It is far easier to test Mersenne numbers for primality than general integers, thanks to the Lucas-Lehmer primality test. Weeding out composite Mersenne numbers by trial factoring is also faster, thanks to a theorem that eliminates most of the candidate factors.
As for numbers of the form 2^n+1, it's easy to show that if 2^n+1 is prime, then n must be a power of two. Such numbers are known as Fermat numbers, and there is also a fast primality test (Pepin's test) for numbers of this form. But because of the power of two in the exponent, each Fermat number is double the size (in terms of the number of digits) of the previous one. It doesn't take long before you exceed current computing capabilities. And so far, all the Fermat numbers we've managed to test have turned out to be composite, apart from the first five. Furthermore, it is conjectured that there are only finitely many Fermat primes, so it's possible we've already found them all. On the other hand, it is conjectured that there are infinitely many Mersenne primes.
The article is already out of date. The round 1 candidates were announced back on December 11. Since that time, 11 candidates have been broken. For the latest information, I recommend visiting the SHA-3 Zoo.
Also, the article suggests that candidates will continue to be broken quickly, but I doubt this will happen. The weak hashes will be broken quickly, but there are likely to be many strong candidates which will not be broken during the contest. Other factors (speed, simplicity, etc.) will determine the ultimate winner.
Somebody ought to cross ball point pens with coat hangers so that the pens will multiply instead of disappear.