Ramanujan's Papers
On the number of divisors of a number
Journal of the Indian Mathematical Society, VII, 1915, 131 – 133
1. If $\delta$ be a divisor of $N$, then there is a conjugate divisor $\delta'$
such that $\delta \delta' = N$. Thus we see that
\begin{equation}
\mbox{the number of divisors from 1 to } \sqrt{N}
\mbox{ is equal to the number of divisors from } \sqrt{N} \mbox{ to } N.
\end{equation}
From this it evidently follows that
\begin{equation}
d(N) \lt 2 \sqrt{N},
\end{equation}
where $d(N)$ denotes the number of divisors of $N$ (including unity and the
number itself). This is only a trivial result, as all the numbers from 1 to
$\sqrt{N}$ cannot be divisors of $N$. So let us try to find the best possible
superior limit for $d(N)$ by using purely elementary reasoning.
2. First let us consider the case in which all the prime divisors of $N$ are
known. Let
$$N = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdots p_n^{a_n},$$
where $p_1, p_2, p_3 \ldots p_n$ are a given set of $n$ primes . Then it is
easy to see that
\begin{equation}
d(N) = (1 + a_1) (1+ a_2) (1+a_3) \cdots (1+a_n).
\end{equation}
But
\begin{eqnarray}
\frac{1}{n} &&\{(1+a_1) \log p_1 + (1+a_2) \log p_2 + \cdots + (1+a_n)
\log p_n \}\notag\\
&& > \{(1 + a_1) (1+a_2) \cdots (1+a_n) \log p_1 \log p_2 \cdots \log
p_n\}^{\frac{1}{n}},
\end{eqnarray}
since the arithmetic mean of unequal positive numbers is always greater than
their geometric mean. Hence
$$ \frac{1}{n} \{\log p_1 + \log p_2 + \cdots + \log p_n + \log N \} \gt \{\log
p_1 \log p_2 \cdots \log p_n \cdot d(N)\}^{\frac{1}{n}}.$$
In other words
\begin{equation}
d(N) \lt \frac{\left\{ \frac{1}{n} \log (p_1 p_2 p_3 \cdots p_n
N)\right\}^n}
{\log p_1 \log p_2 \log p_3 \cdots \log p_n},
\end{equation}
for all values of $N$ whose prime divisors are $p_1, p_2, p_3, \ldots ,
p_n$.
3. Next let us consider the case in which only the number of prime divisors of
$N$ is known. Let
$$ N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_n^{a_n},$$
where $n$ is a given number; and let
$$N' = 2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \cdots p^{a_n},$$
where $p$ is the natural $n$th prime. Then it is evident that
\begin{equation}
N' \leq N;
\end{equation}
and
\begin{equation}
d(N') = d(N).
\end{equation}
But
\begin{equation}
d(N') \lt \frac{\left\{\frac{1}{n} \log (2\cdot 3\cdot 5\cdots p\cdot
N')\right\}^n} {\log 2 \log 3 \log 5 \cdots \log p} ,
\end{equation}
by virtue of
(5). It follows from
(6) to
(8) that, if $p$ be the natural
$n$th prime, then
\begin{equation}
d(N) \lt \frac{\left\{\frac{1}{n} \log(2\cdot 3\cdot 5 \cdots p\cdot
N)\right\}^n}{\log 2 \log 3 \log 5 \cdots \log p},
\end{equation}
for all values of $N$ having $n$ prime divisors.
4. Finally, let us consider the case in which nothing is known about $N$. Any
integer $N$ can be written in the form
$$2^{a_2} \cdot 3^{a_3} \cdot 5^{a_5} \cdots , $$
where $a_\lambda \geq 0$. Now let
\begin{equation}
x^h = 2,
\end{equation}
where $h$ is any positive number. Then we have
\begin{equation}
\frac{d(N)}{N^h} = \frac{1+a_2}{2^{ha_2}} \cdot \frac{1+a_3}{3^{ha_3}}
\cdot \frac{1+a_5}{5^{ha_5}}
\end{equation}
But from
(10) we see that, if $q$ be any prime greater than $x$, then
\begin{equation}
\frac{1+a_q}{q^{ha_q}} \leq \frac{1+a_q}{x^{ha_q}} = \frac{1+a_q}{2^{a_q}}
\leq 1.
\end{equation}
It follows from
(11) and
(12) that, if $p$ be the largest prime not exceeding
$x$, then
\begin{align}
\frac{d(N)}{N^h} &\leq \frac{1+a_2}{2^{ha_2}} \cdot \frac{1+a_3}{3^{ha_3}}
\cdot \frac{1+a_5}{5^{ha_5}} \cdots \frac{1+a_p}{p^{ha_p}}\notag\\
&\leq \frac{1+a_2}{2^{ha_2}} \cdot \frac{1+a_3}{2^{ha_3}} \cdot
\frac{1+a_5}{2^{ha_5}} \cdots \frac{1+a_p}{2^{ha_p}}.
\end{align}
But it is easy to shew that the maximum value of $(1+ a) 2^{-ha}$ for the
variable $a$ is $\frac{2^h}{he \log 2}$. Hence
\begin{equation}
\frac{d(N)}{N^h} \leq \left(\frac{2^h}{he \log 2}\right)^{\omega(x)},
\end{equation}
where $\omega(x)$ denotes the number of primes not exceeding $x$. But from
(10) we
have
$$ h = \frac{\log 2}{\log x}.$$
Substituting this in
(14), we obtain
\begin{equation}
d(N) \leq N \frac{\log 2}{\log x} \left\{ \frac{2^{\frac{\log 2}{\log x}}
\log x}{e(\log 2)^2} \right\}^{\omega(x)} .
\end{equation}
But it is easy to verify that, if $x \geq 6.05,$ then
$$2^h \lt e (\log 2)^2.$$
From this and (15) it follows that, if $x \geq 6.05, $ then
\begin{equation}
d(N) \lt 2^{(\log N)/(\log x)} (\log x)^{\omega(x)}
\end{equation}
for all values of $N$, $\omega(x)$ being the number of primes not exceeding $x$.
5. The symbol ``$O$'' is used in the following sense:
$$ \phi (x) = O \{(\Psi (x)\}$$
means that there is a positive constant $K$ such that
$$ \left| \frac{\phi (x)}{\Psi (x)} \right| \leq K $$
for all sufficiently large values of $x$ (see Hardy, Orders of
Infinity, pp. 5 et seq.). For example:
$$5x =O(x) ; \frac{1}{2} x = O(x); x \sin x = O(x) ; \sqrt{x} = O(x); \log x =
O(x);$$
but $$x^2 \neq O(x) ; x \log x \neq O (x).$$
Hence it is obvious that
\begin{equation}
\omega(x) = O(x).
\end{equation}
Now, let us suppose that
$$x = \frac{\log N}{(\log \log N)^2} $$
in (16). Then we have
$$ \log x = \log \log N + O (\log \log \log N);$$
and so
\begin{equation}
\frac{\log N}{\log x} = \frac{\log N}{\log \log N} + O \left\{\frac{\log
N \log \log \log N}{(\log \log N)^2}\right\}.
\end{equation}
Again
\begin{equation}
\omega(x) \log \log x=O (x \log \log x) = O \left\{\frac{\log N \log \log \log
N}{(\log \log N)^2} \right\} .
\end{equation}
It follows from (16), (18) and (19), that
\begin{equation}
\log d (N) \lt \frac{\log 2 \log N}{\log \log N} + O \left\{\frac{\log N
\log \log \log N}{(\log \log N)^2} \right\}
\end{equation}
for all sufficiently large values of $N$.
