#### 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. )