Perfect Number is Sum of Successive Odd Cubes except 6

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

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

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 $r$.

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

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

for some $r$.

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


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

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

and it follows that:
 * $\displaystyle n = \sum_{k \mathop = 1}^{2^{r - 2} } \left({2 k - 1}\right)^3$

and hence the result.

When $n = 6$ we have:
 * $6 = 2^1 \left({2^2 - 1}\right)$

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