Some n-Color Compositions

Yu-hong Guo^{1} Department of Mathematics Hexi University Gansu, Zhangye, 734000 P. R. China

                        gyh7001@163.com

Abstract

An n-color odd composition is defined as an n-color composition with odd parts, and an n-color composition with parts ≠ 1 is an n-color composition whose parts are > 1. In this paper, we get generating functions, explicit formulas and recurrence formulas for n-color odd compositions and n-color compositions with parts ≠ 1.

1 Introduction

In the classical theory of partitions, compositions were first defined by MacMahon [1] as ordered partitions. For example, there are 5 partitions and 8 compositions of 4. The partitions are 4, 31, 22, 21^{2}, 1^{4} and the compositions are 4, 31, 13, 22, 21^{2}, 121, 1^{2}2, 1^{4}.

Agarwal and Andrews [2] defined an n-color partition as a partition in which a part of size n can come in n different colors. They denoted different colors by subscripts: n_{1}, n_{2}, …, n_{n}. Analogous to MacMahon’s ordinary compositions Agarwal [3] defined an n-color composition as an n-color ordered partition. Thus, for example, there are 21 n-color compositions of 4, viz.,

4_{1}, 4_{2}, 4_{3}, 4_{4},
3_{1}1_{1}, 3_{2}1_{1}, 3_{3}1_{1}, 1_{1}3_{1}, 1_{1}3_{2}, 1_{1}3_{3},
2_{1}2_{1}, 2_{1}2_{2}, 2_{2}2_{2}, 2_{2}2_{1},
2_{1}1_{1}1_{1}, 2_{2}1_{1}1_{1}, 1_{1}2_{1}1_{1}, 1_{1}1_{1}2_{1}, 1_{1}2_{2}1_{1}, 1_{1}1_{1}2_{2},
1_{1}1_{1}1_{1}1_{1}.

1This work is supported by the Fund of the Education Department of Gansu Province (No. 200809-04)

and the fund of Hexi University.

More properties of n-color compositions were found in [4, 5]. In 2006, G. Narang and Agarwal [6, 7] also defined an n-color self-inverse composition and gave some properties. In 2010, Guo [8] defined an n-color even self-inverse composition and proved some properties.

In this paper, we shall study some n-color compositions. We first give the following definitions.

Definition 1. An n-color odd composition is an n-color composition with odd parts.

Thus, for example, there are 7 n-color odd compositions of 4, viz.,

3_{1}1_{1}, 3_{2}1_{1}, 3_{3}1_{1},
1_{1}3_{1}, 1_{1}3_{2}, 1_{1}3_{3}, 1_{1}1_{1}1_{1}1_{1}.

Definition 2. An n-color composition with parts ≠ 1 is an n-color composition whose parts are > 1.

For example, there are 17 n-color compositions with parts ≠ 1 of 5, viz.,

5_{1}, 5_{2}, 5_{3}, 5_{4}, 5_{5},
2_{1}3_{1}, 2_{1}3_{2}, 2_{1}3_{3}, 2_{2}3_{1}, 2_{2}3_{2}, 2_{2}3_{3},
3_{1}2_{1}, 3_{2}2_{1}, 3_{3}2_{1}, 3_{1}2_{2}, 3_{2}2_{2}, 3_{3}2_{2}.

In section 2 we shall give generating functions, recurrence formulas and explicit formulas for n-color compositions above.

Agarwal [3] proved the following theorem.

Theorem 3. ([3]) Let C(m, q) and C(q) denote the enumerative generating functions for C(m, ν) and C(ν), respectively, where C(m, ν) is the number of n-color compositions of ν into m parts and C(ν) is the number of n-color compositions of ν. Then

C(m, q) = \frac{q^{m}}{(1 − q)^{2m}} ,  (1)
C(q) = \frac{q}{1 − 3q + q^{2}} ,  (2)

C(m, ν) = \binom{ν + m − 1}{2m − 1},  (3)
C(ν) = F_{2ν}.  (4)

2 Main results

We denote the number of n-color odd compositions of ν by C(o, ν) and the number of n-color odd compositions of ν into m parts by C(m, o, ν), respectively. In this section, we first prove the following theorem.

Theorem 4. Let C(m, o, q) and C(o, q) denote the enumerative generating functions for C(m, o, ν) and C(o, ν), respectively. Then

C(m, o, q) = \frac{q^{m}(1 + q^{2})^{m}}{(1 − q^{2})^{2m}} ,  (5)

C(o, q) = \frac{q + q^{3}}{1 − q − 2q^{2} − q^{3} + q^{4}} ,  (6)

C(m, o, ν) = ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j},  (7)

C(o, ν) = ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}.  (8)

where (ν − m) is even, and (ν − m) ≥ 0; 0 ≤ i, j are integers.

Proof. Similar to the proof of Agarwal [3], we have

C(m, o, q) = ∑_{ν=1}^{∞} C(m, o, ν)q^{ν} = (q + 3q^{3} + ⋯ +)^{m} = \frac{q^{m}(1 + q^{2})^{m}}{(1 − q^{2})^{2m}} .

This proves (5).

C(o, q) = ∑_{m=1}^{∞} C(m, o, q) = ∑_{m=1}^{∞} \frac{q^{m}(1 + q^{2})^{m}}{(1 − q^{2})^{2m}} = \frac{q + q^{3}}{1 − q − 2q^{2} − q^{3} + q^{4}} .

We get (6).

On equating the coefficients of q^{ν} in (5), we have

C(m, o, ν) = ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}.

Since ν is even if m is even, and ν is odd if m is odd, then ν − m is even. This proves (7).

Obviously m ≤ ν, so (8) is also proven.

We complete the proof of this theorem.

In this section, we also prove the following recurrence formula.

Theorem 5. Let O_{ν} denote the number of n-color odd compositions of ν. Then

O_{1} = 1, O_{2} = 1, O_{3} = 4, O_{4} = 7

and

O_{ν} = O_{ν−1} + 2O_{ν−2} + O_{ν−3} − O_{ν−4}, for ν > 4.

Proof. (Combinatorial) To prove that O_{ν} = O_{ν−1} +2O_{ν−2} +O_{ν−3} −O_{ν−4}, we split the n-color compositions enumerated by O_{ν} + O_{ν−4} into four classes:

(A) enumerated by O_{ν} with 1_{1} on the right.

(B) enumerated by O_{ν} with 3_{3} on the right.

(C) enumerated by O_{ν} with h_{t} on the right, h > 1, 1 ≤ t ≤ h − 2 (where, h is odd). (D) enumerated by O_{ν} with h_{t} on the right, h > 1,h − 1 ≤ t ≤ h except 3_{3} and those

enumerated by O_{ν−4}.

We transform the n-color odd compositions in class (A) by deleting 1_{1} on the right. This produces n-color compositions enumerated by O_{ν−1}. Conversely, for any n-color composition enumerated by O_{ν−1} we add 1_{1} on the right to produce the elements of the class (A). In this way we prove that there are exactly O_{ν−1} elements in the class (A).

Similarly, we can produce O_{ν−3} n-color odd compositions in the class (B) by deleting 3_{3} on the right.

Next, we transform the n-color odd compositions in class (C) by subtracting 2 from h, that is, replacing h_{t} by (h − 2)_{t}. This transformation also establishes the fact that there are exactly O_{ν−2} elements in class (C). This correspondence being one to one.

Finally, we transform the elements in class (D) as follows: Subtract 2_{2} from h_{t} on the right when h > 3, h − 1 ≤ t ≤ h, that is, replace h_{t} by (h − 2)_{(t−2)}; in this way we will get n-color odd compositions of ν − 2 with part ht′^{′} on the right, where, h′ > 1, t′ ≥ h′ − 1. After that we replace h_{t} by (h − 2)_{(t−1)} when h = 3, t = 2. This produces n-color odd compositions of ν − 2 with part 1_{1} on the right. To get the remaining n-color odd compositions from O_{ν−4}, we add 2 to the right parts, that is, replace h_{t} by (h + 2)_{t} to get the n-color odd compositions of (ν − 2) with part ht′^{′} on the right, where, h′ > 1, 1 ≤ t′ ≤ h′ − 2. We see that the number of n-color odd compositions in class (D) is also equal to O_{ν−2}. Hence, O_{ν} + O_{ν−4} = O_{ν−1} + 2O_{ν−2} + O_{ν−3}. viz.,O_{ν} = O_{ν−1} + 2O_{ν−2} + O_{ν−3} − O_{ν−4}.

Thus, we complete the proof.

We also give another proof of Theorem 5.

Proof. We have

O_{ν} = ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

= ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + (i − 1) − 1}{2m − 1}\binom{m}{j} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + (i − 1) − 1}{2m − 2}\binom{m}{j}

(by the binomial identity\binom{n}{m} = \binom{n − 1}{m} + \binom{n − 1}{m − 1})

= ∑_{m≤ν−2} ∑_{i+j= \frac{(ν−2)−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j} + \binom{2ν − 2}{2ν − 1}\binom{ν}{0}

+ ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j} − ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 2}{2m − 1}\binom{m}{j}

= O_{ν−2} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

− ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + (i − 2) − 1}{2m − 1}\binom{m}{j} − ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 3}{2m − 2}\binom{m}{j}

= O_{ν−2} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

− ∑_{m≤ν−4} ∑_{i+j= \frac{(ν−4)−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j} − \binom{2ν − 2 − 1}{2ν − 1}\binom{ν}{0}

− \binom{2(ν − 2) − 2 − 1}{2(ν − 2) − 1}\binom{ν − 2}{1} − \binom{2(ν − 2) − 1 − 1}{2(ν − 2) − 1}\binom{ν − 2}{0}

− ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 3}{2m − 2}\binom{m}{j}

= O_{ν−2} − O_{ν−4} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + (i − 1) − 1}{2m − 1}\binom{m}{j}

+ ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 2}{2m − 2}\binom{m}{j} − ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 3}{2m − 2}\binom{m}{j}

= 2O_{ν−2} − O_{ν−4} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 3}{2m − 3}\binom{m}{j}
= 2O_{ν−2} − O_{ν−4} + ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2(m − 1) + i − 1}{2(m − 1) − 1}\binom{m − 1}{j}

+ ∑_{m≤ν} ∑_{i+j= \frac{ν−m}{2}} \binom{2(m − 1) + i − 1}{2(m − 1) − 1}\binom{m − 1}{j − 1}

= 2O_{ν−2} − O_{ν−4} + ∑_{m≤ν−1} ∑_{i+j= \frac{(ν−1)−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

+ ∑_{m≤ν−3} ∑_{i+j= \frac{(ν−3)−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

= O_{ν−1} + 2O_{ν−2} + O_{ν−3} − O_{ν−4}.

So we have O_{ν} = O_{ν−1} + 2O_{ν−2} + O_{ν−3} − O_{ν−4}. From recurrence formula above we have the following corollary easily.

Corollary 6. If ν > 4, then

∑_{m≤ν−4}( ∑_{i+j= \frac{ν−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j} − ∑_{i+j= \frac{ν−1−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

− 2 ∑_{i+j= \frac{ν−2−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j} − ∑_{i+j= \frac{ν−3−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}

+ ∑_{i+j= \frac{ν−4−m}{2}} \binom{2m + i − 1}{2m − 1}\binom{m}{j}) = 0.

Next, we shall study n-color compositions with parts ≠ 1. We denote the number of n-color compositions with parts ≠ 1 of ν by C_{≠1}(ν) and the number of n-color compositions with parts ≠ 1 of ν into m parts by C_{≠1}(m, ν), respectively. In this section, we present the following theorem.

Theorem 7. Let C_{≠1}(m, q) and C_{≠1}(q) denote the enumerative generating functions for C_{≠1}(m, ν) and C_{≠1}(ν), respectively. Then

C_{≠1}(m, q) = \frac{q^{2m}(2 − q)^{m}}{(1 − q)^{2m}} ,  (9)

C_{≠1}(q) = \frac{2q^{2} − q^{3}}{1 − 2q − q^{2} + q^{3}} ,  (10)

C_{≠1}(m, ν) = ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j},  (11)

C_{≠1}(ν) = ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}.  (12)

where (ν − 2m) is an integer, and (ν − 2m) ≥ 0; 0 ≤ i, j are integers.

Proof. Similar to the proof of Agarwal [3], we have

C_{≠1}(m, q) = ∑_{ν=1}^{∞} C_{≠1}(m, ν)q^{ν} = (2q^{2} + 3q^{3} + ⋯ +)^{m} = \frac{q^{2m}(2 − q)^{m}}{(1 − q)^{2m}} .

This proves (9).

C_{≠1}(q) = ∑_{m=1}^{∞} C_{≠1}(m, q) = ∑_{m=1}^{∞} \frac{q^{2m}(2 − q)^{m}}{(1 − q)^{2m}} = \frac{2q^{2} − q^{3}}{1 − 2q − q^{2} + q^{3}} .

This proves (10).

On equating the coefficients of q^{ν} in (9), we have

C_{≠1}(m, ν) = ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}.

Since ν ≥ 2m, then ν − 2m ≥ 0, i + j ≥ 0, and 0 ≤ i, j are integers. This proves (11). Obviously m ≤ \frac{ν}{2}, therefore (12) is also proven. We complete the proof of this theorem.

In this section, we also prove the following recurrence formula.

Theorem 8. Let C_{≠1}(ν) denote the number of n-color compositions with parts ≠ 1 of ν. Then

C_{≠1}(2) = 2, C_{≠1}(3) = 3, C_{≠1}(4) = 8,

and

C_{≠1}(ν) = 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2) − C_{≠1}(ν − 3) for ν > 4.

Proof. (Combinatorial) To prove that C_{≠1}(ν) = 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2) − C_{≠1}(ν − 3), we split the n-color compositions enumerated by C_{≠1}(ν) + C_{≠1}(ν − 3) into three classes:

(A) enumerated by C_{≠1}(ν) with 2_{1} on the right.

(B) enumerated by C_{≠1}(ν) with h_{t} on the right, h > 2,1 ≤ t ≤ h − 1.

(C) enumerated by C_{≠1}(ν) with h_{h} on the right, h ≥ 2 and those enumerated by C_{≠1}(ν − 3).

We transform the n-color compositions in class (A) by deleting 2_{1} on the right. This produces n-color compositions enumerated by C_{≠1}(ν − 2). Conversely, for any n-color composition enumerated by C_{≠1}(ν − 2) we add 2_{1} on the right to produce the elements of the class (A). In this way we prove that there are exactly C_{≠1}(ν − 2) elements in the class (A).

Next, we transform the n-color compositions in class (B) by subtracting 1 from h, that is, replacing h_{t} by (h − 1)_{t}; this transformation also establishes the fact that there are exactly C_{≠1}(ν − 1) elements in class (B). This correspondence being one to one.

Finally, we transform the elements in class (C) as follows: Subtract 1_{1} from h_{h} on the right when h > 2, that is, replace h_{h} by (h − 1)_{(h−1)}; in this way we will get n-color compositions of ν − 1 with part hh′^{′}(h′ > 1) on the right. We also replace h_{h} by (h − 1)_{(h−1)} when h = 2. This produces n-color compositions of ν − 1 with part 1_{1} on the right. Now we delete 1_{1} and add 1 to the preceding part of it. For example, 2_{1}2_{2}2_{2}⟶2_{1}2_{2}1_{1}⟶2_{1}3_{2}; 4_{1}2_{2}⟶4_{1}1_{1}⟶5_{1}. Then we have n-color compositions of ν − 1 with part ht^{′} on the right, where, h′ > 2, 1 ≤ t ≤ h′ − 1. To get the remaining n-color compositions from C_{≠1}(ν − 3), we set 2_{1} on the right. This produces n-color compositions with parts ≠ 1 of ν − 1 with 2_{1} on the right. We see that the number of n-color compositions in class (C) is also equal to C_{≠1}(ν − 1). Hence, C_{≠1}(ν) + C_{≠1}(ν − 3) = 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2). viz., C_{≠1}(ν) = 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2) − C_{≠1}(ν − 3).

Thus, we complete the proof.

We also give another proof of Theorem 8.

Proof. We have

C_{≠1}(ν) = ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}

= ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + (i − 1) − 1}{2m − 1}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 2}\binom{m}{j}

(by the binomial identity\binom{n}{m} = \binom{n − 1}{m} + \binom{n − 1}{m − 1})

= ∑_{m≤ \frac{ν−1}{2}} ∑_{i+j=(ν−1)−2m}(−1)^{j}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 3}\binom{m}{j}

= C_{≠1}(ν − 1) + ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 3}\binom{m − 1}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2(m − 1) + i − 1}{2(m − 1) − 1}\binom{m − 1}{j − 1}

= C_{≠1}(ν − 1) + ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2}{2m − 2}\binom{m − 1}{j}

− ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m − 1}{j}

+ ∑_{m≤ \frac{(ν−3)}{2}} ∑_{i+j=(ν−3)−2m}(−1)^{j+1}2^{m−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}

= C_{≠1}(ν − 1) − C_{≠1}(ν − 3)

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m − 1}{j − 1}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m − 1}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 3}\binom{m − 1}{j}

= C_{≠1}(ν − 1) − C_{≠1}(ν − 3)

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 2}\binom{m}{j}

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2(m − 1) + i − 1}{2(m − 1) − 1}\binom{m − 1}{j}

= C_{≠1}(ν − 1) − C_{≠1}(ν − 3)

+ ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2}{2m − 1}\binom{m}{j}

− ∑_{m≤ \frac{ν}{2}} ∑_{i+j=ν−2m}(−1)^{j}2^{m−j}\binom{2m + i − 2 − 1}{2m − 1}\binom{m}{j}

+ ∑_{m≤ \frac{ν−2}{2}} ∑_{i+j=(ν−2)−2m}(−1)^{j}2^{m+1−j}\binom{2m + i − 1}{2m − 1}\binom{m}{j}

= C_{≠1}(ν − 1) − C_{≠1}(ν − 3) + C_{≠1}(ν − 1) − C_{≠1}(ν − 2) + 2C_{≠1}(ν − 2)
= 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2) − C_{≠1}(ν − 3).

Thus we have C_{≠1}(ν) = 2C_{≠1}(ν − 1) + C_{≠1}(ν − 2) − C_{≠1}(ν − 3).

3 Acknowledgement

The author would like to thank the referee for his/her suggestions and comments which have improved the quality of this paper.

References

[1] P. A. MacMahon, Combinatory Analysis, AMS Chelsea Publishing, 2001.

[2] A. K. Agarwal and G. E. Andrews, Rogers-Ramanujan identities for partitions with “n copies of n”, J. Combin. Theory Ser. A 45 (1987), 40–49.

[3] A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (2000), 1421–1427.

[4] A. K. Agarwal, An analogue of Euler’s identity and new combinatorial properties of n-colour compositions, J. Comput. Appl. Math. 160 (2003), 9–15.

[5] Yu-Hong Guo, Some identities between partitions and compositions, Acta Math. Sinica (Chin. Ser.) 50 (2007), 707–710.

[6] G. Narang and A. K. Agarwal, n-colour self-inverse compositions, Proc. Indian Acad. Sci. Math. Sci. 116 (2006), 257–266.

[7] G. Narang and A. K. Agarwal, Lattice paths and n-color compositions, Discrete Math. 308 (2008), 1732–1740.

[8] Yu-Hong Guo, n-colour even self-inverse compositions, Proc. Indian Acad. Sci. Math. Sci. 120 (2010), 27–33.

[9] G. E. Andrews, The Theory of Partitions, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, 1998.

2010 Mathematics Subject Classification: Primary 05A17. Keywords: n-color odd composition, n-color composition, generating function, explicit formula, recurrence formula.

Received June 12 2011 revised version received November 27 2011. Published in Journal of Integer Sequences, December 26 2011.

Return to Journal of Integer Sequences home page.