Perfectly Perfect

Every even perfect number is given by $\left( 2^p - 1 \right) \cdot 2^{p-1}$ , where $2^p-1$ is a Mersenne prime .

We can rewrite this as

$\left( 2^p - 1 \right) \cdot 2^{p-1} = \left( 2^p-1 \right) \left( \frac {2^p}2 \right) = \frac {2^p \left( 2^p-1 \right) } 2$

If we now substitute $n = 2^{p-1}$ we get

$\frac {2n (2n-1)} 2$

which is exactly the formula for the $n$ -th hexagonal number.