studying for finals and enountered this problem on a practice test.
If n is a natural number such that n = 2^k(2^(k+1) - 1) with 2^(k+1) -1 prime then 2n = sum of the divisors.
What are the divisors of $\displaystyle n=2^k\left(2^{k+1}-1\right)$? Let's write them down:
** $\displaystyle 1,2,2^2,\ldots,2^k$
** $\displaystyle 2^{k+1}-1$
** $\displaystyle 2^{k+2}-2,2^{k+3}-2^2,\ldots,2^{2k+1}-2^k$
Well, let's sum up the bastards above:
$\displaystyle \sum\limits_{i=0}^k2^i+\sum\limits_{j=0}^k\left(2^ {k+1+j}-2^k\right)$ $\displaystyle =2^{k+1}\sum\limits_{j=0}^k2^j=2^{k+1}\,\frac{2^{k +1}-1}{2-1}=2^{k+1}(2^{k+1}-1)$ ...
Tonio