + (\binom{2}{0}) (\binom{3}{1}) (\binom{3}{1}) (\binom{5}{0}) (\binom{5}{0}) \\ + (\binom{2}{2}) (\binom{3}{0}) (\binom{3}{1}) (\binom{5}{0}) (\binom{5}{0}) + (\binom{2}{1}) (\binom{3}{2}) (\binom{3}{0}) (\binom{5}{0}) (\binom{5}{0}) + (\binom{2}{3}) (\binom{3}{1}) (\binom{3}{0}) (\binom{5}{0}) (\binom{5}{0}) \\ + (\binom{2}{5}) (\binom{3}{0}) (\binom{3}{0}) (\binom{5}{0}) (\binom{5}{0}) \\ = 5 + 10 + 9 + 3 + 6 + 0 + 0 = 33 . \end{matrix}

Theorem 7.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} \sum_{j = 0}^{t_{k}} (Q L_{0} (j) - Q L_{1} (j)) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{m \cdot k}{t_{k}}) . \end{matrix}

The case

m = 1

of Theorem 7 reads as follows.

Corollary 4.

For

n ⩾ 0

,

\begin{matrix} \sum_{k = 0}^{n} (Q L_{0} (k) - Q L_{1} (k)) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} (\binom{2}{t_{2}}) (\binom{3}{t_{3}}) \dots (\binom{n}{t_{n}}) . \end{matrix}

By (7), we see that

\sum_{k = 0}^{5} (Q L_{0} (k) - Q L_{1} (k)) = 1 - 1 - 2 - 1 + 0 + 4 = 1 .

According to Corollary 4, we can write:

\begin{matrix} \sum_{k = 0}^{5} (Q L_{0} (k) - Q L_{1} (k)) = - (\binom{2}{0}) (\binom{3}{0}) (\binom{4}{0}) (\binom{5}{1}) + (\binom{2}{1}) (\binom{3}{1}) (\binom{4}{0}) (\binom{5}{0}) = - 5 + 6 = 1 . \end{matrix}

The remainder of our paper is organized as follows. In Section 2, we consider the elementary symmetric functions and introduce Lemma 1. This result allows us to provide analytic proofs of Theorems 3–7. In Section 3, we consider the complete homogeneous symmetric functions and introduce Lemma 2. This result allows us to obtain new expressions for the partition function

Q L_{0} (n) \pm Q L_{1} (n)

, the first differences of

Q L_{0} (n) \pm Q L_{1} (n)

, and the partial sums of

Q L_{0} (n) \pm Q L_{1} (n)

as sums over partitions of n in terms of binomial coefficients. In the last section, we consider a sum over all the partitions of n in order to provide a new expression for the generating function of

Q L_{0} (n) \pm Q L_{1} (n)

. Finding a combinatorial interpretation in terms of n-color partitions for this sum over all the partitions of n remains an open problem.

2. Proof of Theorems

The results presented in the previous section follow directly from the following lemma.

Lemma 1.

For

| q | < 1

, we have

\begin{matrix} \prod_{n = 1}^{\infty} {(1 \pm q^{n})}^{α_{n}} = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{j = 1}^{n} (\binom{α_{j}}{t_{j}}), \end{matrix}

(8)

where

{α_{n}}_{n ⩾ 1}

is a sequence of non-negative integers.

Proof.

It is well known that the elementary symmetric functions [17]

e_{k} (x_{1}, x_{2}, \dots, x_{n}) = \sum_{1 ⩽ i_{1} < i_{2} < \dots < i_{n} ⩽ n} x_{i_{1}} x_{i_{2}} \dots x_{i_{n}}

are characterized by the following formal power series identity in t:

\sum_{k = 0}^{\infty} e_{k} (x_{1}, x_{2}, \dots, x_{n}) t^{k} = \prod_{i = 1}^{n} (1 + x_{i} t) .

(9)

Taking into account the elementary symmetric functions, we can write

\begin{matrix} \prod_{k = 1}^{n} {(1 + x_{k} z)}^{α_{k}} & = \prod_{k = 1}^{n} \prod_{j = 1}^{α_{k}} (1 + x_{k} z) \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} e_{j} (\underset{α_{k} times}{\underset{⏟}{x_{k}, \dots, x_{k}}}) z^{j}) \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} x_{k}^{j} e_{j} (\underset{α_{k} times}{\underset{⏟}{1, \dots, 1}}) z^{j}) \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} (\binom{α_{k}}{j}) x_{k}^{j} z^{j}) \\ = \sum_{k = 0}^{\infty} (\sum_{t_{1} + t_{2} + \dots + t_{n} = k} \prod_{j = 1}^{n} (\binom{α_{j}}{t_{j}}) x_{j}^{t_{j}}) z^{k}, \end{matrix}

where we have invoked the well-known Cauchy multiplications of power series.

By this identity, with

x_{k}

replaced by

q^{k - 1}

and z replaced by

\pm q

, we obtain

\begin{matrix} \prod_{k = 1}^{n} {(1 \pm q^{k})}^{α_{k}} & = \sum_{k = 0}^{\infty} {(\pm q)}^{k} \sum_{t_{1} + t_{2} + \dots + t_{n} = k} q^{t_{2} + 2 t_{3} + \dots + (n - 1) t_{n}} \prod_{j = 1}^{n} (\binom{α_{j}}{t_{j}}) \\ = \sum_{k = 0}^{\infty} \sum_{t_{1} + t_{2} + \dots + t_{n} = k} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} q^{t_{1} + 2 t_{2} + 3 t_{3} + \dots + n t_{n}} \prod_{j = 1}^{n} (\binom{α_{j}}{t_{j}}) . \end{matrix}

The limiting case

n \to \infty

of this identity reads as

\begin{matrix} \prod_{k = 1}^{\infty} {(1 \pm q^{k})}^{α_{k}} = \sum_{k = 0}^{\infty} q^{k} \sum_{t_{1} + 2 t_{2} + \dots + k t_{k} = k} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{k}} \prod_{j = 1}^{k} (\binom{α_{j}}{t_{j}}) . \end{matrix}

This concludes the proof. □

2.1. Proof of Theorem 3

Taking into account Lemma 1, with

α_{j}

replaced by

m \cdot j

, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} (Q L_{0} (t_{j}) \pm Q L_{1} (t_{j})) \\ = {(\sum_{n = 0}^{\infty} (Q L_{0} (n) \pm Q L_{1} (n)) q^{n})}^{m} \\ = \prod_{n = 1}^{\infty} {(1 \pm q^{n})}^{n m} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{j = 1}^{n} (\binom{m \cdot j}{t_{j}}) . \end{matrix}

The proof follows easily by equating the coefficients of

q^{n}

in this relation.

2.2. Proof of Theorem 4

Considering the generating function for the first differences of

Q L (n)

, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} (Q L (t_{j}) - Q L (t_{j} - 1)) \\ = {(\sum_{n = 0}^{\infty} (Q L (n) - Q L (n - 1)) q^{n})}^{m} \\ = {((1 - q) \sum_{n = 0}^{\infty} Q L (n) q^{n})}^{m} \\ = {(1 - q)}^{m} \prod_{n = 1}^{\infty} {(1 + q^{n})}^{n m} . \end{matrix}

Taking into account

\begin{matrix} 1 = (1 - q) \prod_{n = 0}^{\infty} (1 + q^{2^{n}}), \end{matrix}

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we obtain

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} (Q L (t_{j}) - Q L (t_{j} - 1)) \\ = {(\prod_{n = 1}^{\infty} \frac{{(1 + q^{n})}^{n}}{1 + q^{2^{n - 1}}})}^{m} \\ = \prod_{n = 1}^{\infty} {(1 + q^{n})}^{m \cdot a_{n}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{j = 1}^{n} (\binom{m \cdot a_{j}}{t_{j}}) . \end{matrix}

by Lemma 1 with

α_{j}

replaced by

m \cdot a_{j}

.

The proof follows easily taking into account

a_{1} = 0

.

2.3. Proof of Theorem 5

In order to prove Theorem 5, we take into account the following generating function:

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} ((Q L_{0} (t_{j}) - Q L_{1} (t_{j})) - (Q L_{0} (t_{j} - 1) - Q L_{1} (t_{j} - 1))) \\ = {(\sum_{n = 0}^{\infty} ((Q L_{0} (n) - Q L_{1} (n)) - (Q L_{0} (n - 1) - Q L_{1} (n - 1))) q^{n})}^{m} \\ = {((1 - q) \sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = {(1 - q)}^{2 m} \prod_{n = 2}^{\infty} {(1 - q^{n})}^{n \cdot m} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} (\binom{2 m}{t_{1}}) \prod_{j = 2}^{n} (\binom{m \cdot j}{t_{j}}) \end{matrix}

by Lemma 1 with

α_{1} = 2 m

and

α_{j} = m \cdot j

for

j ⩾ 2

.

2.4. Proof of Theorem 6

Taking into account the generating function for the partial sums of

Q L (n)

, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} \sum_{k = 0}^{t_{j}} Q L (k) \\ = {(\sum_{n = 0}^{\infty} \sum_{k = 0}^{n} Q L (k) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \sum_{n = 0}^{\infty} Q L (n) q^{n})}^{m} \\ = \prod_{n = 1}^{\infty} {(1 + q^{2^{n - 1}})}^{m} {(1 + q^{n})}^{n \cdot m} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{j = 1}^{n} (\binom{m \cdot b_{j}}{t_{j}}) \end{matrix}

by Lemma 1 with

α_{j} = m \cdot b_{j}

.

2.5. Proof of Theorem 7

Taking into account the generating function for the partial sums of

Q L_{0} (n) - Q L_{1} (n)

, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{j = 1}^{m} \sum_{k = 0}^{t_{j}} (Q L_{0} (n) - Q L_{1} (n)) \\ = {(\sum_{n = 0}^{\infty} \sum_{k = 0}^{n} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = \frac{1}{{(1 - q)}^{m}} \prod_{n = 1}^{\infty} {(1 - q^{n})}^{n \cdot m} \\ = {(1 - q)}^{0} \prod_{n = 2}^{\infty} {(1 - q^{n})}^{n \cdot m} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} (\binom{0}{t_{1}}) \prod_{j = 2}^{n} (\binom{m \cdot j}{t_{j}}) \end{matrix}

by Lemma 1 with

α_{j} = m \cdot j

for

j ⩾ 2

\begin{matrix} = \sum_{n = 0}^{\infty} q^{n} \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{j = 2}^{n} (\binom{m \cdot j}{t_{j}}) . \end{matrix}

3. Identities of the Second Kind

In order to derive new formulas involving n-color partitions into distinct parts, we take into account the following lemma.

Lemma 2.

For

| q | < 1

, we have

\begin{matrix} \prod_{n = 1}^{\infty} \frac{1}{{(1 \pm q^{n})}^{α_{n}}} = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(\mp 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{j = 1}^{n} (\binom{α_{j} - 1 + t_{j}}{t_{j}}), \end{matrix}

where

{α_{n}}_{n ⩾ 1}

is a sequence of non-negative integers.

Proof.

The complete homogeneous symmetric functions [17]

h_{k} (x_{1}, x_{2}, \dots, x_{n}) = \sum_{1 ⩽ i_{1} ⩽ i_{2} ⩽ \dots ⩽ i_{k} ⩽ n} x_{i_{1}} x_{i_{2}} \dots x_{i_{k}}

are characterized by the following formal power series identity in z:

\sum_{k = 0}^{\infty} h_{k} (x_{1}, x_{2}, \dots, x_{n}) z^{k} = \prod_{i = 1}^{n} \frac{1}{1 - x_{i} z} .

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Taking into account the complete homogeneous symmetric functions, we can write

\begin{matrix} \prod_{k = 1}^{n} \frac{1}{{(1 - x_{k} z)}^{α_{k}}} & = \prod_{k = 1}^{n} \prod_{j = 1}^{α_{k}} \frac{1}{1 - x_{i} z} \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} h_{j} (\underset{α_{k} times}{\underset{⏟}{x_{k}, \dots, x_{k}}}) z^{j}) \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} x_{k}^{j} h_{j} (\underset{α_{k} times}{\underset{⏟}{1, \dots, 1}}) z^{j}) \\ = \prod_{k = 1}^{n} (\sum_{j = 0}^{\infty} (\binom{α_{k} - 1 + j}{j}) x_{k}^{j} z^{j}) \\ = \sum_{k = 0}^{\infty} (\sum_{t_{1} + t_{2} + \dots + t_{n} = k} \prod_{j = 1}^{n} (\binom{α_{j} - 1 + t_{j}}{t_{j}}) x_{j}^{t_{j}}) z^{k}, \end{matrix}

where we have invoked the well-known Cauchy multiplications of power series.

By this identity, with

x_{k}

replaced by

q^{k - 1}

and z replaced by

\pm q

, we obtain

\begin{matrix} \prod_{k = 1}^{n} \frac{1}{{(1 \pm q^{k})}^{α_{k}}} & = \sum_{k = 0}^{\infty} {(\mp q)}^{k} \sum_{t_{1} + t_{2} + \dots + t_{n} = k} q^{t_{2} + 2 t_{3} + \dots + (n - 1) t_{n}} \prod_{j = 1}^{n} (\binom{α_{j} - 1 + t_{j}}{t_{j}}) \\ = \sum_{k = 0}^{\infty} \sum_{t_{1} + t_{2} + \dots + t_{n} = k} {(\mp 1)}^{t_{1} + t_{2} + \dots + t_{n}} q^{t_{1} + 2 t_{2} + 3 t_{3} + \dots + n t_{n}} \prod_{j = 1}^{n} (\binom{α_{j} - 1 + t_{j}}{t_{j}}) . \end{matrix}

The limiting case

n \to \infty

of this identity reads as

\begin{matrix} \prod_{k = 1}^{\infty} \frac{1}{{(1 \pm q^{k})}^{α_{k}}} = \sum_{k = 0}^{\infty} q^{k} \sum_{t_{1} + 2 t_{2} + \dots + k t_{k} = k} {(\mp 1)}^{t_{1} + t_{2} + \dots + t_{k}} \prod_{j = 1}^{k} (\binom{α_{j} - 1 + t_{j}}{t_{j}}) . \end{matrix}

This concludes the proof. □

In analogy with Theorem 3, we introduce Theorems 8 and 9. In Theorem 8, for any positive integer n we denote by

c_{n}

the number of compositions of n in two parts with at least one even part. It is not difficult to prove that

c_{n} = \{\begin{matrix} n - 1, & for n odd, \\ n / 2 - 1, & for n even . \end{matrix}

For example,

c_{7} = 6

and

c_{8} = 3

because the compositions in question are

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and

(2, 6), (4, 4), (6, 2) .

We remark that the sequence

{(c_{n})}_{n ⩾ 1} = (0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8, 18, 9, 20, \dots)

is known and can be seen in the On-Line Encyclopedia of Integer Sequence ([15] A065423).

Theorem 8.

Let m be a positive integer. For

n ⩾ 0

,

\sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} Q L (t_{k}) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 1}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) .

Proof.

Taking into account the generating function (5), we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} Q L (t_{k}) \\ = {(\sum_{n = 0}^{\infty} Q L (n) q^{n})}^{m} \\ = \prod_{n = 1}^{\infty} {(1 + q^{n})}^{n \cdot m} \\ = \prod_{n = 1}^{\infty} {(\frac{1 - q^{2 n}}{1 - q^{n}})}^{n \cdot m} \\ = \prod_{n = 1}^{\infty} \frac{1}{{(1 - q^{2 n - 1})}^{(2 n - 1) \cdot m} {(1 - q^{2 n})}^{n \cdot m}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 1}^{⌈ n / 2 ⌉} (\binom{m (2 k - 1) - 1 + t_{2 k - 1}}{t_{2 k - 1}}) (\binom{m \cdot k - 1 + t_{2 k}}{t_{2 k}}) . \end{matrix}

by Lemma 2 with

α_{2 n - 1} = m (2 n - 1)

and

α_{2 n} = m \cdot n

.

This concludes the proof. □

The case

m = 1

of Theorem 8 provides a new decomposition of

Q L (n)

in terms of binomial coefficients as a sum over all the partitions of n.

Corollary 5.

For

n ⩾ 0

,

Q L (n) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 1}^{n} (\binom{c_{k} + t_{k}}{t_{k}}) .

We remark that the terms in the right-hand side of th formula given by Theorem 8 are all positive. In addition, the first two factors of the product

\prod_{k = 1}^{n} (\binom{c_{k} + t_{k}}{t_{k}})

are always equal to 1 (because

c_{1} = c_{2} = 0

). According to Corollary 5, for

n = 5

we can write

\begin{matrix} Q L (5) \\ = (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 1}{1}) + (\binom{2 + 0}{0}) (\binom{1 + 1}{1}) (\binom{4 + 0}{0}) + (\binom{2 + 1}{1}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ + (\binom{2 + 1}{1}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ + (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ = 5 + 2 + 3 + 3 + 1 + 1 + 1 = 16 . \end{matrix}

In order to present the following result, for a positive integer n we denote by

s_{n}

the sum of divisors d of n such that

n / d

is a power of two. For example, the divisors of six are one, two, three and six. Since

6 / 3 = 2^{1}

and

6 / 6 = 2^{0}

, we have

s_{6} = 3 + 6 = 9

. We remark that the sequence

{(s_{n})}_{n ⩾ 1} = (1, 3, 3, 7, 5, 9, 7, 15, 9, 15, 11, 21, 13, 21, 15, 31, 17, 27, \dots)

is known and can be seen in the On-Line Encyclopedia of Integer Sequence ([15] A129527).

Theorem 9.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m \cdot s_{k} - 1 + t_{k}}{t_{k}}) . \end{matrix}

Proof.

Taking into account the generating function (5), Lemma 2, and the identity

1 = (1 - q^{2 n}) \prod_{k = 1}^{\infty} (1 + q^{2^{k} n}),

we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) \\ = {(\sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = \prod_{n = 1}^{\infty} {(1 - q^{n})}^{n \cdot m} \\ = \prod_{n = 1}^{\infty} {(\frac{1 - q^{2 n}}{1 + q^{n}})}^{n \cdot m} \\ = \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{n \cdot m}} \prod_{k = 1}^{\infty} \frac{1}{{(1 + q^{2^{k} n})}^{n \cdot m}} \\ = \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{m \cdot s_{n}}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m \cdot s_{k} - 1 + t_{k}}{t_{k}}) . \end{matrix}

This concludes the proof. □

The case

m = 1

provides the following identity.

Corollary 6.

For

n ⩾ 0

,

Q L_{0} (n) - Q L_{1} (n) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{s_{k} - 1 + t_{k}}{t_{k}}) .

We remark that the first factor of the product

\prod_{k = 1}^{n} (\binom{s_{k} - 1 + t_{k}}{t_{k}})

is always equal to 1 (because

s_{1} - 1 = 0

). The case

n = 5

of Corollary 6 reads as follows:

\begin{matrix} Q L_{0} (5) - Q L_{1} (5) \\ = - (\binom{2 + 0}{0}) (\binom{2 + 0}{0}) (\binom{6 + 0}{0}) (\binom{4 + 1}{1}) + (\binom{2 + 0}{0}) (\binom{2 + 0}{0}) (\binom{6 + 1}{1}) (\binom{4 + 0}{0}) \\ + (\binom{2 + 1}{1}) (\binom{2 + 1}{1}) (\binom{6 + 0}{0}) (\binom{4 + 0}{0}) - (\binom{2 + 0}{0}) (\binom{2 + 1}{1}) (\binom{6 + 0}{0}) (\binom{4 + 0}{0}) \\ - (\binom{2 + 0}{0}) (\binom{2 + 2}{2}) (\binom{6 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{2 + 1}{1}) (\binom{2 + 0}{0}) (\binom{6 + 0}{0}) (\binom{4 + 0}{0}) \\ - (\binom{2 + 0}{0}) (\binom{2 + 0}{0}) (\binom{6 + 0}{0}) (\binom{4 + 0}{0}) \\ = - 5 + 7 + 9 - 3 - 6 + 3 - 1 = 4 . \end{matrix}

In analogy with Theorem 4, we have the following result.

Theorem 10.

Let m be a positive integer. For

n ⩾ 0

,

\sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L (t_{k}) - Q L (t_{k} - 1)) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) .

Proof.

By the proof of Theorem 8, we see that

\begin{matrix} \sum_{n = 0}^{\infty} Q L (n) q^{n} = \prod_{n = 1}^{\infty} \frac{1}{{(1 - q^{2 n - 1})}^{2 n - 1} {(1 - q^{2 n})}^{n}} . \end{matrix}

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Taking into account Lemma 2, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L (t_{k}) - Q L (t_{k} - 1)) \\ = {(\sum_{n = 0}^{\infty} (Q L (n) - Q L (n - 1)) q^{n})}^{m} \\ = {((1 - q) \sum_{n = 0}^{\infty} Q L (n) q^{n})}^{m} \\ = {(1 - q)}^{m} \prod_{n = 1}^{\infty} \frac{1}{{(1 - q^{2 n - 1})}^{m (2 n - 1)} {(1 - q^{2 n})}^{m \cdot n}} \\ = \frac{1}{{(1 - q)}^{0}} \prod_{n = 2}^{\infty} \frac{1}{{(1 - q^{n})}^{m (c_{n} + 1)}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{- 1 + t_{1}}{t_{1}}) \prod_{k = 2}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

This concludes the proof. □

Corollary 7.

For

n ⩾ 0

,

Q L (n) - Q L (n - 1) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{c_{k} + t_{k}}{t_{k}}) .

We remark that the first factor of the product

\prod_{k = 2}^{n} (\binom{c_{k} + t_{k}}{t_{k}})

is always equal to 1 (because

c_{2} = 0

). The case

n = 5

of Corollary 7 reads as follows:

\begin{matrix} Q L (5) - Q L (4) & = (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 1}{1}) + (\binom{2 + 1}{1}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) = 5 + 3 = 8 . \end{matrix}

In analogy with Theorem 5, we have the following result, where

{r_{n}}_{n ⩾ 1}

is a sequence of positive integers defined as follows:

r_{n} = \{\begin{matrix} s_{n}, & for n = 2^{m}, m = 0, 1, 2, \dots, \\ s_{n} - 1, & otherwise . \end{matrix}

Theorem 11.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) - (Q L_{0} (t_{k} - 1) - Q L_{1} (t_{k} - 1)) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m (r_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

Proof.

By the proof of Theorem 9, we see that

\begin{matrix} \sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n} = \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{s_{n}}} . \end{matrix}

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Taking into account the generating function (5), the identity (10), and Lemma 2, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) - (Q L_{0} (t_{k} - 1) - Q L_{1} (t_{k} - 1)) \\ = {(\sum_{n = 0}^{\infty} ((Q L_{0} (n) - Q L_{1} (n)) - (Q L_{0} (n - 1) - Q L_{1} (n - 1))) q^{n})}^{m} \\ = {((1 - q) \sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = {(1 - q)}^{m} \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{m \cdot s_{n}}} \\ = \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{m \cdot s_{n}} {(1 + q^{2^{n - 1}})}^{m}} \\ = \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{m (r_{n} + 1)}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m (r_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

This concludes the proof. □

Corollary 8.

For

n ⩾ 0

,

\begin{matrix} (Q L_{0} (n) - Q L_{1} (n)) - (Q L_{0} (n - 1) - Q L_{1} (n - 1)) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{r_{k} + t_{k}}{t_{k}}) . \end{matrix}

The case

n = 5

of Corollary 8 reads as follows:

\begin{matrix} (Q L_{0} (5) - Q L_{1} (5)) - (Q L_{0} (4) - Q L_{1} (4)) \\ = - (\binom{1 + 0}{0}) (\binom{3 + 0}{0}) (\binom{2 + 0}{0}) (\binom{7 + 0}{0}) (\binom{4 + 1}{1}) + (\binom{1 + 1}{1}) (\binom{3 + 0}{0}) (\binom{2 + 0}{0}) (\binom{7 + 1}{1}) (\binom{4 + 0}{0}) \\ + (\binom{1 + 0}{0}) (\binom{3 + 1}{1}) (\binom{2 + 1}{1}) (\binom{7 + 0}{0}) (\binom{4 + 0}{0}) - (\binom{1 + 2}{2}) (\binom{3 + 0}{0}) (\binom{2 + 1}{1}) (\binom{7 + 0}{0}) (\binom{4 + 0}{0}) \\ - (\binom{1 + 1}{1}) (\binom{3 + 2}{2}) (\binom{2 + 0}{0}) (\binom{7 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{1 + 3}{3}) (\binom{3 + 1}{1}) (\binom{2 + 0}{0}) (\binom{7 + 0}{0}) (\binom{4 + 0}{0}) \\ - (\binom{1 + 5}{5}) (\binom{3 + 0}{0}) (\binom{2 + 0}{0}) (\binom{7 + 0}{0}) (\binom{4 + 0}{0}) \\ = - 5 + 16 + 12 - 9 - 20 + 16 - 6 = 4 . \end{matrix}

In analogy with Theorem 6, we have the following result.

Theorem 12.

Let m be a positive integer. For

n ⩾ 0

,

\sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} \sum_{j = 0}^{t_{k}} Q L (j) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{2 m - 1 + t_{1}}{t_{1}}) \prod_{k = 3}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) .

Proof.

Taking into account the generating function (12) and Lemma 2, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} \sum_{j = 0}^{t_{k}} Q L (j) \\ = {(\sum_{n = 0}^{\infty} \sum_{k = 0}^{n} Q L (k) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \sum_{n = 0}^{\infty} Q L (n) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \prod_{n = 1}^{\infty} \frac{1}{{(1 - q^{2 n - 1})}^{2 n - 1} {(1 - q^{2 n})}^{n}})}^{m} \\ = \frac{1}{{(1 - q)}^{2 m}} \prod_{n = 2}^{\infty} \frac{1}{{(1 - q^{n})}^{m (c_{n} + 1)}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{2 m - 1 + t_{1}}{t_{1}}) (\binom{t_{2}}{t_{2}}) \prod_{k = 3}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

This concludes the proof. □

Corollary 9.

For

n ⩾ 0

,

\sum_{k = 0}^{n} Q L (k) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{1 + t_{1}}{t_{1}}) \prod_{k = 3}^{n} (\binom{c_{k} + t_{k}}{t_{k}}) .

The case

n = 5

of Corollary 9 reads as follows:

\begin{matrix} Q L (0) + Q L (1) + Q L (2) + Q L (3) + Q L (4) + Q L (5) \\ = (\binom{1 + 0}{0}) (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 1}{1}) + (\binom{1 + 1}{1}) (\binom{2 + 0}{0}) (\binom{1 + 1}{1}) (\binom{4 + 0}{0}) \\ + (\binom{1 + 0}{0}) (\binom{2 + 1}{1}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{1 + 2}{2}) (\binom{2 + 1}{1}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ + (\binom{1 + 1}{1}) (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) + (\binom{1 + 3}{3}) (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ + (\binom{1 + 5}{5}) (\binom{2 + 0}{0}) (\binom{1 + 0}{0}) (\binom{4 + 0}{0}) \\ = 5 + 4 + 3 + 9 + 2 + 4 + 6 = 33 . \end{matrix}

In analogy with Theorem 7, we have the following result, where

{v_{n}}_{n ⩾ 2}

is a sequence of positive integers defined as follows:

v_{n} = \{\begin{matrix} s_{n} - 2, & for n = 2^{m}, m = 1, 2, 3, \dots, \\ s_{n} - 1, & otherwise . \end{matrix}

Theorem 13.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) \\ = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{m (v_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

Proof.

Taking into account the identity (10), the generating function (13), and Lemma 2, we can write

\begin{matrix} \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + t_{2} + \dots + t_{m} = n} \prod_{k = 1}^{m} (Q L_{0} (t_{k}) - Q L_{1} (t_{k})) \\ = {(\sum_{n = 0}^{\infty} ((Q L_{0} (n) - Q L_{1} (n)) - (Q L_{0} (n - 1) - Q L_{1} (n - 1))) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \sum_{n = 0}^{\infty} (Q L_{0} (n) - Q L_{1} (n)) q^{n})}^{m} \\ = {(\frac{1}{1 - q} \prod_{n = 1}^{\infty} \frac{1}{{(1 + q^{n})}^{s_{n}}})}^{m} \\ = \prod_{n = 1}^{\infty} \frac{{(1 + q^{2^{n - 1}})}^{m}}{{(1 + q^{n})}^{m \cdot s_{n}}} \\ = \frac{1}{{(1 - q)}^{0}} \prod_{n = 2}^{\infty} \frac{1}{{(1 + q^{n})}^{m (v_{n} + 1)}} \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} (\binom{- 1 + t_{1}}{t_{1}}) \prod_{k = 2}^{n} (\binom{m (v_{k} + 1) - 1 + t_{k}}{t_{k}}) \\ = \sum_{n = 0}^{\infty} q^{n} \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{m (v_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

This concludes the proof. □

Corollary 10.

For

n ⩾ 0

,

\begin{matrix} \sum_{k = 0}^{n} (Q L_{0} (n) - Q L_{1} (n)) q^{n} = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{v_{k} + t_{k}}{t_{k}}) . \end{matrix}

The case

n =

of Corollary 10 reads as follows:

\begin{matrix} \sum_{k = 0}^{5} (Q L_{0} (k) - Q L_{1} (k)) & = - (\binom{1 + 0}{0}) (\binom{2 + 0}{0}) (\binom{5 + 0}{0}) (\binom{4 + 1}{1}) \\ + (\binom{1 + 1}{1}) (\binom{2 + 1}{1}) (\binom{5 + 0}{0}) (\binom{4 + 0}{0}) \\ = - 5 + 6 = 1 . \end{matrix}

4. Concluding Remarks

A collection of identities involving

Q L_{0} (n) \pm Q L_{1} (n)

has been introduced in this paper by considering specializations of complete and elementary symmetric functions. In this context, as consequences of these results, we remark some identities involving binomial coefficients.

The following identity follows from Theorems 3 and 8.

Corollary 11.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 1}^{n} (\binom{m \cdot k}{t_{k}}) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 1}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

The following identity follows from Theorems 3 and 9.

Corollary 12.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m \cdot k}{t_{k}}) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m \cdot s_{k} - 1 + t_{k}}{t_{k}}) . \end{matrix}

The following identity follows from Theorems 4 and 10.

Corollary 13.

Let m be a positive integer. For

n ⩾ 0

,

\sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m \cdot a_{k}}{t_{k}}) = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) .

The following identity follows from Theorems 5 and 11.

Corollary 14.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + t_{3} + \dots + t_{n}} (\binom{2 m}{t_{1}}) \prod_{k = 2}^{n} (\binom{m \cdot k}{t_{k}}) \\ = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} {(- 1)}^{t_{1} + t_{2} + \dots + t_{n}} \prod_{k = 1}^{n} (\binom{m (r_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

The following identity follows from Theorems 6 and 12.

Corollary 15.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \prod_{k = 2}^{n} (\binom{m \cdot b_{k}}{t_{k}}) = \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} (\binom{2 m - 1 + t_{1}}{t_{1}}) \prod_{k = 3}^{n} (\binom{m (c_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

The following identity follows from Theorems 7 and 13.

Corollary 16.

Let m be a positive integer. For

n ⩾ 0

,

\begin{matrix} \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{m \cdot k}{t_{k}}) \\ = \sum_{2 t_{2} + 3 t_{3} + \dots + n t_{n} = n} {(- 1)}^{t_{2} + t_{3} + \dots + t_{n}} \prod_{k = 2}^{n} (\binom{m (v_{k} + 1) - 1 + t_{k}}{t_{k}}) . \end{matrix}

On the other hand, from (11), with

x_{j}

replaced by

q^{j - 1}

for each

j \in {1, 2, \dots, n}

, we obtain a well-known identity known as Rothe’s q-binomial theorem ([18] Theorem 12).

Theorem 14

(Cauchy). If n is any non-negative integer and

| q |

and

| t |

are both less than one, then

\sum_{k = 0}^{n} [\binom{n}{k}] q^{(\binom{k}{2})} t^{k} = {(- t; q)}_{n} .

The limiting case

n \to \infty

of Theorem 14 is given by the following theorem of Euler ([18] Theorem 27).

Theorem 15

(Euler). If

| q | < 1

and

| t | < 1

, then

\sum_{k = 0}^{\infty} \frac{q^{(\binom{k}{2})} t^{k}}{{(q; q)}_{k}} = {(- t; q)}_{\infty} .

By this theorem, with t replaced by q, we obtain a well-known expression for the generating function of

Q (n)

, the number of partitions of n into distinct parts, i.e.,

\prod_{n = 1}^{\infty} (1 + q^{n}) = \sum_{n = 0}^{\infty} \frac{q^{(\binom{n}{2}) + n}}{{(q; q)}_{n}} .

We remark an analogous result for the generating function of

Q L (n)

.

Theorem 16.

For

| q | < 1

,

\begin{matrix} \prod_{n = 1}^{\infty} {(1 \pm q^{n})}^{n} = \sum_{n = 0}^{\infty} \sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \frac{{(\pm 1)}^{t_{1} + t_{2} + \dots + t_{n}} q^{(\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{n}}{2}) + n}}{{(q; q)}_{t_{1}} {(q; q)}_{t_{2}} \dots {(q; q)}_{t_{n}}} . \end{matrix}

Proof.

Taking into account the q-binomial coefficients

[\binom{n}{k}] = {[\binom{n}{k}]}_{q} = \{\begin{matrix} \frac{{(q; q)}_{n}}{{(q; q)}_{k} {(q; q)}_{n - k}}, & n, k integers, 0 ⩽ k ⩽ n, \\ 0, & otherwise, \end{matrix}

as specializations of complete homogeneous symmetric functions, namely

e_{k} (1, q, \dots, q^{n - 1}) = [\binom{n}{k}] q^{(\binom{k}{2})},

(14)

we can write

\begin{matrix} \prod_{i = 1}^{n} {(1 + q^{i - 1} z)}^{i} & = \prod_{i = 1}^{n} (\prod_{j = i}^{n} (1 + q^{j - 1} z)) \\ = \prod_{i = 1}^{n} (\sum_{k = 0}^{i} e_{k} (q^{i - 1}, q^{i}, \dots, q^{n - 1}) z^{k}) \\ = \prod_{i = 1}^{n} (\sum_{k = 0}^{i} {(q^{i - 1})}^{k} e_{k} (1, q, \dots, q^{n - i}) z^{k}) \\ = \prod_{i = 1}^{n} (\sum_{k = 0}^{i} q^{(i - 1) k + (\binom{k}{2})} [\binom{n - i + 1}{k}] z^{k}) \\ = \sum_{k = 0}^{n (n + 1) / 2} (\sum_{t_{1} + t_{2} + \dots + t_{n} = k} \prod_{i = 1}^{n} q^{(i - 1) t_{i} + (\binom{t_{i}}{2})} [\binom{n - i + 1}{t_{i}}]) z^{k} . \end{matrix}

Replacing z with q, we obtain

\begin{matrix} \prod_{i = 1}^{n} {(1 + q^{i})}^{i} \\ = \sum_{k = 0}^{n (n + 1) / 2} \sum_{t_{1} + t_{2} + \dots + t_{n} = k} q^{t_{1} + 2 t_{2} + \dots + n t_{n} + (\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{n}}{2})} [\binom{n}{t_{1}}] [\binom{n - 1}{t_{2}}] \dots [\binom{1}{t_{n}}] . \end{matrix}

The limiting case

n \to \infty

of this equation can be written as

\begin{matrix} \prod_{i = 1}^{\infty} {(1 + q^{i})}^{i} = \sum_{N = 0}^{\infty} \sum_{t_{1} + 2 t_{2} + \dots + N t_{N} = N} \frac{q^{(\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{N}}{2}) + N}}{{(q; q)}_{t_{1}} {(q; q)}_{t_{2}} \dots {(q; q)}_{t_{N}}} . \end{matrix}

In a similar way, replacing z with

- q

, we obtain

\begin{matrix} \prod_{i = 1}^{\infty} {(1 - q^{i})}^{i} = \sum_{N = 0}^{\infty} \sum_{t_{1} + 2 t_{2} + \dots + N t_{N} = N} \frac{{(- 1)}^{t_{1} + t_{2} + \dots + t_{N}} q^{(\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{N}}{2}) + N}}{{(q; q)}_{t_{1}} {(q; q)}_{t_{2}} \dots {(q; q)}_{t_{N}}} . \end{matrix}

This concludes the proof. □

Relevant to Theorem 16, it would be very appealing to have combinatorial interpretations for

\sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \frac{q^{(\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{n}}{2}) + n}}{{(q; q)}_{t_{1}} {(q; q)}_{t_{2}} \dots {(q; q)}_{t_{n}}}

and

\sum_{t_{1} + 2 t_{2} + \dots + n t_{n} = n} \frac{{(- 1)}^{t_{1} +_{2} + \dots + t_{n}} q^{(\binom{t_{1}}{2}) + (\binom{t_{2}}{2}) + \dots + (\binom{t_{n}}{2}) + n}}{{(q; q)}_{t_{1}} {(q; q)}_{t_{2}} \dots {(q; q)}_{t_{n}}}

in terms of n-color partitions.

Author Contributions

Conceptualization, M.M. and E.S.; writing—original draft, M.M. and E.S.; writing—review and editing, M.M. and E.S. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflict of interest.

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Merca, M.; Simion, E. n-Color Partitions into Distinct Parts as Sums over Partitions. Symmetry 2023, 15, 2067. https://doi.org/10.3390/sym15112067

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Merca M, Simion E. n-Color Partitions into Distinct Parts as Sums over Partitions. Symmetry. 2023; 15(11):2067. https://doi.org/10.3390/sym15112067

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Merca, Mircea, and Emil Simion. 2023. "n-Color Partitions into Distinct Parts as Sums over Partitions" Symmetry 15, no. 11: 2067. https://doi.org/10.3390/sym15112067

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n-Color Partitions into Distinct Parts as Sums over Partitions

Abstract

1. Introduction

2. Proof of Theorems

2.1. Proof of Theorem 3

2.2. Proof of Theorem 4

2.3. Proof of Theorem 5

2.4. Proof of Theorem 6

2.5. Proof of Theorem 7

3. Identities of the Second Kind

4. Concluding Remarks

Author Contributions

Funding

Data Availability Statement

Conflicts of Interest

References

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