{{AimForCont}} $n$ 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: {{begin-eqn}} {{eqn | l = \tau \left({3 n / 2}\right)       | o = \le       | r = \tau \left({n}\right)       | c =  }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3^{b + 1} \times 5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | o = \le       | r = \tau \left({2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | c =  }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3^{b + 1} }\right) \times \tau \left({5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | o = \le       | r = \tau \left({2^a \times 3^b}\right) \times \tau \left({5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3^{b + 1} }\right)       | o = \le       | r = \tau \left({2^a \times 3^b}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a - 1}\right) + 1}\right) \left({\left({b + 1}\right) + 1}\right)       | o = \le       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = a \left({b + 2}\right)       | o = \le       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = simplifying }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \tau \left({4 n / 3}\right)       | o = \le       | r = \tau \left({n}\right)       | c =  }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a + 2} \times 3^{b - 1} \times 5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | o = \le       | r = \tau \left({2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | c =  }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a + 2} \times 3^{b - 1} }\right) \times \tau \left({5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | o = \le       | r = \tau \left({2^a \times 3^b}\right) \times \tau \left({5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a + 2} \times 3^{b - 1} }\right)       | o = \le       | r = \tau \left({2^a \times 3^b}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a + 2}\right) + 1}\right) \left({\left({b - 1}\right) + 1}\right)       | o = \le       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = \left({a + 3}\right) b       | o = \le       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = simplifying }} {{end-eqn}} This leads to: {{begin-eqn}} {{eqn | l = a       | o = \le       | r = b + 1       | c =  }} {{eqn | l = 2 b - 1       | o = \le       | r = a       | c =  }} {{eqn | ll= \leadsto       | l = 2 b - 1       | o = \le       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \le       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \le       | r = 3       | c =  }} {{end-eqn}} It has already been established that: :$a \ge 3$ :$b \ge 2$ {{begin-eqn}} {{eqn | l = a       | o = \ge       | r = 3 }} {{eqn | l = b       | o = \le       | r = 2 }} {{end-eqn}} so it is now possible to state: {{begin-eqn}} {{eqn | l = a       | r = 3 }} {{eqn | l = b       | r = 2 }} {{end-eqn}} Suppose: :$(1): \quad f \ge 1$ Then: {{begin-eqn}} {{eqn | l = 2^5 \times 3^3 \times 5^c \times 7^d \times 11^e \times 13^{f - 1} \times r       | o = <       | r = 2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r       | c = because $12 = 2^2 \times 3 < 13$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^3 \times 5^c \times 7^d \times 11^e \times 13^{f - 1} \times r}\right)       | o = <       | r = \tau \left({2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r}\right)       | c = as $2^3 \times 3^2 \times 5^c \times 7^d \times 11^e \times 13^f \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^3 \times 13^{f - 1} }\right) \times \tau \left({5^c \times 7^d \times 11^e \times r}\right)       | o = <       | r = \tau \left({2^3 \times 3^2 \times 13^f}\right) \times \tau \left({5^c \times 7^d \times 11^e \times r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^3 \times 13^{f - 1} }\right)       | o = <       | r = \tau \left({2^3 \times 3^2 \times 13^f}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({5 + 1}\right) \left({3 + 1}\right) \left({\left({f - 1}\right) + 1}\right)       | o = <       | r = \left({3 + 1}\right) \left({2 + 1}\right) \left({f + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 24 f       | o = <       | r = 12 \left({f + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = f       | o = <       | r = 1       | c = which is a contradiction of $(1)$ }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = 2^5 \times 3^2 \times 5 \times 7^d \times 11^e       | o = <       | r = 2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e       | c = because $4 = 2^2 < 5$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^2 \times 5 \times 7^d \times 11^e}\right)       | o = <       | r = \tau \left({2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e}\right)       | c = as $2^3 \times 3^2 \times 5^2 \times 7^d \times 11^e$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 5}\right) \times \tau \left({3^2 \times 7^d \times 11^e}\right)       | o = <       | r = \tau \left({2^3 \times 5^2}\right) \times \tau \left({3^2 \times 7^d \times 11^e}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 5}\right)       | o = <       | r = \tau \left({2^3 \times 5^2}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({5 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({3 + 1}\right) \left({2 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 12       | o = <       | r = 12       | c = which is an absurdity }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \tau \left({27 \, 720}\right)       | r = \tau \left({2^3 \times 3^2 \times 5 \times 7 \times 11}\right)       | c =  }} {{eqn | r = \left({3 + 1}\right) \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 96       | c =  }} {{end-eqn}} 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 $\tau \left({n}\right) < 96$. But: {{begin-eqn}} {{eqn | l = \tau \left({45 \, 360}\right)       | r = \tau \left({2^4 \times 3^4 \times 5 \times 7}\right)       | c =  }} {{eqn | r = \left({4 + 1}\right) \left({4 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 100       | c =  }} {{end-eqn}} 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. {{qed}}