Multiply Perfect Number of Order 6

Theorem
The number defined as:
 * $n = 2^{36} \times 3^8 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
 * $\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
 * $\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$

is multiply perfect of order $6$.

Proof
From Sigma Function is Multiplicative, we may take each prime factor separately and form $\sigma \left({n}\right)$ as the product of the $\sigma$ function of each.

Each of the prime factors which occur with multiplicity $1$ will be treated first.

A prime factor $p$ contributes towards the combined $\sigma$ a factor $p + 1$.

Hence we have:

The remaining factors are treated using Sigma Function of Power of Prime:
 * $\map \sigma {p^k} = \dfrac {p^{k + 1} - 1} {p - 1}$

Thus:

Gathering up the prime factors, we have:


 * $\map \sigma n = 2^{37} \times 3^9 \times 5^5 \times 7^7 \times 11 \times 13^2 \times 19 \times 31^2$
 * $\times \ 43 \times 61 \times 83 \times 223 \times 331 \times 379 \times 601 \times 757 \times 1201$
 * $\times \ 7019 \times 112 \, 303 \times 898 \, 423 \times 616 \, 318 \, 177$

By inspection of the multiplicities of the prime factors of $n$ and $\map \sigma n$, it can be seen that they match for all except for $2$ and $3$.

It follows that $\map \sigma n = 2 \times 3 \times n = 6 n$.

Hence the result.