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arXiv 1301.0948math.NT
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具有限制素数的最优本原集

Optimal primitive sets with restricted primes

William D. Banks, Greg Martin

摘要:

一个自然数集合是本原的,如果这个集合中没有一个元素能整除另一个元素。Erdberg证明了,如果S是任何本原集,则\sum_{n\in S} 1/(n log n)\le \sum_{n\in ¶} 1/(p log p),其中¶表示素数集。在本文中,我们取得了进展,这一猜想限制设置为较小的素数集。设P表示<$的任意子集,N(P)表示所有素因子都在P中的自然数集合。如果不等式\sum_{n\in S} 1/(n log n)\le \sum_{n\in P} 1/(p log p)对包含在N(P)中的每个本原集S成立,则称P是N(P)的本原子集中的Erdens-best。证明了如果P的元素倒数之和足够小,则P是N(P)的本原子集中的Erdens-best。作为应用,我们证明了大于3的孪生素数的集合在相应的本原集合中是Erdens-best的。这个问题原来是有关的一个类似的问题,涉及乘法权重。对于任意实数t>1,如果不等式\sum_{n\in S} n^{-t} \le \sum_{n\in P} p^{-t}对包含在N(P)中的每个本原集S成立,则称P是N(P)的本原子集中的t-最佳的。我们证明了,如果这个不等式右边的和足够小,那么P是N(P)的本原子集中的t-最好的。

英文摘要:

A set of natural numbers is primitive if no element of the set divides another. Erdős conjectured that if S is any primitive set, then \sum_{n\in S} 1/(n log n) \le \sum_{n\in ¶} 1/(p log p), where ¶denotes the set of primes. In this paper, we make progress towards this conjecture by restricting the setting to smaller sets of primes. Let P denote any subset of ¶, and let N(P) denote the set of natural numbers all of whose prime factors are in P. We say that P is Erdős-best among primitive subsets of N(P) if the inequality \sum_{n\in S} 1/(n log n) \le \sum_{n\in P} 1/(p log p) holds for every primitive set S contained in N(P). We show that if the sum of the reciprocals of the elements of P is small enough, then P is Erdős-best among primitive subsets of N(P). As an application, we prove that the set of twin primes exceeding 3 is Erdős-best among the corresponding primitive sets. This problem turns out to be related to a similar problem involving multiplicative weights. For any real number t>1, we say that P is t-best among primitive subsets of N(P) if the inequality \sum_{n\in S} n^{-t} \le \sum_{n\in P} p^{-t} holds for every primitive set S contained in N(P). We show that if the sum on the right-hand side of this inequality is small enough, then P is t-best among primitive subsets of N(P).

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