DIVISIBILITY PROPERTIES OF THE 5-REGULAR AND 13-REGULAR PARTITION FUNCTIONS

Transcription

1 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 DIVISIBILITY PROPERTIES OF THE 5-REGULAR AND 13-REGULAR PARTITION FUNCTIONS Neil Calkin Department of Mathematical Sciences, Clemson University, Clemson, SC 9634 Nate Drake Department of Mathematical Sciences, Clemson University, Clemson, SC 9634 Kevin James Department of Mathematical Sciences, Clemson University, Clemson, SC 9634 Shirley Law Department of Mathematics, North Carolina State University, Raleigh, NC 7695 Philip Lee Department of Electrical & Computer Engineering, Illinois Institute of Technology, Chicago, IL David Penniston Department of Mathematics, University of Wisconsin Oshkosh, Oshkosh, WI Jeanne Radder Stafford, VA 554 Received: 5/19/08, Revisde: 11/9/08, Accepted: 11/0/08, Published: 1/3/08 Abstract The function b k (n) is defined as the number of partitions of n that contain no summand divisible by k. In this paper we study the -divisibility of b 5 (n) and the - and 3-divisibility of b 13 (n). In particular, we give exact criteria for the parity of b 5 (n) and b 13 (n). 1. Introduction A partition of a positive integer n is a non-increasing sequence of positive integers whose sum is n. In other words, n = λ 1 + λ + + λ t with λ 1 λ λ t 1. For instance, the partitions of 4 are

2 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 4, 3 + 1, +, , and We denote the number of partitions of n by p(n). So, as shown above, p(4) = 5. Note that p(n) = 0 if n is not a nonnegative integer, and we adopt the convention that p(0) = 1. The generating function for the partition function is then given by the infinite product p(n)q n = 1 (1 q n ) = 1 + q + q + 3q 3 + 5q 4 + 7q 5 +. Let k be a positive integer. We say that a partition is k-regular if none of its summands is divisible by k, and denote the number of k-regular partitions of n by b k (n). For example, b 3 (4) = 4 because the partition has a summand divisible by 3 and therefore is not 3-regular. Adopting the convention that b k (0) = 1, the generating function for the k-regular partition function is then b k (n)q n = k n 1 (1 q n ) = (1 q kn ) (1 q n ). (1) Note that b (n) equals the number of partitions of n into odd parts, which Euler proved is equal to the number of partitions of n into distinct parts. The partition function satisfies the famous Ramanujan congruences p(5n + 4) 0 (mod 5), p(7n + 5) 0 (mod 7), p(11n + 6) 0 (mod 11) for every n 0. Ono [7] proved that such congruences for p(n) exist modulo every prime 5, and Ahlgren [1] extended this to include every modulus coprime to 6. Given these facts, for a positive integer m it is natural to wonder for which values of n we have that p(n) is divisible by m, or simply how often p(n) is divisible by m. By the results cited above, lim inf X #{1 n X p(n) 0 (mod m)}/x > 0 for any m coprime to 6. The m = and m = 3 cases, meanwhile, have proven elusive. The state of knowledge for k-regular partition functions is better. For example, Gordon and Ono [4] have shown that if p is prime, p v k and p v k, then for any j 1 the arithmetic density of positive integers n such that b k (n) is divisible by p j is one. In certain cases one can find even more specific information. As an illustration we consider the parity of b (n). Noting that b (n)q n = (1 q n ) (1 q n ) (1 q n ) (1 q n ) (1 q n ) (mod )

3 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 3 by Euler s Pentagonal Number Theorem it follows that b (n)q n q l(3l+1)/ (mod ), l= and so b (n) is odd if and only if n = l(3l + 1)/ for some l Z. Thus, in contrast to the case of p(n) we have a complete answer for the -divisibility of b (n) (see [6] and [3] for analogous results for the k-divisibility of b k (n) for k {3, 5, 7, 11}). Now consider the m-divisibility of b k (n) when (m, k) = 1. In [] Ahlgren and Lovejoy prove that if p 5 is prime, then for any j 1 the arithmetic density of positive integers n such that b (n) 0 (mod p j ) is at least p+1 (they also prove that b p (n) satisfies Ramanujantype congruences modulo p j ). In [9] Penniston extended this to show that for distinct primes k and p with 3 k 3 and p 5, the arithmetic density of positive integers n for which b k (n) 0 (mod p j ) is at least p+1 if p k 1, and at least p 1 if p k 1 (in [11] and p p [1] Treneer has shown that divisibility and congruence results such as these hold for general k). The latter result indicates that a special role may be played by the prime divisors of k 1, and we consider this here. Upon numerically investigating the m-divisibility of b k (n) for small values of k and m not covered by the results above, the most striking and regular patterns we found occurred for k = 5, m = and for k = 13 and m {, 3}. Theorem 1. Let n be a nonnegative integer. Then b 5 (n) is odd if and only if n = l(3l + 1) for some l Z. That is, b 5 (n)q n q l(3l+1) (mod ). l= Remark. By Euler s Pentagonal Number Theorem, Theorem 1 is equivalent to b 5 (n)q n (1 q n ) 4 (mod ). () Theorem. Let n be a nonnegative integer. Then b 13 (n) is odd if and only if n = l(l + 1) or n = 13l(l + 1) + 3 for some nonnegative integer l. That is, b 13 (n)q n q l(l+1) + l=0 q 6l(l+1)+6 (mod ). Remark. Jacobi s triple product formula yields (1 q n ) 3 = ( 1) l (l + 1)q l(l+1)/, l=0 and hence Theorem is equivalent to b 13 (n)q n (1 q 4n ) 3 + q 6 (1 q 5n ) 3 (mod ). (3) l=0

4 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 4 Theorems 1 and yield infinitely many Ramanujan-type congruences modulo for b 5 (n) and b 13 (n) in even arithmetic progressions. It turns out that our proof of Theorem 1 yields two congruences for b 5 (n) in odd arithmetic progressions. Theorem 3. For every n 0, b 5 (0n + 5) 0 (mod ) and b 5 (0n + 13) 0 (mod ). Finally, we make the following conjecture regarding the 3-divisibility of b 13 (n). Conjecture 1. For any l, ( b 13 3 l n + 5 ) 3l (mod 3) for every n 0. It turns out (see Proposition below) that one can reduce the verification of each of the congruences in Conjecture 1 to a finite computation. We have verified the conjecture for each l 6 (one can easily check that the conjecture does not hold for l = 1).. Modular Forms We begin with some background on the theory of modular forms. Given a positive integer N, let {( ) } a b Γ 0 (N) := SL c d (Z) c 0 (mod N). ( ) a b Let H := {z C I(z) > 0} be the complex upper half plane, and for γ = c d SL (Z) and z H define γz := az+b. Throughout, we let q := cz+d eπiz. Suppose k is a positive integer, f : H C is holomorphic and χ is a Dirichlet character modulo N. Then f is said to be a modular form of weight k on Γ 0 (N) with character χ if f(γz) = χ(d)(cz + d) k f(z) (4) ( ) a b for all γ = Γ c d 0 (N) and f is holomorphic at the cusps of Γ 0 (N). The modular forms of weight k on Γ 0 (N) with character χ form a finite-dimensional complex vector space which we denote by M k (Γ 0 (N), χ) (we will omit χ from our notation when it is the trivial character). For instance, if we denote by θ(z) the classical theta function θ(z) := q n = 1 + q + q 4 + q 9 +, n=

5 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 5 then θ 4 (z) M (Γ 0 (4)) (see, for example, [5]). A theorem of Sturm [10] provides a method to test whether two modular forms are congruent modulo a prime. If f(z) = a(n)qn has integer coefficients and m is a positive integer, let ord m (f(z)) be the smallest n for which a(n) 0 (mod m) (if there is no such n, we define ord m (f(z)) := ). Theorem 4. (Sturm) Suppose p is prime and f(z), g(z) M k (Γ 0 (N), χ) Z[[q]]. If ord p (f(z) g(z)) > k 1 [SL (Z) : Γ 0 (N)], then f(z) g(z) (mod p), i.e., ord p (f(z) g(z)) =. We note here that [SL (Z) : Γ 0 (N)] = N ( ) l+1 l, where the product is over the prime divisors of N. Hecke operators play a crucial role in the proofs of our results. If f(z) = a(n)qn Z[[q]] and p is prime, then the action of the Hecke operator T p,k,χ on f(z) is defined by (f T p,k,χ )(z) := (a(pn) + χ(p)p k 1 a(n/p))q n (we follow the convention that a(x) = 0 if x / Z). Notice that if k > 1, then (f T p,k,χ )(z) a(pn)q n (mod p). (5) Moreover, if f(z) M k (Γ 0 (N), χ), then (f T p,k,χ )(z) M k (Γ 0 (N), χ). When k and χ are clear from context, we will write T p := T p,k,χ. The next proposition follows directly from (5) and the definition of T p,k,χ. Proposition 1. Suppose p is prime, g(z) Z[[q]], h(z) Z[[q p ]] and k > 1. Then (gh T p,k,χ )(z) (g T p,k,χ )(z) h(z/p) (mod p). We will construct modular forms using Dedekind s eta function, which is defined by η(z) := q 1 4 for z H. A function of the form (1 q n ) f(z) = η r δ (δz), (6) δ N where r δ Z and the product is over the positive divisors of N, is called an eta-quotient.

6 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 6 and From ([8], p. 18), if f(z) is the eta-quotient (6), k := 1 δr δ 0 (mod 4) ( ( 1) k s d δ N N δ N r δ δ 0 (mod 4), δ N r δ Z, then f(z) satisfies ) the transformation property (4) for all γ Γ 0 (N). Here χ is given by χ(d) :=, where s := δ N δr δ. Assuming that f satisfies these conditions, then since η(z) is analytic and does not vanish on H, we have that f(z) M k (Γ 0 (N), χ) if f(z) is holomorphic at the cusps of Γ 0 (N). By ([8], Theorem 1.65) we have that if c and d are positive integers with (c, d) = 1 and d N, then the order of vanishing of f(z) at the cusp c d is N 4d(d, N ) d δ N (d, δ) r δ. δ 3. Proof of the Main Results Proof of Theorem 1. We begin with the modular forms and f(z) := η5 (5z) η(z) g(z) := η 4 (z)η 4 (5z) = q = q + q + q 3 + 3q 4 + 5q 5 + (1 q n ) 4 (1 q 5n ) 4. (7) Define the character χ m by χ m (d) := ( m d ). Using the results on eta-quotients cited above we find that f(z) M (Γ 0 (5), χ 5 ) and g(z) M 4 (Γ 0 (5)). Next, recall that Notice that (θ 4 (z)) M 4 (Γ 0 (0)). From (1) we have θ 4 (z) = 1 + 8q + 4q + 3q 3 + M (Γ 0 (4)). f(z) = η(5z) η(z) η4 (5z) = q5/4 (1 q5n ) q 1/4 (1 qn ) q0/4 (1 q 5j ) 4 (8) b 5 (n)q n+1 (1 q 0j ) (mod ). (9)

7 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 7 It follows from Proposition 1 that (f T )(z) b 5 (n + 1)q n+1 (1 q 10j ) (mod ), (10) and hence h(z) := f(z) (f T )(z) b 5 (n)q n+1 (1 q 0j ) (mod ). (11) Note that f(z) and (f T )(z) are in M (Γ 0 (10), χ 5 ), and hence h(z) lies in this space as well. It follows that h (z)θ 8 (z) M 8 (Γ 0 (0)). Now, g (z) M 8 (Γ 0 (0)), and one can check that the forms h (z)θ 8 (z) and g (z) are congruent modulo out to their q 4 terms. By Sturm s theorem we conclude that these forms are congruent modulo. Since θ(z) 1 (mod ), we have that h (z) g (z) (mod ), and hence h(z) g(z) (mod ). Then by (11) and (7), b 5 (n)q n (1 q 0j ) Since (1 q 5n ) 4 1 q 0n (mod ), () now follows from (1). (1 q n ) 4 (1 q 5n ) 4 (mod ). (1) Proof of Theorem. To begin, we define u(z) := η13 (13z) η(z) M 6 (Γ 0 (13), χ 13 ). We will also use the following two forms in M 1 (Γ 0 (13)): and v(z) := η 1 (z)η 1 (13z) = q 7 (1 q n ) 1 (1 q 13n ) 1 (13) w(z) := η 4 (13z) = q 13 (1 q 13n ) 4. (14) From (1) we have that u(z) b 13 (n)q n+7 (1 q 5j ) 3 (mod ). Then (u T )(z) b 13 (n + 1)q n+4 (1 q 6j ) 3 (mod ),

8 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 8 and hence m(z) := u(z) (u T )(z) b 13 (n)q n+7 (1 q 5j ) 3 (mod ). (15) Note that since u(z) and (u T )(z) lie in M 6 (Γ 0 (6), χ 13 ), so does m(z). Then since θ 4 (z) M 1 (Γ 0 (5)), we have that m (z)θ 4 (z) M 4 (Γ 0 (5)). Note that v (z), w (z) M 4 (Γ 0 (5)) as well, and one can check that the forms m (z)θ 4 (z) and v (z) + w (z) are congruent modulo out to their q 168 terms. By Sturm s theorem we conclude that m (z)θ 4 (z) v (z) + w (z) (mod ), and therefore m(z)θ 1 (z) v(z) + w(z) (mod ). Since θ(z) 1 (mod ), we find that m(z) v(z) + w(z) (mod ). Then (15), (13) and (14) give b 13 (n)q n+7 (1 q 13j ) 1 q 7 (1 q n ) 1 (1 q 13n ) 1 which implies (3). +q 13 (1 q 13n ) 4 (mod ), Proof of Theorem 3. We prove only the first congruence, as the second can be proved in a similar way. Sturm s theorem gives that f(z) and (f T )(z) are congruent modulo, which by (10) yields b 5 (n + 1)q n+1 (1 q 10j ) q (1 q 5n ) 5 (1 q n ) (mod ). Then b 5 (n + 1)q n+1 (1 q 10j ) q (1 q 5n ) (1 q n ) (1 q 5j ) 4 (mod ), and hence b 5 (n + 1)q n b 5 (l)q l (1 q 10j ) (mod ). (16) l=0 Note that n + 1 has the form 0m + 5 if and only if n (mod 10). Since the infinite product on the right hand side of (16) only produces powers of q that are 0 modulo 10, it suffices to show that b 5 (10n + ) 0 (mod ) (17) for all n 0. One can easily check that the congruence 6l + l (mod 10) has no solution, and so (17) follows from Theorem 1.

9 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 9 With regard to Conjecture 1, we have the following elementary proposition. Proposition. Let l. If the congruence ( b 13 3 l n + 5 ) 3l (mod 3) holds for all 0 n 7 3 l 1 3, then it holds for all n 0. Proof. The idea of our proof is to repeatedly apply the T 3 operator to the modular form P l (z) := η(13z) η(z) η e (13z), where e := 4 3 l. By the criteria for eta-quotients cited above, P l (z) M e (Γ 0(13), χ 13 ). Then For each 1 t l let Note that P l (z) P l (z) = δ t := 13 3t b 13 (n)q n+δl (1 q 13j ) e. b 13 (n)q n+δl (1 q 3l 13j ) 4 (mod 3). Using Proposition 1 and the fact that δ t (mod 3) for t l, an easy induction argument gives that (P l T3 s )(z) is congruent modulo 3 to ( )) 3 b 13 (3 s s 1 n + q n+δ l s (1 q 3l s 13j ) 4 for any 1 s l 1. In particular, ( )) 3 (P l T3 l 1 )(z) b 13 (3 l 1 l 1 1 n + q n+7 Then ( 3 (P l T3)(z) l b 13 (3 l 1 l 1 1 (3n + ) + ( b 13 3 l n + 5 3l 1 1 ) q n+3 (1 q 39j ) 4 (mod 3). )) q (3n+)+7 3 (1 q 13j ) 4 (1 q 13j ) 4 (mod 3). Since (P l T3)(z) l M e (Γ 0(13), χ 13 ), by Sturm s theorem we have that if ord 3 ((P l T3)(z)) l > 7 3 l 1, then (P l T3)(z) l 0 (mod 3). Therefore, if the congruence ( b 13 3 l n + 5 ) 3l (mod 3) holds for all 0 n 7 3 l 1 3, then it holds for all n 0.

10 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (008), #A60 10 Acknowledgements All of the authors would like to thank the National Science Foundation for their support through grants DMS and DMS References [1] S. Ahlgren, Distribution of the partition function modulo composite integers M. Math. Ann. 318 (4) (000), [] S. Ahlgren and J. Lovejoy, The arithmetic of partitions into distinct parts, Mathematika 48 (001), [3] B. Dandurand and D. Penniston, l-divisibility of l-regular partition functions, Ramanujan J., in press. [4] B. Gordon and K. Ono, Divisibility of certain partition functions by powers of primes, Ramanujan J. 1 (1997), [5] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer-Verlag New York Inc., New York, New York, [6] J. Lovejoy and D. Penniston, 3-regular partitions and a modular K3 surface, Contemp. Math. 91 (001), [7] K. Ono, Distribution of the partition function modulo m, Ann. of Math. () 151(1) (000), [8] K. Ono, The Web of Modularity, American Math. Society, Providence, RI, pp , 004. [9] D. Penniston, Arithmetic of l-regular partition functions, Int. J. Number Theory 4() (008), [10] J. Sturm, On the congruence of modular forms, Springer Lecture Notes in Math 140 (1984), [11] S. Treneer, Congruences for the coefficients of weakly holomorphic modular forms, Proc. London Math. Soc. 93(3) (006), [1] S. Treneer, Quadratic twists and the coefficients of weakly holomorphic modular forms, J. Ramanujan Math. Soc., in press.