Ratio between Consecutive Highly Composite Numbers Greater than 2520 is Less than 2

Theorem
The ratio between $2$ consecutive highly composite numbers both greater than $2520$ is less than $2$.

Proof
$n$ and $m$ are consecutive highly composite numbers such that:
 * $2520 < n < m$
 * $m / n \ge 2$

By definition of highly composite:
 * $\tau \left({m}\right) > \tau \left({n}\right)$

and, by hypothesis, $m$ is the smallest such integer.

We have that:
 * $\tau \left({2 n}\right) > \tau \left({n}\right)$

so it follows that $m \le 2 n$, otherwise $m$ would not be the smallest such integer.

So from $m / n \ge 2$ and $m \le 2 n$, it follows that $m = 2 n$.

We have that $2520$ is a special highly composite number.

The prime decomposition of $2520$ s given by:
 * $2520 = 2^3 \times 3^2 \times 5 \times 7$

We have that $n$ is a highly composite number such that $n > 2520$.

As $2520$ is a special highly composite number, $2520$ is a divisor of $n$.

Thus $n$ can be expressed as:
 * $n = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r$

where:
 * $a \ge 3$
 * $b \ge 2$
 * $c \ge 1$
 * $d \ge 1$
 * $r$ is a possibly vacuous product of prime numbers strictly greater than $13$.

We have that $n$ and $m = 2 n$ are consecutive highly composite numbers.

Hence it follows that:
 * $\tau \left({3 n / 2}\right) \le \tau \left({n}\right)$

and:
 * $\tau \left({4 n / 3}\right) \le \tau \left({n}\right)$

otherwise $3 n / 2$ or $4 n / 3$ would be highly composite numbers between $n$ and $2 n$.

Then:

and:

This leads to:

It has already been established that:


 * $a \ge 3$
 * $b \ge 2$

so it is now possible to state:

Suppose:
 * $(1): \quad f \ge 1$

Then:

So $f = 0$ and so by Prime Decomposition of Highly Composite Number $r = 1$.

Thus:
 * $n = 2^3 \times 3^2 \times 5^c \times 7^d \times 11^e$

where $c = 1$ or $c = 2$.

Suppose $c = 2$.

Then:

So $c = 1$ and so by Prime Decomposition of Highly Composite Number:


 * $n = 2^3 \times 3^2 \times 5 \times 7$

or:
 * $n = 2^3 \times 3^2 \times 5 \times 7 \times 11$

But $2^3 \times 3^2 \times 5 \times 7 = 2520$.

As $n > 2520$ it follows that:
 * $n = 2^3 \times 3^2 \times 5 \times 7 \times 11 = 27 \, 720$

Then:

But then:

So $45 \, 360$ has a higher $\tau$ than $27 \, 720$ and so the next higher highly composite number than $27 \, 720$ is less than twice it.

The result follows by Proof by Contradiction.