Multiply Perfect Number of Order 8

Theorem
The number defined as:
 * $n = 2^{65} \times 3^{23} \times 5^9 \times 7^{12} \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29^2 \times 31^2$
 * $\times 37 \times 41 \times 53 \times 61 \times 67^2 \times 71^2 \times 73 \times 83 \times 89 \times 103 \times 127 \times 131$
 * $\times 149 \times 211 \times 307 \times 331 \times 463 \times 521 \times 683 \times 709 \times 1279 \times 2141 \times 2557 \times 5113$
 * $\times 6481 \times 10 \, 429 \times 20 \, 857 \times 110 \, 563 \times 599 \, 479 \times 16 \, 148 \, 168 \, 401$

is multiply perfect of order $8$.

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 of Power of Prime:
 * $\sigma \left({p^k}\right) = \dfrac {p^{k + 1} - 1} {p - 1}$

Thus:

Gathering up the prime factors, we have:


 * $\sigma \left({n}\right) = 2^{68} \times 3^{23} \times 5^9 \times 7^{12} \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29^2 \times 31^2$
 * $\times 37 \times 41 \times 53 \times 61 \times 67^2 \times 71^2 \times 73 \times 83 \times 89 \times 103 \times 127 \times 131$
 * $\times 149 \times 211 \times 307 \times 331 \times 463 \times 521 \times 683 \times 709 \times 1279 \times 2141 \times 2557 \times 5113$
 * $\times 6481 \times 10 \, 429 \times 20 \, 857 \times 110 \, 563 \times 599 \, 479 \times 16 \, 148 \, 168 \, 401$

By inspection of the multiplicities of the prime factors of $n$ and $\sigma \left({n}\right)$, it can be seen that they match for all except for $2$.

It follows that $\sigma \left({n}\right) = 2^3 \times n = 8 n$.

Hence the result.