An elementary proof of a theorem of Hardy and Ramanujan

Abstract

Let Q(n) denote the number of integers $1 \le q \le n$ whose prime factorization $q= \prod ^{t}_{i=1}p^{a_i}_i$ satisfies $a_1\ge a_2\ge \cdots \ge a_t$. Hardy and Ramanujan proved that

$$\begin{aligned} \log Q(n) \sim \frac{2\pi }{\sqrt{3}} \sqrt{\frac{\log (n)}{\log \log (n)}}\;. \end{aligned}$$

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.

Appendix  Proofs of (7) and (9)

We first prove (7). By Lagrange’s remainder theorem, for every $r>1$ and t such that $0\le t\le r$ there is $r-t\le c\le r$ such that

$$\begin{aligned} f(r-t)&=f(r)- tf'(r)+t^2f''(c)/2\;. \end{aligned}$$

Noting that $f''(z)=\frac{3-\log ^2(z)}{4z^{3/2}\log ^{5/2}(z)}$ is negative for all $z>e^{\sqrt{3}}$, we obtain (7) for $r>e^{\sqrt{3}}$ and $t>0$ with $r-t>e^{\sqrt{3}}$.

We now prove (9). For every r and $t\le r$, we let

$$\begin{aligned} g_r(t){:}{=}f(r-t)-f(r)+tf'(r)-t^2f''(r)\;. \end{aligned}$$

Note that $g_r(0)=g_r'(0)=0$. Hence, proving that for all $r\ge 10^3$ and $0\le t\le r/10$, we have $g''_r(t)>0$ implies (9). Indeed, let $r\ge 10^3$ and let $t=\varepsilon r$ with $0\le \varepsilon \le 1/10$. We have

$$\begin{aligned} g''_r(\varepsilon r)&=\frac{3-\log ^2((1-\varepsilon )r)}{4(1-\varepsilon )^{3/2}r^{3/2}\log ^{5/2}((1-\varepsilon )r)} -\frac{3-\log ^2(r)}{2r^{3/2}\log ^{5/2}(r)}\\&\ge \frac{1}{4r^{3/2}}\left( \frac{1}{(1-\varepsilon )^{3/2}}\cdot \frac{3-\log ^{2}(9r/10)}{\log ^{5/2}(9r/10)}- 2\cdot \frac{3-\log ^{2}(r)}{\log ^{5/2}(r)}\right) \\&\ge \frac{1}{4r^{3/2}}\cdot {\frac{2(1-1/10)-(1-\varepsilon )^{-3/2}(1+1/10)}{\sqrt{\log (r)}}}>0 \end{aligned}$$

where the first inequality holds as $\frac{3-\log ^2(x)}{\log ^{5/2}(x)}$ is monotone increasing for $x>e^{\sqrt{15}}$ and as $(1-\varepsilon )r\ge 900\ge e^{\sqrt{15}}$, and the second inequality holds as $r>10^3$ which implies both $\frac{3-\log ^2(9r/10)}{\log ^{5/2}(9r/10)}\ge \frac{-(1+1/10)}{\sqrt{\log (r)}}$ and $\frac{3-\log ^2{r}}{\log ^{5/2}(r)}\le \frac{-(1-1/10)}{\sqrt{\log (r)}}$, and the last inequality holds as $\varepsilon \le 1/10$.