Perfect numbers

"'I'll show you one more thing about perfect numbers," he said..."You can express them as the sum of consecutive natural numbers." Yoko Ogawa, The Housekeeper and the Professor, translated by Stephen Snyder

The examples that immediately follow the statement in the novel show that the perfect numbers 6, 28, and 496 are triangular numbers, that is, can be expressed as a sum of the form $1+2+3+\cdots+n$.

This got me thinking about perfect numbers. For example,

Theorem: The reciprocals of the divisors of any perfect number sum to 2.
Proof: If the divisors of the perfect number $N$ are $1, d_2, d_3, \ldots, d_k$, and $N$, then the sum of the reciprocals would be
$1/1 + 1/d_2+ 1/d_3 \cdots + 1/d_k+ 1/N$. If we call this finite sum $s$, then
$$Ns=N/1 + N/d_2+ N/d_3 \cdots + N/d_k+ N/N$$
$$= N + d_k +\cdots +d_3+ d_2 + 1$$
$$= N +(\text{proper divisors of N})  = N+N = 2N$$ Thus $s=2$

Theorem: Every even perfect number has the form $N=(2^k -1)\cdot 2^{k-1}$, where $2^k - 1$ is a Mersenne prime.
Proof:  Let $N$ be an even perfect number. Then $N$ can be written in the form $N=2^{k-1} m$, where $k>1$and $m$ is odd.
Define $\sigma(n)$ to be the sum of all positive divisors of $n$.  In particular, $\sigma(n)=2n$  whenever $n$ is perfect.
When $a$ and $b$ are relatively prime, $\sigma(ab)=\sigma(a)\sigma(b)$  because every divisor of $ab$ can be uniquely written as the product of a divisor of $a$ times a divisor of $b$, so summing the divisors of $ab$ can be accomplished by first computing $\sigma(a)$  and multiplying it by each of the divisors of $b$, and summing those products.

Now $\sigma(N)=2N=2^k m$ because $N$ is perfect. By adding a finite geometric series with common ratio 2, we see that $\sigma(2^{k-1})=2^k-1$, and we have

$2^k m=  \sigma(2^{k-1}m) = \sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m)$  

Solving for $\sigma(m)$, we get $\sigma(m)=m+ \frac{m}{2^k-1}$

From the definition of $\sigma(m)$, $m+ \frac{m}{2^k-1}$  must represent the sum of all divisors of  $m$, and in particular the fraction must be an integer.  But then $\frac{m}{2^k-1}$ must itself divide $m$, and as $m$ clearly divides itself, $m$ and $\frac{m}{2^k-1}$  must be all the divisors of $m$.  Thus $m=2^k-1$ must be prime.

Conversely, Theorem: Each Mersenne prime $2^k -1$ gives the perfect number $N=(2^k -1)\cdot 2^{k-1}$.
Proof: If $2^k - 1$ is prime, then the divisors of $N=(2^k -1)\cdot 2^{k-1}$ are $1, 2, 2^2, \ldots, 2^{k-1},$ and also the product of any of those powers with the prime $2^k - 1$. Summing all the proper divisors is the sum of two geometric series, each with common ratio 2. We get
$$\left(1+2+…+2^{k-1}\right) + \left(2^{k}-1\right) \left(1+2+…+2^{k-2}\right)$$
$$= \left(2^k -1\right) + \left(2^{k}-1\right) \left(2^{k-1}-1\right)$$
$$= \left(2^k -1\right) \left(1+ 2^{k-1}-1\right) = N$$
We can see that every even perfect number is a triangular number, because $N=(2^k -1)\cdot2^{k-1}$ has the form $\frac{(n-1)n}{2}$, where $n = 2^k$.

Euler evidently knew everything about even perfect numbers, but as far as I know, neither he nor anyone else has proven whether or not any odd perfect number exists.

I don't know whether an odd perfect number would need to be a triangular number. But the Professor only asserted that perfect numbers can be expressed as a sum of consecutive natural numbers. And of course any odd number $O = 2n+1$ --including any odd perfect number-- can be expressed as the sum of two consecutive natural numbers: $n + (n+1)$.