Are Perfect Numbers Bigger than Six initial sums of odd cubes (answered)

(NONE of this is my work. In fact some of it is on Wikipedia.)

In my last blog I noticed that

28 = 1³  + 3³

496= 1³ + 3³ + 5³ + 7³

noting that 28 and 496 are the 2nd and 3rd perfect numbers.

I asked if 8128, the next perfect number is also an initial sum of odd cubes. It is!

8128 = 1³ + 3³ + ... + 15³

I also asked if there was something interesting going on .The answer is YES but not that interesting.

All of the math with proofs are  here. I sketch below.

Known Theorem  1: n is an even perfect number iff n is of the form (2^p-1)(2^p- 1) where 2^p-1 is prime.

Known Theorem  2: 1³ + 3³ + 5³ + ... + (2(m-1)+1)³ = m²(2m²-1).

Interesting theorem: if n is an even perfect number larger than 6 and p is the p from Known Theorem 1 then n is the sum of the first  2^(p-1)/2 odd cubes.

Why this is less interesting: The proof does not use that n is perfect. It holds for any number of the form 2^p-1(2^p-1) where p is odd.

So the theorem has nothing to do with perfect numbers. Oh well.