Perfect Number is Sum of Successive Odd Cubes except 6
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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.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $6$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $28$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $6$
- Note that this result is incorrectly stated.
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $28$