Integer factorization using polynomial whose roots are prime factors Let $n$ be a square-free positive integer, let $n=p_{1}p_{2}\ldots p_{k}$
be the prime factorization of $n$ into $k$ distinct primes $p_{i}$.
For such $n$, define $F_{n}(x)\triangleq\prod_{i=1}^{k}(x-p_{i})$.
That is, $F_{n}(x)\in\mathbb{Z}[x]$ is the monic polynomial whose
roots are distinct prime factors of $n$. If we know $F_{n}(x)$,
we can obviously factorize $n$ in polynomial time. Now suppose that
we are given some partial information on $F_{n}(x)$ instead of complete
$F_{n}(x)$. We would like to know 


  What kind of partial information on $F_{n}(x)$ would allow factorization
  of $n$ in polynomial time? 


For example, it was demonstrated in SUMS OF DIVISORS, PERFECT NUMBERS
AND FACTORING that knowing just the sum of absolute values of coefficients
of $F_{n}(x)$ is enough to factorize $n$. What other partial information
on $F_{n}(x)$ allows factorization of $n$ in polynomial time?

After seeing comments, I am adding for kind of "partial information" we should be looking for.  As pointed out by @joro, knowing $F_n(1)$ or $F_n(-1)$ allows factorization of $n$. First question is that, does knowing value of $F_n(x)$ at any non-zero integer allow factorization of $n$? Second question is that, does knowing only constant many coefficients of $F_n(x)$ allow factorization of $n$?
Knowing $F_n(1)$ or $F_n(-1)$ gives good randomized polynomial algorithm for the factorization of $n$.

We have $|F_n(1)|=\phi(n)$ and $|F_n(-1)|=\sigma(n)$ by definition.

$\phi(n)$ is Euler's totient function and $\sigma(n)$ is the sum of divisors function.

The paper you link to proves the case for $\sigma(n)$.

The same paper cites that the case for $\phi(n)$ is known
and is derandomized assuming ERH (p. 1).

I am pretty sure both algorithms will be quite fast in practice.