Special Highly Composite Number/Examples/6

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Example of Special Highly Composite Number

$6$ 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, $6$ is highly composite.


Aiming for a contradiction, suppose $n > 6$ is a highly composite number which is not divisible by $6$.

We have that $2$ is a special highly composite number.

Therefore $2$ is a divisor of $n$.

As $6$ is not a divisor of $n$, it follows that $3$ is also not a divisor of $n$.


By Prime Decomposition of Highly Composite Number, that means $n = 2^k$ for some $k \ge 3$.

Then:

\(\ds 2^{k - 2} \times 3\) \(<\) \(\ds 2^k\) because $3 < 2^2 = 4$
\(\ds \leadsto \ \ \) \(\ds \map {\sigma_0} {2^{k - 2} \times 3}\) \(<\) \(\ds \map {\sigma_0} {2^k}\) as $2^k$ is highly composite
\(\ds \leadsto \ \ \) \(\ds \paren {k - 1} \paren {1 + 1}\) \(<\) \(\ds k + 1\) Definition of Divisor Count Function
\(\ds \leadsto \ \ \) \(\ds k\) \(<\) \(\ds 3\) after algebra


But this contradicts our deduction that $n = 2^k$ where $k \ge 3$.

The result follows by Proof by Contradiction.

$\blacksquare$


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