A proof of a partition conjecture of Bateman and Erdős

Transcription

1 proof of a partition conjecture of Bateman and Erdős Jason P. Bell Department of Mathematics University of California, San Diego La Jolla C, US jbell@math.ucsd.edu 1

2 Proposed Running Head: Bateman-Erdős conjecture ddress: Jason Bell Department of Mathematics University of California, San Diego La Jolla C, US 2

3 bstract Bateman and Erdős found necessary and sufficient conditions on a set for the k th differences of the partitions of n with parts in, (n), to eventually be positive; moreover, they showed that when these conditions occur p (k+1) tends to infinity. (n)/ (n) tends to zero as n Bateman and Erdős conjectured that the ratio p (k+1) (n)/ (n) = O(n 1/2 ). We prove this conjecture. Key words: symptotic Enumeration, partitions. 1 Introduction Let be a non-empty set of positive integers. Throughout this paper, p (n) will denote the number of partitions of n with parts from and (n) will denote the k th difference of p (n). That is to say (n) = [xn ](1 x) k a (1 x a ) 1. We shall say, as do Bateman and Erdős, that a subset of the natural numbers has property P k, if there are more than k elements in, and if any subset of k elements is removed from, the remaining elements have gcd one. Bateman and Erdős say that if k < 0, then any non-empty set of positive numbers has property P k. Bateman and Erdős [1] showed that (n) is eventually positive if and only if has property P k. t the end of their paper, Bateman and Erdős made a conjecture about the behavior of the k th differences of p (n). We now state this conjecture. 3

4 Conjecture 1 (Bateman-Erdős) If a set of positive integers has property P k, then p (k+1) (n)/ (n) = O(n 1/2 ). Bateman and Erdős prove (see Theorem 3 of [1]) the conjecture when is a finite set; in fact, when is a finite set with property P k they show that p (k+1) (n)/ (n) = O(1/n). We shall show their conjecture is true when is infinite. The best known conditions under which the Bateman-Erdős conjecture has been verified are due to Richmond [3], who, using the saddle point method, obtained asymptotics of certain partition functions that allowed him to prove the Bateman-Erdős conjecture for certain sets. Bateman and Erdős observed (see page 12 of [1]) that if the conjecture were true, then it would be the best possible result. To see this, take = {1, 2, 3,...}. By Rademacher s exact formula for p (n) (see [2]) it follows that for all k. p (k+1) (n)/ (n) = π(6n) 1/2 (1 + o(1)) Before we begin the proof we introduce some notation. Notation 1 Given a set of positive integers, we define π (x) to be the number of elements of that are less than or equal to x. 2 cknowledgments The author would like to thank Bruce Richmond for his careful reading of this paper. 4

5 3 Proofs We require the following straightforward lemma. Lemma 1 Suppose we have three power series (x) = i α i x i, B(x) = i β i x i, and C(x) = i γ i x i, where B(x) = (1 x d ) 1 (x) and C(x) = (1 x) 1 B(x). Suppose also α n, β n, γ n are eventually positive and that α n = O(γ n /n). Then β n = O(γ n n 1/2 ). Proof. Take N 0 such that α n, β n, γ n > 0 for all n N 0. Notice that γ n = i n β i and hence γ j γ k 0 whenever j k N 0 (3.1) Choose a C such that α n < Cγ n /n for all n greater than N 0. Suppose β n O(γ n n 1/2 ). Then there exists some m such that β m > (C +2)γ m m 1/2, m d m 1/2 1 > N 0, and m 1/2 ( m 1/2 + 1)/2(m d m 1/2 ) < 1. (3.2) Notice α m = β m β m d < Cγ m /m. Hence β m d > (C + 2)γ m m 1/2 Cγ m /m. Similarly, α m d = β m d β m 2d < Cγ m d /(m d). 5

6 Thus β m 2d > (C +2)γ m m 1/2 Cγ m /m Cγ m d /(m d). nd by induction we have that if rd < m N 0, r 1 β m rd (C + 2)γ m m 1/2 Cγ m jd /(m jd) j=0 r 1 (C + 2)γ m m 1/2 Cγ m /(m rd) (by (3.1)) j=0 (C + 2)γ m m 1/2 Crγ m /(m rd). (3.3) Notice if we take q to be the greatest integer less than or equal to m 1/2, then we have γ m = β m + β m β m qd + γ m qd 1 q i=0 β m id q ( ) (C + 2)γ m m 1/2 Ciγ m /(m id) i=0 q ( ) (C + 2)γ m m 1/2 Ciγ m /(m qd) i=0 (by (3.3)) = (C + 2)(q + 1)γ m m 1/2 q(q + 1)Cγ m /2(m qd) (C + 2)γ m Cγ m (by (3.2)) = 2γ m. This is a contradiction. Hence the lemma is true. We now prove a proposition that will allow us to prove the Bateman-Erdős conjecture. 6

7 Proposition 1 Suppose that is a set of positive integers having property P k. Then (n) = O(p(k 1) (n)n 1/2 ). Proof. When is finite this is proven in Theorem 3 of [1]. Hence it suffices to prove the proposition when is infinite. Bateman and Erdős showed (see Lemma 2 of [1]) that has property P k if and only if some finite subset of has property P k. Hence we may choose d, e such that d, e > 1 and 1 := {d} (3.4) and 2 := {d, e} (3.5) have property P k. Notice (1 x e )( j 0 1 (j)x j ) = (1 x e )(1 x) k = (1 x) k = j 0 a 2 (1 x a ) 1 2 (j)x j, a 1 (1 x a ) 1 and hence 2 (n) = 1 (n) 1 (n e), (3.6) where 1 (j) is understood to be zero when j < 0. Let H(x) = i 0 ( h i x i = log (1 x) k (1 x d ) (1 x a ) ). 1 (3.7) a 7

8 and let S(x) = j 0 s j x j := x(1 x e ) 1 H (x). (3.8) Notice xh (x) = kx/(1 x) dx d /(1 x d ) + a ax a /(1 x a ). If we consider just the positive contribution to the coefficients of xh (x), we obtain the following inequality. s n = = = j n j n mod b jh j j n {a :a j} a {a :a n} j n/a {a :a n} {a :a n} a n/a a n = nπ (n) n 2. (3.9) Similarly, if we consider only the negative contribution we find s n = j n j n mod b jh j 8

9 j n k d kn dn/d m n/d = (k + 1)n. (3.10) Combining (3.9) and (3.10), we see s n (k + 1)n 2. (3.11) From Eq. (3.7) we have n 0 Differentiating both sides of this equation we see i 1 i 1 (i)x i 1 = H (x) exp(h(x)) = H (x) i 0 1 (n)x n = exp(h(x)). (3.12) 1 (i)x i (using (3.12)) = S(x) i 0 ( 1 (i) 1 (i b))x i 1 (using (3.8)) = S(x) i 0 2 (i)x i 1 (by (3.6)). (3.13) Comparing the coefficients of x n 1 of the first and last line of (3.13) we find n 1 (n) = n i=0 9 s n i 2 (i). (3.14)

10 Recall that 2 has property P k, and hence there exists some N > 0 such that 2 (j) > 0 for all j N. Using this fact along with (3.9) and (3.11), we rewrite the right hand side of (3.14) as follows. n i=n s n i 2 (i) + s n i 2 (i) i<n = n i=n n i=n (n i) 2 2 (i) + N(k + 1)n 2 (max i<n p(k) 2 (i) ) (n i) 2 2 (i) + O(n 2 ). (3.15) Thus there exists some C > 0 such that n 1 (n) n i=n (n i) 2 2 (i) + Cn 2. (3.16) From Lemma 1 of [1], we know that there exist C 1, C 2 > 0 such that C 1 n 2 < [x n ](1 x) 1 (1 x d ) 1 (1 x e ) 1 < C 2 n 2. (3.17) for all n. Hence we can say that C 1 n 1 (n) n i=n i=n C 1 (n i) 2 2 (i) + C 1 Cn 2 n ) ([x n i ](1 x) 1 (1 x d ) 1 (1 x e ) 1 2 (i) + C 1 Cn 2 10

11 = n ) ([x n i ](1 x) 1 (1 x d ) 1 (1 x e ) 1 i=0 i<n 2 (i) ([x n i ](1 x) 1 (1 x d ) 1 (1 x e ) 1 ) 2 (i) + C 1 Cn 2 n ) ([x n i ](1 x) 1 (1 x d ) 1 (1 x e ) 1 i=0 2 (i) + NC 2 n 2 (max i<n p(k) 2 (i) ) + C 1 Cn 2 = n ) ([x n i ](1 x) 1 (1 x d ) 1 (1 x e ) 1 i=0 2 (i) + O(n 2 ). (3.18) Recall that 2 (i) = [x i ](1 x) k (1 x d )(1 x e ) a (1 x a ) 1. Hence the last line of (3.18) can be replaced by [x n ](1 x) k 1 (1 x a ) 1 + O(n 2 ) = p (k 1) (n) + O(n 2 ). a Hence there exists a D > 0 such that C 1 n 1 (n) p (k 1) (n) + Dn 2. By Theorem 5.i. of [1] we have that n 2 = o(p (k 1) (n)), hence 1 (n) = O(p (k 1) (n)/n). Notice i 0 (i)xi = (1 x d ) 1 1 (i)x i. i 0 11

12 and i 0 p (k 1) (i)x i = (1 x) 1 (i)xi. i 0 pplying Lemma 1, taking α n, β n and γ n to be 1 (n), (n) and p(k 1) (n) respectively, we see that This proves the proposition. (n) = O(p(k 1) (n)n 1/2 ). We now complete the proof of the Bateman-Erdős conjecture. To do this, we will need the following simple lemma. Lemma 2 Suppose {f n } is a sequence of integers that is eventually positive and that there exists a positive integer c such that f n f m > 0 whenever n m > c. Moreover, suppose that f n 1 /f n 1. If then h n = O(f n ). h i x i := (1 x)(1 x d ) 1 ( f k x k ), i 0 k 0 Proof. Notice h n = f n jd f n 1 jd. 0 j n/d 0 j (n 1)/d Choose an integer r such that rd 1 > c. we will say that f k = 0 for all k < 0. Then we have h n = f n + f n d + + f n (r 1)d (f n 1 jd f n jd rd ) 12 0 j r+n/d

13 r+n/d<j (n 1)/d f n 1 jd f n + f n d + + f n (r 1)d + r max j rd 1 f j Similarly, = f n + f n d + + f n (r 1)d + O(1). (3.19) h n f n 1 f n 1 d f n 1 d(r 1) + O(1). (3.20) The fact that {f n } is a sequence of integers with the property that f n > f m whenever n m > 0 shows that f n. Combining this fact with (3.19) and (3.20) and the fact that f n 1 /f n 1, we see that r lim inf n Hence h n = O(f n ) as required. h n /f n lim sup h n /f n r. n We are finally ready to prove the conjecture of Bateman and Erdős. Theorem 1 Conjecture 1 is correct. Proof. Suppose has property P k. s stated in the introduction, we only need to prove the conjecture when is infinite. Thus we assume that is infinite. Choose d such that 1 := {d} has property P k. Since 1 has property P k, we have that 1 (n) is eventually positive. Moreover, by Proposition 1 we have that 1 (n) = O(p (k 1) 1 n 1/2 ). (3.21) 13

14 By Theorem 6 of [1], we have that there exists a c such that 1 (n) 1 (m) > 0 whenever n m > c. lso, 1 (n 1)/ 1 (n) 1 (by Theorem 5.ii. of [1]). Furthermore, we have that i 0 p (k+1) (i)x i = (1 x)(1 x d ) 1 ( 1 (j)x j ). j 0 Hence by an application of the preceding lemma, taking f n and h n to be 1 (n) and p (k+1) (n) respectively, we see that Next notice that (1 x d )(1 x) 1 i 0 p (k+1) (n) = O( 1 (n)). (3.22) (i)xi = p (k 1) 1 (j)x j. j 0 Therefore d 1 p (k 1) 1 (n) = (n j). j=0 By Theorem 5.ii. of [1], we know that for a fixed integer j, as n tends to infinity. Hence (n j)/p(k) (n) 1 d 1 p (k 1) 1 (n)/ (n) 1 = d. j=0 14

15 Combining this fact with (3.21) and (3.22), we see that p (k+1) (n) = O( 1 (n)) = O(p (k 1) 1 (n)n 1/2 ) = O( (n)n 1/2 ). This completes the proof of the Bateman-Erdős conjecture. /qed References [1] P. T. Bateman and P. Erdős, Monotonicity of partition functions, Mathematika 3 (1956), [2] H. Rademacher, On the expansion of the partition function in a series, nn. of Math. (2) 44 (1943), [3] L.B. Richmond, symptotic relations for partitions, Trans. mer. Math. Soc. 219 (1976),