A conjecture about perfect numbers ( - continued )

Conjecture

Show that: if $p$ is odd then
$$2^{p-1}(2^p-1) = \sum_{k=1}^{\frac{p+1}{2}-1} (2k-1)^3$$
( Notice that $2^{p-1}(2^p-1)$ yields a perfect number if $(2^p-1)$ is prime. )

Plan

The plan of the proof is as follows.
- Find expressions for $1^3 + 2^3 + 3^3 + ... + k^3$
- and $2^3 + 4^3 + 6^3 + ... + (2k)^3$
- Subtract both expressions
- Create $f(p)$ by injecting $2^p-1$ into the upper-index

Proof

We use the Pascal Triangle to determine $\sum_{k=1}^{n} k^3$.

$\underline{n}$
0: 1
1: 1 - 1
2: 1 - 2 - 1
3: 1 - 3 - 3 - 1
4: 1 - 4 - 6 - 4 - 1
5: 1 - 5 - 10-10 - 5 - 1

Since $n^3={n \choose 1} + 6{n \choose 2} + 6{n \choose 3}$ we seek the second column, one row down or $\sum_{k=1}^{n} k^3 = {n+1 \choose 2} + 6{n+1 \choose 3} + 6{n+1 \choose 3}$.

Clearly, $\sum_{k=1}^{n} 2k^3 = 8({n+1 \choose 2} + 6{n+1 \choose 3} + 6{n+1 \choose 4})$.

If
$$f(n)=\sum_{k=1}^{n} k^3 = {n+1 \choose 2} + 6{n+1 \choose 3} + 6{n+1 \choose 4}$$
and
$$g(n)=8 \sum_{k=1}^{n} k^3 =8({n+1 \choose 2} + 6{n+1 \choose 3} + 6{n+1 \choose 4})$$
then the function we require is $$s(n) = f(n) - g( \lfloor \frac{n}{2} \rfloor ) \text{ for } n=1,3, \cdots$$ .
$$\begin{array}{ll} \underline{n} & \underline{s(n)}\\ 1 & 1 \\ 3 & 28 \\ 5 & 153 \\ 7 & 496 \\ 9 & 1225 \\ 11 & 2556 \\ 13 & 4753 \\ 15 & 8128 \end{array}$$
Since $n=1,3, \cdots$ anyway, we can remove the difficult to handle floor function by $\frac{n-1}{2}$ giving $$s(n) = f(n) - g(\frac{n-1}{2}) \text{ for } n=1,3, \cdots .$$
Finally we rework
$$s(p)=f(2^{\frac{p+1}{2}}-1) - g(\frac{(2^{\frac{p+1}{2}}-1)-1}{2})$$
to $$s(p)=2^{p-1} \left(2^p-1\right)$$
yielding for $p=1 \cdots 7$
$$\begin{array}{ll} \underline{p} & \underline{s(p)}\\ 1. & 1. \\ 2. & 6. \\ 3. & 28. \\ 4. & 120. \\ 5. & 496. \\ 6. & 2016. \\ 7. & 8128. \end{array}$$
( Since $2^3-1$, $2^5-1$ and $2^7-1$ are prime $28$ and $496$, $8128$ are perfect. )