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.
$\blacksquare$
Sources
- Dec. 1991: Steven Ratering: An Interesting Subset of the Highly Composite Numbers (Math. Mag. Vol. 64, no. 5: pp. 343 – 346) www.jstor.org/stable/2690653