TITLE: Is 641 the Smallest Factor of any Composite Fermat Number?
QUESTION [9 upvotes]: Consider the sequence $a_n = 2^{2^n}+1$ of so-called Fermat numbers. It's well known that $a_5$ isn't prime ($a_5 = 641 \cdot 6700417$, this is due to Euler). What I want to know about this sequence is the smallest factor of any of its composite terms. Is it simply 641, or does a later composite term in the sequence have a smaller factor?
I originally wanted to ask if there had been any effort to solve similar problems for other sequences as well, but that's not something I feel MSE would want. If someone has any info on this, however, they should feel free to comment or add it to their answer!

REPLY [3 votes]: There is a deterministic answer to this, easy computer program. For each prime $p$ of interest, calculate 
$$ 2,4,16,256,\ldots, 2^{\left( 2^n \right)}, \ldots \pmod p  $$ until you find repetition. For the prime being worked on, this tells you whether $2^{\left( 2^n \right)}$ can ever be congruent to $-1 \pmod p.$