It is known that if a Fermat number $F(n) \triangleq 2^{2^n} + 1$ is composite, then every one of its prime factors can be written as
$p = 2^{n+2}k + 1\;,$
for some positive integer $k$. Since $p \leq F(n)$, we can show that $k < 2^{2^{n-1} - n - 2}$.
Now, going backwards, given some prime number of this form, I want to know if there is a Fermat number that "contains" it.
It's possible to write $p$ in $n+1$ ways:
$p = 2^{n+2}k + 1 = 2^{n+1}(2k) + 1 = 2^n(4k) + 1 = \ldots = 2^3(2^{n-1}k) + 1 = 2^2(2^{n}k) + 1\;,$
which suggests $n+1$ possible Fermat numbers to be a multiple of $p$. However, since $k$ is not arbitrarily large, we may eliminate the $r$ smallest ones, where
$\begin{align} r &= \max_{s \in \mathbb{Z}} \left\{2^{s+1}(2^{n-s+1}k) + 1 \not \leq \sqrt{F(s-1)} \right\} =\\ &= \max_{s \in \mathbb{Z}} \{2^{s+1}(2^{n-s+1}k) + 1 \not \leq 2^{2^{s-2}}\} =\\ &= \max_{s \in \mathbb{Z}} \{2^{n-s+1}k \not< 2^{2^{s-2}-s-1}\} =\\ &= \max_{s \in \mathbb{Z}} \{2^{n+2}k \geq 2^{2^{s-2}}\} =\\ &= \max_{s \in \mathbb{Z}} \{s \leq \log_2(n + \log_2k + 2) + 2\}\\ &= \lfloor \log_2(n + \log_2k + 2) + 2 \rfloor\;. \end{align}$
How do I go on and sieve the remaining Fermat numbers, $F(r)$ to $F(n)$?