A proof of a partition conjecture of Bateman and Erd os 1 - PDF document

A proof of a partition conjecture of Bateman and Erd˝ os 1

Observation: If λ 1 ≥ · · · ≥ λ k ≥ 1 satisfies λ 1 + · · · + λ k = n, then we can obtain a partition of n + 1 by simply adding 1 ; that is, λ 1 + · · · + λ k + 1 = n + 1 is a partition of n + 1 . We therefore have p ( n + 1) − p ( n ) ≥ 0 for all n ≥ 0 . 2

TWO NATURAL QUESTIONS: 1. Are the k th differences of the partitions eventually pos- itive? 2. If so, then what happens if we impose restrictions upon the patitions? 3

Bateman and Erd˝ os (1956) answered these questions completely. Their motivation was to improve existing Tauberian theorems. 4

Given a subset A of { 1 , 2 , 3 , · · ·} , let p A ( n ) denote the number of partitions of n with parts from A ; i.e., p A ( n ) = [ x n ] (1 − x a ) − 1 . � a ∈ A Let p ( k ) A ( n ) denote the k th difference of p A ( n ) ; i.e., p ( k ) (1 − x a ) − 1 . A ( n ) = [ x n ](1 − x ) k � a ∈ A For example, p (1) A ( n ) = p A ( n ) − p A ( n − 1) , p (2) p (1) A ( n ) − p (1) A ( n ) = A ( n − 1) = p A ( n ) − 2 p A ( n − 1) + p A ( n − 2) . 5

DEFINITION: We say a set A ⊆ { 1 , 2 , 3 , . . . } has prop- erty P k if: 1. | A | > k ; and 2. gcd( A \ { a 1 , . . . , a k } ) = 1 for any a 1 , . . . , a k ∈ A . 6

Bateman and Erd˝ os showed the following remarkable fact: p ( k ) A ( n ) is eventually positive if and only if A has property P k . Moreover, they showed that if A has property P k , p ( k +1) ( n ) /p ( k ) A ( n ) → 0 as n → ∞ . A 7

When A = { 1 , 2 , 3 , . . . } , Rademacher’s formula for the number of partitions of n gives √ p ( k +1) ( n )/ p ( k ) A ( n ) ∼ π/ 6 n. A It seems therefore reasonable to expect the following conjec- ture. Conjecture. (Bateman-Erd˝ os) If A has property P k , p ( k +1) ( n ) /p ( k ) A ( n ) = O( n − 1 / 2 ) . A 8

The proof of this conjecture appears in the Journal of Number Theory 87 (2001) 144–153. 9

The first step in proving this conjecture is the following lemma. Lemma. Let ∞ f ( n ) x n � F ( x ) = n =0 ∞ g ( n ) x n = (1 − x ) − 1 F ( x ) , and � G ( x ) = n =0 ∞ h ( n ) x n = (1 − x ) − 2 F ( x ) � H ( x ) = n =0 be three power series; moreover, suppose these power series have nonnegative coefficients. Then ⇒ n 1 / 2 g ( n ) = O( h ( n )) . nf ( n ) = O( h ( n )) = 10

This lemma allows us to work in the ring of formal power series. WHY? ∞ xF ′ ( x ) = nf ( n ) x n . � n =0 Thus to prove g ( n ) = O( h ( n ) n − 1 / 2 ) it suffices to show that xF ′ ( x ) ≤ CH ( x ) + p ( x ) , for some constant C and some polynomial p ( x ) , where the inequality is taken coefficient-wise. 11

Let A be a subset of the positive integers. Let ∞ p A ( n ) x n = (1 − x a ) − 1 � � H ( x ) = n =0 a ∈ A ∞ p (1) A ( n ) x n , and � G ( x ) = n =0 ∞ p (2) A ( n ) x n . � F ( x ) = n =0 GOAL: To show p (1) A ( n ) /p A ( n ) = O( n − 1 / 2 ) when A has property P 0 ; i.e., we must show g ( n ) n 1 / 2 = O( h ( n )) when gcd( A ) = 1 . To do this, we use the lemma and compare the coefficients of xF ′ ( x ) to the coefficients of H ( x ) . 12

Recall F ( x ) = (1 − x ) 2 � (1 − x a ) − 1 . a ∈ A We have xF ′ ( x ) = F ( x )( − 2 x/ (1 − x )+ ax a / (1 − x a )) . � a ∈ A 13

ASIDE ∞ ax a / (1 − x a ) = σ ( n ) x n . � � a ≥ 1 n =1 Unfortunately, σ ( n ) is not very well-behaved. Its sequence of partial sums, however, is very well-behaved. σ (1) + σ (2) + · · · + σ ( n ) ∼ Cn 2 . We have ∞ (1 − x ) − 1 ( σ ( n ) x n ) � n =1 σ (1) x + ( σ (1) + σ (2)) x 2 = + ( σ (1) + σ (2) + σ (3)) x 3 + · · · . 14

We have ∞ (1 − x ) − 1 � ax a / (1 − x a ) ≤ Cn 2 x n , � n =1 a ∈ A for some C . Therefore xF ′ ( x ) ax a / (1 − x a )) � = F ( x )( − 2 x/ (1 − x ) + a ∈ A ∞ F ( x )(1 − x )( − 2 x/ (1 − x ) 2 + Cn 2 x n ) � ≤ n =1 ∞ ( Cn 2 − 2 n ) x n ) � = F ( x )(1 − x )( n =1 15

Now [ x n ]2 C/ (1 − x ) 3 = C ( n +2)( n +1) ≥ ( Cn 2 − 2 n ) ∞ ( Cn 2 − 2 n ) x n xF ′ ( x ) � ≤ F ( x )(1 − x )( n =1 F ( x )(1 − x )(2 C/ (1 − x ) 3 ) ≤ 2 CF ( x )(1 − x ) − 2 = = 2 CH ( x ) . Taking the coefficient of x n we see nf ( n ) = O( h ( n )) , or equivalently, np (2) A ( n ) = O( p A ( n )) p (1) A ( n ) = O( p A ( n ) n − 1 / 2 ) . = ⇒ 16