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arXiv 2103.09933math.COmath.NT
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由多重划分函数的正奇密度导出$p(n)$的正奇密度:一个无条件证明

Deducing the positive odd density of $p(n)$ from that of a multipartition function: An unconditional proof

Fabrizio Zanello

摘要:

著名的Parkin-Shanks猜想预言$p(n)$是密度为$1/2$的奇数。尽管在过去几十年里做了大量的工作,但是,即使证明这种密度是积极的,似乎也是遥不可及的。在2018年与Judge合作的一篇论文中,我们引入了一种不同的方法,并证明了一个“惊人”的事实,即如果对于任何$A \equiv \pm 1\ (\bmod 6)$,多重划分函数$p_A(n)$具有正奇密度,那么$p(n)$也是如此。类似地,任何具有$A\equiv 3\ (\bmod 6)$的$p_{A}(n)$的正奇密度将意味着$p_3(n)$的正奇密度。我们的猜想被证明是一个推论的早期猜想的同一文件。在这个简短的说明中,我们提供了一个无条件的证明。一个重要的工具将是陈的最近突破的一个特殊情况下,我们以前的猜想。

英文摘要:

A famous conjecture of Parkin-Shanks predicts that $p(n)$ is odd with density $1/2$. Despite the remarkable amount of work of the last several decades, however, even showing this density is positive seems out of reach. In a 2018 paper with Judge, we introduced a different approach and conjectured the "striking" fact that, if for any $A \equiv \pm 1\ (\bmod 6)$ the multipartition function $p_A(n)$ has positive odd density, then so does $p(n)$. Similarly, the positive odd density of any $p_{A}(n)$ with $A\equiv 3\ (\bmod 6)$ would imply that of $p_3(n)$. Our conjecture was shown to be a corollary of an earlier conjecture of the same paper. In this brief note, we provide an unconditional proof of it. An important tool will be Chen's recent breakthrough on a special case of our earlier conjecture.

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