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:
 * $\map \tau m > \map \tau n$

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

We have that:
 * $\map \tau {2 n} > \map \tau n$

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:
 * $\map \tau {3 n / 2} \le \map \tau n$

and:
 * $\map \tau {4 n / 3} \le \map \tau n$

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

Thus it has been established that these are the only possible values of $n$ greater than $2520$ which may fit the criteria for $n$ and $2 n$ to be consecutive highly composite numbers.

But the first of these is $2^3 \times 3^2 \times 5 \times 7 = 2520$ which fails through not being greater than $2520$.

Thus we consider:
 * $n = 2^3 \times 3^2 \times 5 \times 7 \times 11 = 27 \, 720$

We have that:

It is now necessary to show that there are no highly composite numbers between $27 \, 720$ and $2 \times 27 \, 720 = 55 \, 440$.

That is, all numbers $n$ such that $27 \, 720 < n < 55 \, 440$ are to be shown to have $\map \tau n < 96$.

But:

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.