Perfect Number is Sum of Successive Odd Cubes except 6

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

Then:
 * $\ds n = \sum_{k \mathop = 1}^m \paren {2 k - 1}^3 = 1^3 + 3^3 + \cdots + \paren {2 m - 1}^3$

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 $r$.

Proof
From Sum of Sequence of Odd Cubes:
 * $1^3 + 3^3 + 5^3 + \cdots + \paren {2 m − 1}^3 = m^2 \paren {2 m^2 − 1}$

By the Theorem of Even Perfect Numbers:
 * $n = 2^{r - 1} \paren {2^r - 1}$

for some $r$.

Setting $m = 2^{r - 2}$:


 * $m^2 = 2^{r - 1}$

and so:
 * $2 m^2 = 2^r$

and it follows that:
 * $\ds n = \sum_{k \mathop = 1}^{2^{r - 2} } \paren {2 k - 1}^3$

and hence the result.

When $n = 6$ we have:
 * $6 = 2^1 \paren {2^2 - 1}$

leading to $r = 2$ and thence $m^2 = 2$, at which point the formula fails to work.