Special Highly Composite Number/Examples/12

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Example of Special Highly Composite Number

$12$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.


Proof

By inspection of the sequence of highly composite numbers, $12$ is highly composite.


Aiming for a contradiction, suppose $n > 12$ is a highly composite number which is not divisible by $12$.

We have that $6$ is a special highly composite number.

Therefore $6$ is a divisor of $n$.

As $12$ is not a divisor of $n$, it follows that the multiplicity of $2$ in $n$ is $1$.


From Prime Decomposition of Highly Composite Number, that means:

$n = 2 \times 3 \times 5 \times r$

where $r$ is a possibly vacuous square-free product of prime numbers strictly greater than $5$.

Then:

\(\ds 2^3 \times 3 \times r\) \(<\) \(\ds 2 \times 3 \times 5 \times r\) because $4 = 2^2 < 5$
\(\ds \leadsto \ \ \) \(\ds \map {\sigma_0} {2^3 \times 3 \times r}\) \(<\) \(\ds \map {\sigma_0} {2 \times 3 \times 5 \times r}\) as $2 \times 3 \times 5 \times r$ is highly composite
\(\ds \leadsto \ \ \) \(\ds \map {\sigma_0} {2^3 \times 3} \times \map {\sigma_0} r\) \(<\) \(\ds \map {\sigma_0} {2 \times 3 \times 5} \times \map {\sigma_0} r\) Divisor Count Function is Multiplicative
\(\ds \leadsto \ \ \) \(\ds \map {\sigma_0} {2^3 \times 3}\) \(<\) \(\ds \map {\sigma_0} {2 \times 3 \times 5}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {3 + 1} \paren {1 + 1}\) \(<\) \(\ds \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds 8\) \(<\) \(\ds 8\) which is a falsehood


The result follows by Proof by Contradiction.

$\blacksquare$


Sources

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