On the enumeration of partitions with summands in arithmetic progression

Transcription

1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 8 (003), Pages On the enumeration of partitions with summans in arithmetic progression M. A. Nyblom C. Evans Department of Mathematics an Statistics RMIT University, GPO Box 476V Melbourne, Victoria 3001 Australia Abstract Enumerating formulae are constructe which count the number of partitions of a positive integer into positive summans in arithmetic progression with common ifference D. These enumerating formulae (enote p D (n)) which are given in terms of elementary ivisor functions together with auxiliary arithmetic functions (to be efine) are then use to establish a known characterisation for an integer to possess a partition of the form in question. 1 Introuction In recent times there has been some interest in the problem of representing a positive integer as the sum of at least two consecutive terms of an arithmetic progression of positive integers with a prescribe common ifference. It is known ([], [3, p. 85], [4]) that the number n can be expresse as a sum of consecutive positive integers provie it is not a power of an that the number of such representations is one less than the number of o ivisors of n. A more general result in this irection has been foun ([1]) which gives a necessary an sufficient conition for a positive integer to possess a partition with summans in arithmetic progression. If n = h s with s o, an n>1, then n is the sum of positive integers in arithmetic progression with common ifference D if an only if (1) when D is o, n isnotapowerofaneithers>d( h+1 1) or n> Dp(p 1) where p is the smallest o prime factor of n; 1 () when D is even, either n is even an n>dor n is o an n> 1 Dp(p 1) where again p is the smallest o prime factor of n.

2 150 M.A. NYBLOM AND C. EVANS In this paper we will show how the above characterisation can, for D>, be erive as a corollary of two new formulae which count the number of partitions of the esire type an which epen on the parity of D. These enumerating functions, enote p D (n), like those of Jacobi for representations of a number as the sum of two, four, six or eight squares, are given in terms of elementary ivisor functions, but together with auxiliary arithmetic functions, f(n) ang(n), which are efine later. Although these latter functions o not possess a close form expression for general n, we are able to fin specific conitions uner which f(n), g(n) assume the value 0, thereby allowing close form expressions for p D (n) in those instances. Before eriving these enumerating functions in 3 we will, for completeness, etermine in a close form expression for p (n). Inee, we shall show that p (n) = 1 ( ) (n) + ( 1)(n)+1 +1, (1) where (n) is the number of ivisors of n. In aition, as a consequence of (1), we shall erive an enumerating function for the number of representations of n as a ifference of two squares. Partition formula for D = In what follows i (n) enotes the number of ivisors of n with i(mo ), that is, 0 (n) an 1 (n) are the number of even an o ivisors of n respectively, an (n) = 0 (n)+ 1 (n)isthetotalnumberofivisorsofn. In aition, let N enote the set of non-negative integers. We procee now to establish a close form expression for p (n) via the use of generating functions. Theorem.1 For any integer n>1, the number of partitions of n with summans in arithmetic progression having common ifference is given by p (n) = 1 ( ) (n) + ( 1)(n) () Proof: Recall that a +(a +)+ +(a +(n 1)) = n(n + a 1) an for the partitions in question a, n N with a 1ann. Thus we see that the generating function of p (n) isgivenby f(q) = p (n)q n q n = 1 q = n n= n= n=0 k=0 q n(n+k). It follows that p (N) is the number of representations of N = n(n + k) withn an k 0. As n ann + k n our task is reuce to etermining the number

3 ENUMERATION OF PARTITIONS 151 of ivisors of N such that 1,N an N. If N is not a square then (N) is even. Excluing the ivisors 1, N we see after grouping the remaining (N) ivisors into pairs of the form (, N ) that there are precisely ((N) )/ ivisors that satisfy the above conition. On the other han if N is a square then (N) is o. After excluing the ivisors 1, N, N an pairing, we see that there are ((N) 3)/ ivisors with < N, an, incluing N,thereare((N) 1)/ ivisors with 1,N an N.Soineithercase, p D (N) = 1 ((N) +1 +( 1)(N)+1 ). Corollary.1 An integer n>1 is representable as a sum of positive integers in arithmetic progression with common ifference if an only if n is not prime. Proof: For prime p, (p) =,sop (p) =0. Conversely,ifp (n) =0then (n)+ ( 1)(n)+1 +1 =. However if n>1, (n), so the only solution to the above equation is (n) =, an n is prime. We now examine an unexpecte consequence of Theorem.1. Corollary. The number s(n) of representations of an integer n>1, asaifference of squares of two non-negative integers is given by s(n) = 1 ( ) 0 (n)+( 1) n+1 1 (n)+ ( 1)(n) (3) Proof: We begin by making the simple observation that the partitions of n counte by p (n) havesummansthatareeitherallooralleven. Ifweenotebyφ(n), σ(n) the number of partitions with consecutive even an o summans respectively we have p (n) =φ(n)+σ(n). Now for n> an even, there are p 1 ( n)= 1( n) 1= 1(n) 1 partitions of n of the form n = p r=m r with p>m. Consequently there are 1(n) 1 partitions of n of the form n = p r=m r, ansoφ(n) = 1(n) 1. Of course, when n is o, φ(n) =0so φ(n) = ( 1)n +1 ( 1 (n) 1). Thus from the ecomposition of p (n) above an () we fin σ(n) = 1 ( ) (n) + ( 1)(n)+1 +1 ( 1)n +1 ( 1 (n) 1) = 1 ( 0 (n)+( 1) n+1 1 (n)+ ( 1)(n)+1 +1 ) + ( 1)n 1, (4)

4 15 M.A. NYBLOM AND C. EVANS wherewehavemaeuseofthefactthat(n) = 0 (n)+ 1 (n). Recalling that n is equal to the sum of the first n consecutive o integers, it is clear that each partition counte by σ(n) correspons to a unique representation of n in the form x y with x, y N. Since by efinition each partition counte by σ(n) contains at least two summans, we have x y>1. However, when n =r +1forsomer N, oneof the representations counte by s(n) isn =(r +1) r,ansos(n) =σ(n) +1. On the other han, if n =r then n is not the ifference of consecutive squares an s(n) =σ(n). Thus we may set s(n) =σ(n)+ ( 1)n This together with (4) yiels (3). Finally, observe that (3) also hols for n =. Remark.1 Clearly for any positive integer n, s(n ) 1 gives the number of Pythagorean trias with n as a sie. 3 Partition formulae for D> So far we have manage to prouce a close form expression for p (n) intermsof thenumberofivisors(n), while it is well-known that p 1 (n) = 1 (n) 1. In this section we shall erive two further formulae for p D (n) base on the parity of D. We shall establish these enumerating formulae via purely combinatorial arguments. In what follows we nee only consier integers n D +, since clearly n =1+(1+D) is the smallest number with a partition of the esire form. We begin with case D o. Theorem 3.1 Suppose D>1 Nis o with n D +. Then the number of partitions of n into positive integers in arithmetic progression with common ifference D is given by { 1 (n) f(n) if n = D m(m+1) for some m>1 p D (n) = 1 (n) 1 f(n) otherwise where f(n) = A n with A n = { n : o, <D(n ), n <D( 1)}. Proof: The argument will be split into two main steps. In the first step, we emonstrate that the number of ways of expressing n as a finite sum of integers, some possibly negative, in arithmetic progression with the require common ifference, is 1 (n). In the secon step, we show how to count those arithmetic progressions with positive terms only, which will lea to the construction of the esire enumerating functions.

5 ENUMERATION OF PARTITIONS 153 Step 1: Suppose that n is representable as a sum of integers in arithmetic progression with common ifference D, n = a +(a + D)+(a +D)+ +(a + rd), forsomepair(a, r) Z Z. Then clearly we have For the given n an D consier the set n =(r + 1)(a + Dr). (5) S D (n) ={(a, r) Z Z :n =(r + 1)(a + Dr)}, which we now show contains exactly 1 (n) istinct elements. By recalling that D is o, observe from the equality (r +1)+(a 1+(D 1)r) =a + Dr, that the terms r+1 an a+dr are of opposite parity. Thus to solve the Diophantine equation in (5) it suffices to consier the system of simultaneous equations r +1 = x a + Dr = y where (x, y) =(, n )or(n,) for a positive o ivisor of n. If we enote the solutions (a, r) arising from these right han sies by (a 1 (),r 1 ()) an (a (),r ()) respectively, we fin that ( ) 1 (a 1 (),r 1 ()) = (n D( 1)), 1 an ( 1 (a (),r ()) = ( D(n 1)), n ) 1. As n an both an D are o, a simple parity check establishes that both solutions are orere pairs of integers. Thus the set of integer solutions (a, r) to(5) can be recast in the form S D (n) = I, o, n where I = {(a 1 (),r 1 ()), (a (),r ())}. To show that there is no repetition (or uplication) of any orere pairs, it will suffice to emonstrate that the secon components of all orere pairs in S D (n) areistinct. Nowasr 1 an r are clearly of opposite parity we have r 1 () r ( ) for any two o, possibly equal, ivisors, of n. Moreover, r i () =r i ( )fori =1, if an only if =. Consequently I I is empty when an so S D (n) is a finite union of mutually isjoint

6 154 M.A. NYBLOM AND C. EVANS sets, each containing two ifferent elements. Thus S D (n) contains 1 (n) istinct elements, which is the number of integer arithmetic progressions, as require. Step : Clearly the partitions we seek correspon to those arithmetic progressions of n in Step 1 which consist of at least two terms, all of which are strictly positive. Consequently we wish to count those orere pairs (a, r) S D (n) wherea 1anr 1. With this is min it is convenient to consier the following two cases separately. Case 1: n D m(m+1) for all m>1. In this instance, no orere pair (a, r) S D (n)hasa = 0, since otherwise as n D+ we woul have n = r i=1 id = D r(r+1) for some r>1. Now to etermine the number of orere pairs (a, r) S D (n) witha 1anr 1, we examine the elements in I for every o ivisor of n. Clearly I 1 contributes no such orere pairs as r 1 (1) = 0, while a (1) = 1 D(n 1) < 0. In the remaining solution set S D (n)\i 1,observe that since 3, r 1 () = 1 anr () = n 1 1as n 1. Thus we nee only concentrate on fining those orere pairs (a, r) S D (n)\i 1 with a>0. To this en, consier the sum (a 1 ()+a ()) = ( ) n (1 D) + +D (1 D)5 + D = 5 3D, noting here that the inequality hols since n 1an 3. Now, 5 3D 4 as D 3ansoa 1 () +a () < 0. Consequently, in each set I for 3, a 1 () an a () are not both positive. That is, a 1 () ana () are both negative or are of opposite sign. Thus if we extract from S D (n)\i 1 those sets I with both a 1 () an a () negative, exactly half the remaining orere pairs (a, r) havea>0. By efinition, A n is the set of o ivisors of n for which both a 1 () < 0ana () < 0 an so after extracting the f(n) orere pairs (a, r) witha<0from S D (n)\i 1 (noting here that 1 A n ) we fin p D (n) = 1 ( 1(n) f(n)) = 1 (n) 1 f(n). Case : n = D m(m+1) for some m>1. In this case, one representation of n is n =0+D + + md an so there exists an o ivisor > 1ofn such that either a 1 ( )=0ora ( ) = 0 (noting here that > 1 since again I 1 contributes no partition of the require form). Furthermore we have n = D + +(D +(m 1)D), that is, (D, m 1) S D (N)\I 1 an this orere pair rather than (0,m)canbe consiere as corresponing to one of the require partitions of n. Moreover as a 1 ( )+a ( ) < 0 we see that the remaining orere pairs (a, r) I have a<0, an so (D, m 1) I,sinceD>0. Consequently the number of esire partitions

7 ENUMERATION OF PARTITIONS 155 of n is equal to the number of orere pairs (a, r) S D (n)\(i 1 I )witha>0. Thus as in Case 1, after extracting from this set the f(n) orere pairs (a, r) witha<0, precisely half the remaining orere pairs have a>0 (noting here that 1, A n ). Hence as require. p D (n) = 1 ( 1(n) 4 f(n)) = 1 (n) f(n), Using the above formulation for p D (n), we can now establish the characterisation, prove in [1], for a number to be representable as a sum of positive integers in arithmetic progression with o common ifference D>1. Corollary 3.1 Anumbern = r s D + with s o is a sum of positive integers in arithmetic progression with o common ifference D>1 if an only if n is not a power of an either s>d( r+1 1) or n> 1 Dp(p 1) where p is the smallest o prime factor of n. Proof: Suppose n satisfies the above conition. It suffices to show that p D (n) 1 when n D m(m+1), since if n = D m(m+1) for some m>1thenn = D+D+ +md an p D (n) 1. We note first that 1 A n as n>0anso0 f(n) 1 (n) 1, since A n has at most 1 (n) 1 elements. Now if s>d( r+1 1) it is clear that the inequality <D(n ) fails for = s while if n> 1 Dp(p 1) it is clear that the inequality n <D( 1) fails for = p (noting that s, p > 1). So A n fails to contain another o ivisor of n. Thus A n has at most 1 (n) elements. Hence the function f(n) oes not attain its maximum value, 1 (n) 1, an so p D (n) 1. Establishing the converse is equivalent to showing that if n isapoweroforif both s D( r+1 1) an n 1Dp(p 1) then p D(n) = 0. Now if n = r then the only o ivisor of n is 1, an as 1 A n, clearly A n is empty an p D (n) =1 1 0=0. Now suppose n isnotapowerof. Ifn D m(m+1) then for any o ivisor >1 of n we have n< 1Dp(p 1) 1 D( 1) (noting here that the strict inequality hols since n D p(p 1) ). Furthermore, s<d( r+1 1), since if s = D( r+1 1) then (a (s),r (s)) = (0, 1) an so n<d+, a contraiction. Consequently, for any o ivisor >1ofn we have s<d( r+1 1) D( n 1) as n r. That is, <D(n ). Thus there are 1 (n) 1 o ivisors of n containe in A n,an so f(n) = 1 (n) 1anp D (n) =0. Ifn = D m(m+1) then since n 1 Dp(p 1) we have m p 1. However, from the minimality of p we have m = p 1. So for any o ivisor >pof n we have n = 1Dp(p 1) < 1 D( 1). That is, precisely 1 (n) o ivisors of n satisfy the inequality n < D( 1). Moreover, since s<d( r+1 1) we see that all o ivisors >1ofn satisfy the inequality <D(n ). Thus in this case A n has exactly 1 (n) elementsan so f(n) = 1 (n) an again p D (n) =0.

8 156 M.A. NYBLOM AND C. EVANS Clearly for an arbitrary positive integer n it may not be easy to evaluate f(n). In the following corollary we provie an example where f(n) can be etermine explicitly, thereby giving a close form expression for p D (n). Corollary 3. If n = r s D + where s is o an r+1 >D(s 1) then { 1 (n) if n = D m(m+1) for some m>1 p D (n) = 1 (n) 1 otherwise. Proof: The assume inequality can be recast in the form of n >D(s 1). Then s for any o ivisor of n we have n n >D(s 1) D( 1). That is, s n >D( 1), so A n is empty an f(n) =0. By applying a somewhat analogous argument to that use in Theorem 3.1 we can now obtain a formulation for p D (n) inthecased even, which is given in terms of the number of ivisors of n an another auxiliary function. Theorem 3. Suppose D> Nis even an n D +. Then the number of partitions of n into positive integers in arithmetic progression with common ifference D is given by { 1 p D (n) = ((n) 4+ ( 1)(n)+1 +1 g(n)) if n = D m(m+1) for some m>1 1 ((n) + ( 1)(n)+1 +1 g(n)) otherwise where g(n) = B n with B n = { n : n, n <D( 1), <D(n )}. Proof: Once again we split the argument into two main steps. In the first step we emonstrate that the number of ways of expressing n as a finite sum of integers, some possibly negative, in arithmetic progression with the require common ifference, is (n). In the secon step we show how to count those arithmetic progressions with positive terms only, by examining the solution set of a Diophantine equation in two cases base on the parity of (n). Step 1: Suppose that n is representable as a sum of integers in arithmetic progression with common ifference D, n = a +(a + D)+(a +D)+ +(a + rd), for some pair (a, r) Z Z. DenotingS D (n) ={(a, r) Z Z :n =(r + 1)(a + Dr)} it is clear, since a + Dr is even, that to solve the Diophantine equation, it suffices to consier the system of simultaneous equations a + Dr = r +1 = n

9 ENUMERATION OF PARTITIONS 157 where is a positive ivisor of n. Denoting for each such the resulting solution (a, r) by (a(),r()), we have (a(),r()) = ( 1( D( n 1)), n 1). A simple parity check establishes that (a(),r()) is an orere pair of integers. Thus for every ivisor of n there is an integer pair (a, r) corresponing to the equation n =(r + 1)(a + Dr). Moreover, there are exactly (n) orere pairs, since the secon components are istinct for istinct ivisors. Consequently S D (n) has(n) istinct elements which correspon to the integer arithmetic progressions, as require. Step : As before, in orer to etermine p D (n) it suffices to count those orere pairs (a, r) S D (n) witha 1anr 1. To this en it is convenient to consier the following two cases. Case 1: (n) even In this case n is not a square an so n for every ivisor of n. Hence S D(n) can be recast in the form S D (n) = I, n, 1 < n where I = {(a(),r()), (a( n),r( n m(m+1) ))}. Now if n D then as no orere pair (a, r) S D (n) hasa = 0, it suffices to etermine the number of such pairs with a 1anr 1. Clearly I 1 contributes no such orere pairs as a(1) = 1 ( D(n 1)) < 0anr(n) = 0. In the remaining solution set S D(n)\I 1,observe that since, n > 1, r(),r( n ) 1. Thus we nee only concentrate on fining those orere pairs (a, r) S D (n)\i 1 with a>0. To this en it will be necessary to examine the sign of a()+a( n ). First observe from the arithmetic-geometric mean inequality that + n n D + 6, an since + n is a positive integer, + n 5. Consequently (a()+a( n )) = ( D)( + n )+D ( D)5 + D = 10 3D < 0. Thus in each set I with 1 << n, a() ana( n ) aren t both positive. That is, either a()ana( n ) are both negative or they are of opposite sign. If we extract from S D (n)\i 1 those sets I with both a() ana( n ) negative, exactly half the remaining orere pairs (a, r) havea>0. By efinition, B n is the set of ivisors of n with both a() < 0ana( n ) < 0, an so after extracting these g(n) orere pairs (a, r) with a<0 from S D (n)\i 1 (noting here that i B n ), we fin p D (n) = 1 ((n) g(n)). Suppose now n = D m(m+1) for some m>1. Then one of the representations of n is

10 158 M.A. NYBLOM AND C. EVANS of the form n =0+D + + md an so there is a ivisor 1 < < n such that either a( )=0ora( n ) = 0. Furthermore, we also have n = D + +(D +(m 1)D), that is, (D, m 1) S D (n)\i 1 an this orere pair rather than (0,m)canbe consiere to correspon to one of the require partitions of n. Moreoverasa( )+ a( n ) < 0 we see that the remaining (a, r) I have a<0anso(d, m 1) I since D > 0. Consequently the number of esire partitions of n equals the number of orere pairs (a, r) S D (n)\(i 1 I )witha>0. Thus after extracting from this set the g(n) orere pairs (a, r) witha<0, exactly half the remainer have a>0 (noting here that 1, B n ). Hence p D (n) = 1 ((n) 4 g(n)). Case : (n) o In this case n is a square an S D (n) is of the form S D (n) = I, n, 1 n wherewenotethati n = {(a( n),r( n))}. Now if n D m(m+1) then as above we nee only count those orere pairs (a, r) S D (n)\i 1 with a>0. If = n observe that a( )=a( n )= D( 1) < 0asD 4, so as I contains only one element, there are g(n) 1 orere pairs (a, r) S D (n)\i 1 with a<0. After extracting these orere pairs, exactly half the remainer have a>0 (noting here that 1 B n ). Hence p D (n) = 1 ((n) (g(n) 1)) = 1 ((n) 1 g(n)). However if n = D m(m+1) for some m>1 then again there is a ivisor 1 < < n such that either a( )=0ora( n ) = 0. Arguing as in Case 1, we euce that the number of esire partitions of n equals the number of orere pairs (a, r) S D (n)\(i 1 I ) witha>0. After extracting from this set the g(n) 1 orere pairs with a<0, exactly half the remaining have a>0 (noting here that 1, B n ). Hence Thus Theorem 3. is proven. p D (n) = 1 ((n) 4 (g(n) 1)) = 1 ((n) 3 g(n)). Using the above formulation for p D (n), we can now establish the characterisation, prove in [1], for a number to be representable as a sum of positive integers in arithmetic progression with an even common ifference D>1.

11 ENUMERATION OF PARTITIONS 159 Corollary 3.3 Anumbern D + is a sum of positive integers in arithmetic progression with even common ifference D> if an only if either n is even or n is o an n> 1 Dp(p 1) where p is the smallest o prime factor of n. Proof: Suppose n satisfies the above conition. It suffices to show that p D (n) 1 when n D m(m+1), since if n = D m(m+1) for some m>1, n = D +D + + md an p D (n) 1. Recall that the number of ivisors of n with 1 < n is (n) when (n) iseven,an (n) 1 when (n) is o. Consequently since 1 B n,asn>0, we euce that B n contains at most (n) elements when (n) iseven,atmost (n) 1 elements when (n) is o. If n is even, then the inequality n <D( 1) fails to hol for =sincen D +, while if n is o an n> 1 Dp(p 1) then the same inequality will fails for = p. SoB n fails to contain one of, p. Ineithercaseg(n) oesn t attain its maximum value an p D (n) 1. Conversely, assume p D (n) 1. If n is even then n>d,sinced + is the smallest value of n for which p D (n) 0. Ifn is o, suppose n< 1 Dp(p 1) (noting here that n 1 Dp(p 1) as n is o). If >1isaivisorofnthen is o an p. Consequently n< 1Dp(p 1) 1 D( 1) an the inequality n <D( 1) hols for every ivisor >1ofn. However provie n, as n is also a ivisor of n with n > 1 we fin, on substituting n for in the inequality n <D( 1) that <D(n ). Thus all ivisors 1 < n, must be containe in B n an the function g(n) attains its maximum value, an p D (n) = 0, a contraiction. Hence n> 1 Dp(p 1), as require. Acknowlegement. The authors are inebte to the anonymous referee for the many valuable comments an suggestions mae which helpe to improve the presentation of this paper. References [1] R. Cook an D. Sharp, Sums of arithmetic progressions, Fibonacci Quarterly 33 (1995), [] E. E. Guerin, Consecutive integer partitions, Ars Combinatoria 39 (1995), [3] W. J. LeVeque, Topics in Number Theory, Aison-Wesley, (1965). [4] M. A. Nyblom, On the representation of integers as a ifference of nonconsecutive triangular numbers, Fibonacci Quarterly 39 (001), (Receive 3 May 00; revise 7 Feb 003)