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

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Theorem

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


Proof

Aiming for a contradiction, suppose $n$ 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:

\(\ds \map \tau {3 n / 2}\) \(\le\) \(\ds \map \tau n\)
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a - 1} \times 3^{b + 1} \times 5^c \times 7^d \times 11^e \times 13^f \times r}\) \(\le\) \(\ds \map \tau {2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r}\)
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a - 1} \times 3^{b + 1} } \times \map \tau {5^c \times 7^d \times 11^e \times 13^f \times r}\) \(\le\) \(\ds \map \tau {2^a \times 3^b} \times \map \tau {5^c \times 7^d \times 11^e \times 13^f \times r}\) Divisor Count Function is Multiplicative
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a - 1} \times 3^{b + 1} }\) \(\le\) \(\ds \map \tau {2^a \times 3^b}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {\paren {a - 1} + 1} \paren {\paren {b + 1} + 1}\) \(\le\) \(\ds \paren {a + 1} \paren {b + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds a \paren {b + 2}\) \(\le\) \(\ds \paren {a + 1} \paren {b + 1}\) simplifying


and:

\(\ds \map \tau {4 n / 3}\) \(\le\) \(\ds \map \tau n\)
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a + 2} \times 3^{b - 1} \times 5^c \times 7^d \times 11^e \times 13^f \times r}\) \(\le\) \(\ds \map \tau {2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r}\)
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a + 2} \times 3^{b - 1} } \times \map \tau {5^c \times 7^d \times 11^e \times 13^f \times r}\) \(\le\) \(\ds \map \tau {2^a \times 3^b} \times \map \tau {5^c \times 7^d \times 11^e \times 13^f \times r}\) Divisor Count Function is Multiplicative
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^{a + 2} \times 3^{b - 1} }\) \(\le\) \(\ds \map \tau {2^a \times 3^b}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {\paren {a + 2} + 1} \paren {\paren {b - 1} + 1}\) \(\le\) \(\ds \paren {a + 1} \paren {b + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds \paren {a + 3} b\) \(\le\) \(\ds \paren {a + 1} \paren {b + 1}\) simplifying

This leads to:

\(\ds a\) \(\le\) \(\ds b + 1\)
\(\ds 2 b - 1\) \(\le\) \(\ds a\)
\(\ds \leadsto \ \ \) \(\ds 2 b - 1\) \(\le\) \(\ds b + 1\)
\(\ds \leadsto \ \ \) \(\ds b\) \(\le\) \(\ds 2\)
\(\ds \leadsto \ \ \) \(\ds a\) \(\le\) \(\ds 3\)


It has already been established that:

$a \ge 3$
$b \ge 2$
\(\ds a\) \(\ge\) \(\ds 3\)
\(\ds b\) \(\le\) \(\ds 2\)

so it is now possible to state:

\(\ds a\) \(=\) \(\ds 3\)
\(\ds b\) \(=\) \(\ds 2\)


Suppose:

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

Then:

\(\ds 2^5 \times 3^3 \times 5^c \times 7^d \times 11^e \times 13^{f - 1} \times r\) \(<\) \(\ds 2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r\) because $12 = 2^2 \times 3 < 13$
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 3^3 \times 5^c \times 7^d \times 11^e \times 13^{f - 1} \times r}\) \(<\) \(\ds \map \tau {2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r}\) as $2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r$ is highly composite
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 3^3 \times 13^{f - 1} } \times \map \tau {5^c \times 7^d \times 11^e \times r}\) \(<\) \(\ds \map \tau {2^3 \times 3^2 \times 13^f} \times \map \tau {5^c \times 7^d \times 11^e \times r}\) Divisor Count Function is Multiplicative
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 3^3 \times 13^{f - 1} }\) \(<\) \(\ds \map \tau {2^3 \times 3^2 \times 13^f}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {5 + 1} \paren {3 + 1} \paren {\paren {f - 1} + 1}\) \(<\) \(\ds \paren {3 + 1} \paren {2 + 1} \paren {f + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds 24 f\) \(<\) \(\ds 12 \paren {f + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds f\) \(<\) \(\ds 1\) which is a contradiction of $(1)$

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:

\(\ds 2^5 \times 3^2 \times 5 \times 7^d \times 11^e\) \(<\) \(\ds 2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e\) because $4 = 2^2 < 5$
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 3^2 \times 5 \times 7^d \times 11^e}\) \(<\) \(\ds \map \tau {2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e}\) as $2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e$ is highly composite
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 5} \times \map \tau {3^2 \times 7^d \times 11^e}\) \(<\) \(\ds \map \tau {2^3 \times 5^2} \times \map \tau {3^2 \times 7^d \times 11^e}\) Divisor Count Function is Multiplicative
\(\ds \leadsto \ \ \) \(\ds \map \tau {2^5 \times 5}\) \(<\) \(\ds \map \tau {2^3 \times 5^2}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {5 + 1} \paren {1 + 1}\) \(<\) \(\ds \paren {3 + 1} \paren {2 + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds 12\) \(<\) \(\ds 12\) which is an absurdity


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:

\(\ds \map \tau {27 \, 720}\) \(=\) \(\ds \map \tau {2^3 \times 3^2 \times 5 \times 7 \times 11}\)
\(\ds \) \(=\) \(\ds \paren {3 + 1} \paren {2 + 1} \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 96\)


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:

\(\ds \map \tau {45 \, 360}\) \(=\) \(\ds \map \tau {2^4 \times 3^4 \times 5 \times 7}\)
\(\ds \) \(=\) \(\ds \paren {4 + 1} \paren {4 + 1} \paren {1 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 100\)

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.

$\blacksquare$


Sources

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