Integer partitions into arithmetic progressions

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1 Rostock. Math. Kolloq. 64, (009) Subject Classification (AMS) 05A17, 11P81 Saek Bouroubi, Nesrine Benyahia Tani Integer partitions into arithmetic progressions ABSTRACT. Every number not in the form k can be partitione into two or more consecutive parts. Thomas E. Mason has shown that the number of ways in which a number n may be partitione into consecutive parts, incluing the case of a single term, is the number of o ivisors of n. This result is generalize by etermining the number of partitions of n into arithmetic progressions with a common ifference r, incluing the case of a single term. KEY WORDS. Integer partitions, Arithmetic progression, ivisors of an integer. 1 Introuction Let n be an integer. A partition of n is an integral solution of the system: { n = n n k, 1 n 1 n k. The positive integers n 1,..., n k are calle parts, an k is the length of the partition. In partition ientities, we are often intereste in the number of partitions that satisfy some conitions. For example, the number of partitions into o parts, the number of partitions into even parts, the number of partitions into even an o parts [3], the number of partitions into istinct parts, an so on. For more on integer partitions the reaer is hereby invite to see for instance [1], [], [4], [5], [6] an [8]. Thomas E. Mason [7] was intereste in the number of ways in which a number n may be partitione into consecutive parts, incluing the case of a single term, an he has shown that this number equals the number of o ivisors of n. Let n = α p α 1 1 p α p αr r, where the p s are istinct o primes an the α s are the powers for which they occur. Then the number of ways in which a number n may be partitione into consecutive parts, incluing the case of a single term, equals r i=1 (α i + 1), except in the case n = α where the number of ways is 1. In this paper we treat the case of partitions of n into arithmetic progressions.

2 1 S. Bouroubi, N. Benyahia Tani For 1 r n, if n i n i 1 = r, i =,..., k, we say that we have a partition into an arithmetic progression with a common ifference r. Our problem can be formulate in terms of partitions: For a given positive integer n, what is the number of partitions of n into an arithmetic progression with a common ifference r? Partitioning n into an arithmetic progression with a common ifference r means that there exist two positive integers l 1 an m 0, such that n = l + (l + r) + (l + r) + + (l + mr). (1) Let us enote such a partition briefly by π(l, m). From (1), Hence, n = (m + 1)l + The solution of equation () in m yiels m(m + 1) r. rm + (l + r)m + l n = 0. () m = (l + r) + (l r) + 8rn (3) r Since m is an integer, (l r) + 8rn is a perfect square, i.e., (l r) + 8rn = u. (4) Then, Putting, we have A = rn = (u (l r)) u (l r) an B = (u + (l r)) u + (l r), A + B = u, an (5) B A = l r. (6) By (3), (5) an (6) we get r + B A Consequently, r ivies A. Hence an m = A r (7) r n = A r B. (8) Now we iscuss the parity of r.

3 Integer partitions into arithmetic progressions Arithmetic progressions with an o common ifference r By (4) u is o if r is o. Then in view of (5), A an B must have ifferent parity. Case 1. If A is even then B is o an n = A r B. In this case we have r + rn where = B, is an o ivisor of n. an m = n 1, Since l 1, must verify (n )r ( ), i.e., r + (r ) + 8rn. Case. If A is o then A r is o as well an n = A r B. In this case we have r + n r an m = 1, where = A r is an o ivisor of n. Because l 1, must verify ( 1)r (n ), i.e., r + (r ) + 8rn. r For every such o ivisor of n there exists a partition into an arithmetic progression with an o common ifference r, an vice versa. A moment s reflection will show that an o ivisor of n cannot satisfy the inequalities (n )r ( ), ( 1)r (n ) simultaneously, otherwise ( 1)r ( ), a contraiction. 1. Arithmetic progressions with an even common ifference r By (4) u is even, if r is even. Then in view of (5), A an B must have the same parity. Since r ivies A, A an B are even. Hence from (8) n = A r B. From (7) we get r + an m = n 1, where = B is a ivisor of n.

4 14 S. Bouroubi, N. Benyahia Tani Since l 1, must verify (n )r ( 1), i.e., r + (r ) + 8rn. 4 Now we are able to formulate our results as follows: Theorem 1 Let n 3 be a positive integer an let 1 r n be an o integer. Then the number of partitions of n into an arithmetic progression with an o common ifference r, incluing the case of a single term, is the number of o ivisors of n, satisfying ( n ) r ( ) or ( 1)r (n ) an for every such o ivisor, the partition π(l, m) is given by: r + n r an m = 1 if ( 1)r (n ), r + rn an m = n 1 if (n )r ( ). Example 1 An example illustrating the above theorem is the following: Let n = 15 an r = 3. The o ivisors of 15 satisfying (n )r ( ) or ( 1)r (n ) are 1, 3 an 15, so 15 amits three partitions into an arithmetic progression of the common ifference 3, each one is associate with one of these ivisors an this can be shown as follows: = 1 satisfies ( 1)r (n ). We have r + n r Hence π(l, m) = 15. = 3 satisfies ( 1)r (n ). We have r + n r Hence π(l, m) = = 15 an m = 1 = 0. = an m = 1 =. = 15 satisfies ( n ) r ( ). We have r + rn Hence π(l, m) = 6+9. = 6 an m = n 1 = 1. Theorem Let n 3 be a positive integer an let 1 r n be an even integer. Then the number of partitions of n into an arithmetic progression with an even common ifference r, incluing the case of a single term, is the number of ivisors of n, satisfying (n )r ( 1) an for every such ivisor, the partition π(l, m) is given by: r + an m = n 1.

5 Integer partitions into arithmetic progressions 15 Example An example illustrating the above theorem is the following: Let n = 30 an r = 4. The ivisors of 30 satisfying (n )r ( 1) are 10, 15 an 30, so 30 amits three partitions into an arithmetic progression of the common ifference 4, each one is associate with one of these ivisors an this can be shown as follows: = 10 = 15 = 30 r + r + r + = 6 an m = n 1 =, hence π(l, m) = = 13 an m = n 1 = 1, hence π(l, m) = = 30 an m = n 1 = 0, hence π(l, m) = 30. Why oes Theorem 1 generalize Mason s theorem? Theorem 1 is an extension of Mason s theorem [7]. Inee, for r = 1 an for every o ivisor of n, one an only one of the two inequalities ( 1)r (n ), (n ) r ( ) necessarily hols. In fact, r = 1, = p + 1 an n = k, imply an Hence, if p k 1, we get otherwise ( 1) (n ) p k 1, n ( ) p k. 1 + k an m = 1, 1 + k an m = k 1. For example, n = 15 has four o ivisors: 1, 3, 5 an 15: = 1 p = 0 an k = 15, then 1 + k π(l, m) =15. = 3 p = 1 an k = 5, then 1 + k π(l, m) = = 5 p = an k = 3, then 1 + k π(l, m) = = 15 p = 7 an k = 1, then 1 + k π(l, m) =7+8. = 15 an m = 1 = 0, hence = 4 an m = 1 =, hence = 1 an m = 1 = 4, hence = 7 an m = k 1 = 1, hence

6 16 S. Bouroubi, N. Benyahia Tani Acknowlegement The authors woul like to thank the referee for his valuable corrections an comments which have improve the quality of the paper. References [1] Aews, G. E., an Eriksson, K. : Integer partitions. Cambrige University Press, Cambrige, 004 [] Charalambies, A. Charalambos : Enumerative Combinatorics. Chapman & Hall/ CRC, 00 [3] Guo, Y., an Zhang, X. : Partitions of a positive integer as the sum of even an o numbers. J. Univ. Electron. Sci. Technol. China 35, (006) [4] Pak, I. : Partition bijections, a survey. Ramanujan Journal 1, 5 75 (006) [5] Raemacher, H. : On the partition function p(n). Proc. Lonon, Math. Soc. 43, (1937) [6] Stanley, R. P. : Enumerative Combinatorics, Vol. 1. Wasworth 1986 [7] Mason, Thomas E. : On the Representation of an Integer as the Sum of Consecutive Integers. The American Mathematical Monthly, Vol. 19, No. 3., (Mar., 191) [8] Wilf, H. S. : Lectures on integer partitions. (unpublishe), available at wilf receive: June 3, 008 Authors: Saek Bouroubi USTHB, Faculty of Mathematics Department of Operational Research Laboratory LAID3 P.O. Box El-Alia Bab-Ezzouar, Algeria sbouroubi@usthb.z or: bouroubis@yahoo.fr Nesrine Benyahia Tani Algiers University, Faculty of Economics an Management Sciences, Department of Commercial Sciences, Laboratory LAID3 Ahme Wake Street, Dely Brahim, Algiers, Algeria benyahiatani@yahoo.fr