Perfect Number is Sum of Successive Odd Cubes except 6

Theorem
Let $n$ be an even perfect number such that $n \ne 6$.

Then:
 * $\displaystyle n = \sum_{k \mathop = 0}^m \left({2 k + 1}\right)^3 = 1^3 + 3^3 + \cdots + \left({2 m + 1}\right)^3$

for some $m \in \Z_{>0}$.

That is, every even perfect number is the sum of the sequence of the first $r$ odd cubes, for some $r$.