论文标题
Hardy和Ramanujan定理的基本证明
An Elementary Proof of a Theorem of Hardy and Ramanujan
论文作者
论文摘要
令$ q（n）$表示整数的数量$ 1 \ leq q \ leq n $，其主要分解$ q = \ prod^{t} _ {i = 1} p^{a_i} _i} _i $满足$ a_1 \ geq a_1 \ geq a_2 \ geq a_2 \ geq \ geq \ ldots \ geq a _________t $。 Hardy和Ramanujan证明了$$ \ log q（n）\ sim \ frac {2π} {\ sqrt {\ sqrt {3}} \ sqrt {\ frac {\ frac {\ log（n）} {\ log log \ log \ log \ log \ log \ log（n）}}}}}} \;。。。 $$在证明上述精确的渐近公式之前，他们详细研究了可以使用纯粹的基本方法获得有关$ q（n）$的内容，并且只能使用这种方法获得大量的Cruder下限和上限。 在本文中，我们表明，实际上有可能获得纯粹的基本（且短得多的）证明，这是hardy-ramanujan定理。为了实现这一目标，我们首先提出一个简单的组合论点，表明$ q（n）$满足（伪）复发关系。这使我们能够用简短的归纳论证代替原始证明的几乎所有硬分析部分。
Let $Q(n)$ denote the number of integers $1 \leq q \leq n$ whose prime factorization $q= \prod^{t}_{i=1}p^{a_i}_i$ satisfies $a_1\geq a_2\geq \ldots \geq a_t$. Hardy and Ramanujan proved that $$ \log Q(n) \sim \frac{2π}{\sqrt{3}} \sqrt{\frac{\log(n)}{\log\log(n)}}\;. $$ Before proving the above precise asymptotic formula, they studied in great detail what can be obtained concerning $Q(n)$ using purely elementary methods, and were only able to obtain much cruder lower and upper bounds using such methods. In this paper we show that it is in fact possible to obtain a purely elementary (and much shorter) proof of the Hardy--Ramanujan Theorem. Towards this goal, we first give a simple combinatorial argument, showing that $Q(n)$ satisfies a (pseudo) recurrence relation. This enables us to replace almost all the hard analytic part of the original proof with a short inductive argument.
