Enumeration of the Partitions of an Integer into Parts of a Specified Number of Different

Sizes and Especially Two Sizes

Nesrine Benyahia Tani Algiers 3 University Faculty of Economics and Management Sciences 2 Ahmed Waked Street Dely Brahim, Algiers Algeria

                 benyahiatani@yahoo.fr

Sadek Bouroubi^{1} University of Science and Technology Houari Boumediene Faculty of Mathematics P. O. Box 32 16111 El-Alia, Bab-Ezzouar, Algiers Algeria

                  sbouroubi@usthb.dz
                  bouroubis@yahoo.fr

Abstract

A partition of a non-negative integer n is a way of writing n as a sum of a nondecreasing sequence of parts. The present paper provides the number of partitions of an integer n into parts of a specified number of different sizes. We establish new formulas for such partitions with particular interest to the number of partitions of n into parts of two sizes. A geometric application is given at the end of this paper.

1Both authors were supported by the laboratory LAID3.

1 Introduction and definitions

Let n and k be integers. A partition of n into k parts is an integral solution of the system

\begin{cases}n = n_{1} + ⋯ + n_{k}; \\ 1 ≤ n_{1} ≤ ⋯ ≤ n_{k}. \end{cases}

Euler was the first to undertake the problem of counting an integer’s partitions. Since then, mathematicians have been more and more interested in integer partitions and their fascinating properties. In fact, in the theory of integer partitions, various restrictions on the nature of partitions are often considered. One may require that the n_{i}’s be distinct, odd or even, or that n must be split into exactly k parts, etc. More on integer partitions can be found in [1, 2, 3, 4, 5] and [7, 8, 9].

Here, we are interested in partitions of the number n into parts of precisely s different sizes. Extending prior results, we derive several identities linking this kind of partitions to the number of divisors τ(n). In addition, we obtain new recurrence formulas to count the number of such partitions and a new identity to count the number of partitions of an integer into two sizes of parts.

Let t(n, k, s) be the number of partitions of n into k parts of precisely s different sizes, k = s, …, n − \frac{s(s−1)}{2}, it is an integral solution of the system

\begin{cases}n = a_{1} n_{1} + ⋯ + a_{s} n_{s}; \\ 1 ≤ n_{1} < ⋯ < n_{s}; \\ a_{1} + ⋯ + a_{s} = k; \\ a_{1}, …, a_{s} ≥ 1. \end{cases}  (1)

The total number of partitions of n into s different sizes of parts is denoted t(n, s) (see \underline{A002133} for t(n, 2)).

If s is specified, then t(n, k, s) = 0 if k ≤ s − 1 and either k > n − \frac{s(s−1)}{2} or n < max\{k, \frac{s(s+1)}{2}\}. Then we have

t(n, s) = ∑_{k=s}^{\frac{2n−s(s−1)}{2}} t(n, k, s) = ∑_{k≥1} t(n, k, s).  (2)

For instance, if s = 1, then k ≥ 1, n ≥ k, and

t(n, k, 1) = \begin{cases}1, & if n is a multiple of k; \\ 0, & otherwise. \end{cases}  (3)

Therefore

∑_{n≥k}t(n, k, 1) q^{n} = \frac{q^{k}}{1 − q^{k}} ·

Also, it is easy to see that

t(n, 2, 2) = ⌊\frac{n − 1}{2}⌋ ,

where ⌊x⌋ is the greatest integer ≤ x. So, we have

∑_{n≥k}t(n, 2, 2) q^{n} = q^{3} + q^{4} + 2q^{5} + 2q^{6} + 3q^{7} + 3q^{8} + ⋯

= \frac{q^{3}}{1 − q} + \frac{q^{5}}{1 − q} + \frac{q^{7}}{1 − q} + ⋯
= \frac{q^{3}}{(1 − q)(1 − q^{2})} ·

P. A. MacMahon [6] was the first mathematician interested in this kind of partitions. Also, Emeric Deutsch studied the number of partitions of n into exactly two odd sizes of parts (see \underline{A117955}) and the number of partitions of n into exactly two sizes of parts, one odd and one even (see \underline{A117956}).

2 Preliminary results

Throughout the remainder of the paper, let τ(k) and let τ_{d↓}(k) be respectively the number of positive divisors of k and the number of positive divisors of k less than or equal to d.

In this section we state some recurrence formulas involving the number t(n, k, s). The main identity of the present work is based on the following results:

Theorem 1. Let n, k and s be integers. For k ≥ s ≥ 2, n ≥ k + \frac{s(s−1)}{2} and n ≥ max\{k, \frac{s(s+1)}{2}\}, we have

t(n, k, s) = ∑^{⌊ \frac{2n−s(s−1)}{2k} ⌋} ; i=1 ∑^{k−s+1} ; j=1t(n − ki, k − j, s − 1),  (4)

and

t(n, k, 2) = ∑_{i=1}^{⌊ \frac{n−1}{k} ⌋} τ_{k−1↓}(n − ki).  (5)

Proof. Note that every part n_{i} in System (1), for i = 2, …, s, can be written as n_{i} = n_{1} + d_{i}, d_{i} ≥ 1. Considering n_{1} and a_{1} as parameters, System (1) can be rewritten as follows:

\begin{cases}n − kn_{1} = a_{2} d_{2} + ⋯ + a_{s} d_{s}; \\ 1 ≤ d_{2} < ⋯ < d_{s}; \\ a_{2} + ⋯ + a_{s} = k − a_{1}; \\ a_{1}, …, a_{s} ≥ 1. \end{cases}  (6)

Hence, we get

t(n, k, s) = ∑_{n_{1}∈N} ∑_{a_{1}∈A} t(n − kn_{1}, k − a_{1}, s − 1),

where N and A are the sets containing the values of n_{1} and a_{1} respectively.

The smallest values of n_{1} and a_{1} is 1. The largest value of n_{1} is found by setting a_{i+1} = 1 and d_{i+1} = i, for i = 1, …, s − 1, in the first equation of System (6). Then, we get

1 ≤ n_{1} ≤ ⌊\frac{2n − s(s − 1)}{2k}⌋ .

Setting a_{i} = 1, for i = 2, …, s in the third equation of System (6), one can see that the largest value of a_{1} is k − s + 1.

To prove (5) we apply (4) with s = 2,

t(n, k, 2) = ∑_{i=1}^{⌊ \frac{n−1}{k} ⌋} ∑_{j=1}^{k−1} t(n − ki, j, 1).

So by (3) we get

∑_{j=1}^{k−1} t(n − ki, j, 1) = τ_{k−1↓}(n − ki).

This implies (5).

Theorem 1 allows the easy recovery of known identities such as,

t(n, 2, 2) = ⌊\frac{n − 1}{2}⌋ .  (7)

Also, it allows to deduce some new values for t(n, k, 2). For instance, for k = 3 … 6, we have

Corollary 2. For n ≥ 3, we have

t(n, 3, 2) = \begin{cases}\frac{n − 3}{3} + ⌊\frac{n − 3}{6}⌋ , & if n ≡ 0 (mod 3); \\ \frac{n − 1}{3} + ⌊\frac{n − 1}{6}⌋ , & if n ≡ 1 (mod 3); \\ \frac{n − 2}{3} + ⌊\frac{n + 1}{6}⌋ , & if n ≡ 2 (mod 3). \end{cases}

t(n, 4, 2) = \begin{cases}\frac{n − 4}{2} + ⌊\frac{n − 4}{12}⌋ , & if n ≡ 0 (mod 4); \\ \frac{n − 1}{4} + ⌊\frac{n − 1}{12}⌋ , & if n ≡ 1 (mod 4); \\ \frac{n − 2}{2} + ⌊\frac{n + 2}{12}⌋ , & if n ≡ 2 (mod 4); \\ \frac{n − 3}{4} + ⌊\frac{n + 5}{12}⌋ , & if n ≡ 3 (mod 4). \end{cases}

t(n, 5, 2) = \begin{cases}\frac{n − 5}{5} + ⌊\frac{n − 5}{10}⌋ + ⌊\frac{n − 5}{15}⌋ + ⌊\frac{n − 5}{20}⌋ , & if n ≡ 0 (mod 5); \\ \frac{n − 1}{5} + ⌊\frac{n − 1}{10}⌋ + ⌊\frac{n + 4}{15}⌋ + ⌊\frac{n − 1}{20}⌋ , & if n ≡ 1 (mod 5); \\ \frac{n − 2}{5} + ⌊\frac{n + 3}{10}⌋ + ⌊\frac{n − 2}{15}⌋ + ⌊\frac{n + 3}{20}⌋ , & if n ≡ 2 (mod 5); \\ \frac{n − 3}{5} + ⌊\frac{n − 3}{10}⌋ + ⌊\frac{n + 2}{15}⌋ + ⌊\frac{n + 7}{20}⌋ , & if n ≡ 3 (mod 5); \\ \frac{n − 4}{5} + ⌊\frac{n + 1}{10}⌋ + ⌊\frac{n + 1}{15}⌋ + ⌊\frac{n + 11}{20}⌋ , & if n ≡ 4 (mod 5). \end{cases}

t(n, 6, 2) = \begin{cases}\frac{n − 6}{2} + ⌊\frac{n − 6}{12}⌋ + ⌊\frac{n − 6}{30}⌋ , & if n ≡ 0 (mod 6); \\ \frac{n − 1}{6} + ⌊\frac{n − 1}{30}⌋ , & if n ≡ 1 (mod 6); \\ \frac{n − 2}{3} + ⌊\frac{n − 2}{12}⌋ + ⌊\frac{n + 4}{30}⌋ , & if n ≡ 2 (mod 6); \\ \frac{n − 3}{3} + ⌊\frac{n + 9}{30}⌋ , & if n ≡ 3 (mod 6); \\ \frac{n − 4}{3} + ⌊\frac{n + 2}{12}⌋ + ⌊\frac{n + 14}{30}⌋ , & if n ≡ 4 (mod 6); \\ a \\ \frac{n − 5}{6} + ⌊\frac{n + 19}{30}⌋ , & if n ≡ 5 (mod 6). \end{cases}

Proof. The results follow immediately from Theorem 1. Corollary 3. For n ≥ 3 and ⌈\frac{n+1}{2}⌉ ≤ k ≤ n − 1, we have

t(n, k, 2) = τ(n − k).

Proof. On the one hand, the sum in (5) is reduced to one element if 1 ≤ \frac{n−1}{k} < 2, i.e.,

\frac{n − 1}{2} < k ≤ n − 1.

On the other hand, τ_{k−1↓}(n − k) = τ(n − k), if k − 1 ≥ n − k, i.e.,

k ≥ \frac{n + 1}{2} ·

Hence the result follows. Remark 4. From Corollary 3, for n ≥ 2k + 1 and k ≥ 1, we have

t(n, n − k, 2) = τ(k).

For instance, for n ≥ 27, we get

t(n, n − 13, 2) = τ(13) = 2.

3 Main identity

The aim of this section is to derive an explicit formula for t(n, k, 2). Before giving the next Theorem, we introduce some notation. Let

• φ_{i}(j) = \begin{cases}1, & if j ≡ 0 (mod i); \\ 0, & otherwise. \end{cases}

• χ_{k} (i, j) = \begin{cases}0, & if i ≠ 0 and gcd(k, j) ≠ 1 and gcd(i, j) = 1; \\ 1, & otherwise. \end{cases}

• W_{k} = [W_{k}(i, j)], 0 ≤ i ≤ k − 1, 1 ≤ j ≤ k − 1 be a matrix, whose elements are given by

W_{k} (i, j) = \begin{cases}d, & if i ∈ I_{k,j}(d) and χ_{k} (i, j) = 1; \\ j, & otherwise. \end{cases}

where, 0 ≤ d ≤ \frac{j}{gcd(k, j)} − 1 and

I_{k,j}(d) = \{i = (⌊\frac{dk − 1}{j}⌋ + a) j − dk / 1 ≤ a ≤ ⌊\frac{(d + 1)k − 1}{j}⌋ − ⌊\frac{dk − 1}{j}⌋\} .

Remark 5. The construction of matrix W_{k} is special, it is filled column by column as follows:

1. Case χ_{k} (i, j) = 1 Each value of the parameter d generates some values of the parameter a, which in return produce the values of the lines i, this process allows to define the elements of the concerned lines.

2. Case χ_{k} (i, j) = 0 The empty elements are replaced by the number j of the column.

For example, for k = 6, we get

W_{6} = [ \begin{matrix}0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 3 & 4 & 4 \\ 0 & 0 & 3 & 1 & 3 \\ 0 & 2 & 0 & 4 & 2 \\ 0 & 0 & 3 & 0 & 1 \\ 0 & 2 & 3 & 4 & 0 \end{matrix} ] .

The formulas of Corollary 2 are special cases that motivate the following generalization.

Theorem 6. For n ≥ 3, n ≡ i (mod k), 2 ≤ k ≤ n − 1, we have

t(n, k, 2) = \begin{cases}∑_{j=1}^{k−1} ⌊\frac{gcd(k, j)}{kj}(n − k)⌋ , & if i = 0; \\ ∑_{j=1}^{k−1} χ_{k}(i, j) ⌊1 + gcd(k, j)\frac{n − i − k − kW_{k}(i, j)}{kj}⌋ , & otherwise. \end{cases}

Proof. Case 1. n = kl, i.e., i = 0. Using Theorem 1, we get

\begin{aligned}t(n, k, 2) & = & ∑_{h=1}^{l−1} τ_{k−1↓}(kh) \\ & = & ∑_{j=1}^{k−1} ∑_{h=1}^{l−1} φ_{j}(kh). \end{aligned}

The divisors of kh which are multiples of j are of the form \frac{kjd_{h}}{gcd(k, j)}. Then

\begin{aligned}t(n, k, 2) & = & ∑_{j=1}^{k−1} ⌊\frac{gcd(k, j)}{j}(l − 1)⌋ \\ & = & ∑_{j=1}^{k−1} ⌊\frac{gcd(k, j)}{kj}(n − k)⌋ . \end{aligned}

Case 2. n = kl + i, 1 ≤ i ≤ k − 1. Using Theorem 1, we get

\begin{aligned}t(n, k, 2) & = & ∑_{h=0}^{l−1} τ_{k−1↓}(kh + i) \\ & = & ∑_{i=1}^{k−1} ∑_{h=0}^{l−1} φ_{j}(kh + i). \end{aligned}

It is straightforward to verify that the divisors of kh + i that are multiples of j are of the following form

χ_{k}(i, j) (\frac{kjd_{h}}{gcd(k, j)} + i + kW_{k}(i, j)).

Hence,

t(n, k, 2) = ∑_{j=1}^{k−1} χ_{k}(i, j) ⌊\frac{j + gcd(k, j)(l − 1 − W_{k}(i, j))}{j}⌋

= ∑_{j=1}^{k−1} χ_{k}(i, j) ⌊\frac{j + gcd(k, j)(\frac{n−i}{k} − 1 − W_{k}(i, j))}{j}⌋

= ∑_{j=1}^{k−1} χ_{k}(i, j) ⌊1 + \frac{gcd(k, j)}{kj}(n − i − k − kW_{k}(i, j))⌋ .

Example 7. k = 6 1. For i = 0, n = 6l, we get

\begin{aligned}t(6l, 6, 2) & = & ∑_{j=1}^{5} ⌊\frac{n − 6}{6j} gcd(6, j)⌋ \\ & = & \frac{n − 6}{2} + ⌊\frac{n − 6}{12}⌋ + ⌊\frac{n − 6}{30}⌋ . \end{aligned}

2. For i = 1, n = 6l + 1 we get

t(6l + 1, 6, 2) = ∑_{j=1}^{5} χ_{6}(1, j) ⌊1 + \frac{gcd(6, j)}{6j}(n − 7 − 6W_{6}(1, j))⌋
= ⌊1 + \frac{n − 7}{6}⌋ + ⌊1 + \frac{n − 7 − 24}{30}⌋
= \frac{n − 1}{6} + ⌊\frac{n − 1}{30}⌋ .

3. For i = 2, n = 6l + 2 we get

\begin{matrix}t(6l + 2, 6, 2) = & ∑_{j=1}^{5} χ_{6}(2, j) ⌊1 + \frac{gcd(6, j)}{6j}(n − 8 − 6W_{6}(2, j))⌋ \\ = & ⌊1 + \frac{n − 8}{6}⌋ + ⌊1 + \frac{2(n − 8)}{12}⌋ + ⌊1 + \frac{2(n − 8 − 6)}{24}⌋ + ⌊1 + \frac{n − 8 − 18}{30}⌋ \\ = & \frac{n − 2}{3} + ⌊\frac{n − 2}{12}⌋ + ⌊\frac{n + 4}{30}⌋ . \end{matrix}

4. For i = 3, n = 6l + 3 we get

t(6l + 3, 6, 2) = ∑_{j=1}^{5} χ_{6}(3, j) ⌊1 + \frac{gcd(6, j)}{6j}(n − 9 − 6W_{6}(3, j))⌋
= ⌊1 + \frac{n − 9}{6}⌋ + ⌊1 + \frac{3(n − 9)}{18}⌋ + ⌊1 + \frac{n − 9 − 12}{30}⌋
= \frac{n − 3}{3} + ⌊\frac{n + 9}{30}⌋ .

5. For i = 4, n = 6l + 4 we get

\begin{matrix}t(6l + 4, 6, 2) = & ∑_{j=1}^{5} χ_{6}(4, j) ⌊1 + \frac{gcd(6, j)}{6j}(n − 10 − 6W_{6}(4, j))⌋ \\ = & ⌊1 + \frac{n − 10}{6}⌋ + ⌊1 + \frac{2(n − 10)}{12}⌋ + ⌊1 + \frac{2(n − 10)}{24}⌋ + ⌊1 + \frac{n − 10 − 6}{30}⌋ \\ = & \frac{n − 4}{3} + ⌊\frac{n + 2}{12}⌋ + ⌊\frac{n + 14}{30}⌋ . \end{matrix}

6. For i = 5, n = 6l + 5 we get

t(6l + 5, 6, 2) = ∑_{j=1}^{5} χ_{6}(5, j) ⌊1 + \frac{gcd(6, j)}{6j}(n − 11 − 6W_{6}(5, j))⌋
= ⌊1 + \frac{n − 11}{6}⌋ + ⌊1 + \frac{n − 11}{30}⌋
= \frac{n − 5}{6} + ⌊\frac{n + 19}{30}⌋ .

Using Theorem 6, we obtain the following table for n ≤ 20.

\begin{matrix}n\k & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & t(n, 2) \\ 3 & 1 & & & & & & & & & & & & & & & & & & 1 \\ 4 & 1 & 1 & & & & & & & & & & & & & & & & & 2 \\ 5 & 2 & 2 & 1 & & & & & & & & & & & & & & & & 5 \\ 6 & 2 & 1 & 2 & 1 & & & & & & & & & & & & & & & 6 \\ 7 & 3 & 3 & 2 & 2 & 1 & & & & & & & & & & & & & & 11 \\ 8 & 3 & 3 & 2 & 2 & 2 & 1 & & & & & & & & & & & & & 13 \\ 9 & 4 & 3 & 2 & 3 & 2 & 2 & 1 & & & & & & & & & & & & 17 \\ 10 & 4 & 4 & 5 & 1 & 3 & 2 & 2 & 1 & & & & & & & & & & & 22 \\ 11 & 5 & 5 & 3 & 4 & 2 & 3 & 2 & 2 & 1 & & & & & & & & & & 27 \\ 12 & 5 & 4 & 4 & 3 & 3 & 2 & 3 & 2 & 2 & 1 & & & & & & & & & 29 \\ 13 & 6 & 6 & 4 & 5 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & & & & & & & 37 \\ 14 & 6 & 6 & 7 & 5 & 5 & 1 & 4 & 2 & 3 & 2 & 2 & 1 & & & & & & & 44 \\ 15 & 7 & 6 & 4 & 3 & 4 & 4 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & & & & & 44 \\ 16 & 7 & 7 & 7 & 5 & 6 & 4 & 3 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & & & & 55 \\ 17 & 8 & 8 & 5 & 7 & 3 & 5 & 3 & 4 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & & & 59 \\ 18 & 8 & 7 & 9 & 6 & 7 & 4 & 5 & 2 & 4 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & & 68 \\ 19 & 9 & 9 & 6 & 7 & 3 & 7 & 3 & 4 & 3 & 4 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & & 71 \\ 20 & 9 & 9 & 9 & 5 & 7 & 5 & 8 & 3 & 3 & 3 & 4 & 2 & 4 & 2 & 3 & 2 & 2 & 1 & 81 \end{matrix} ; Table 1: t(n, k, 2), 2 ≤ k ≤ 19, 3 ≤ n ≤ 20.

Also, Theorem 6 allows us to obtain t(n, 2) for large values of n, the following table is introduced to illustrate a few.

\begin{matrix}n & 100 & 500 & 1000 & 1500 & 2000 & 2500 & 3000 & 3500 & 4000 \\ t(n, 2) & 1135 & 11103 & 28340 & 54652 & 70128 & 91440 & 136790 & 144687 & 169953 \end{matrix}

Table 2: Some values of t(n, 2).

4 Application

Let P_{n} be an n-side regular polygon. We say that an inscribed quadrilateral in P_{n} is proper if none of its sides belongs to P_{n}.

Theorem 8. Let n ≥ 9 be an odd integer and let ♢(n) be the number of inscribed, nonisometric and proper quadrilaterals in P_{n}, using three equal chords. Then we have

♢(n) = \begin{cases}\frac{n − 5}{4} + ⌊\frac{n − 5}{12}⌋ , & if n ≡ 1 (mod 4); \\ \frac{n − 7}{4} + ⌊\frac{n + 1}{12}⌋ , & if n ≡ 3 (mod 4). \end{cases}

Proof. The chords belonging to an inscribed quadrilateral in P_{n}, separate the number of vertices of P_{n} into four parts, which do not include the quadrilateral vertices. In other words, each such quadrilateral generates a partition of n − 4 into four parts, using only two types of parts and vice versa. Then

♢(n) = t(n − 4, 4, 2),

and the result yields from Corollary 2.

Figure 1, illustrates this idea in P_{19}. The first quadrilateral is generated by the partition 15 = 1 + 1 + 1 + 12, the second by 15 = 2 + 2 + 2 + 9, the third by 15 = 3 + 3 + 3 + 6 and the fourth by 15 = 3 + 4 + 4 + 4.

Figure 1: The non-isometric proper quadrilaterals inscribed in P_{19}, using three equal chords.

5 Acknowledgement

The authors are indebted to the unknown referee for his valuable corrections and comments which have improved the quality of the paper. The authors are also grateful to Professors L. Benaissa and N. Hannoun for their help.

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2010 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: integer partitions, partitions into parts of different sizes, partitions into parts of two sizes, number of divisors.

(Concerned with sequences \underline{A002133}, \underline{A117955}, and \underline{A117956}.)

Received May 4 2010; revised version received October 3 2010; January 31 2011; February 28 2011. Published in Journal of Integer Sequences, March 25 2011.

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