TITLE: On the prime-generating polynomial $m^2+m+234505015943235329417$
QUESTION [12 upvotes]: In 2009, J. Waldvogel and Peter Leikauf found the remarkable Euler-like polynomial,
$$F(m)=m^2+m+234505015943235329417$$
which is prime for $m=0\to20$, but composite for $m=21$. Define,
$$F(m)=m^2+m+A$$
such that $F(m)$ is prime for $m=0\to n-1$, but composite for $m=n$. Then the least $\color{brown}{A>41}$ (compare to A164926) are, 
$$\begin{array}{|c|c|l|}
\hline
n&\lceil\log_{10}A\rceil&A\\
\hline
1 &2&43 \\
2 &2 &59 \\
3 &3 &107 \\
4 &3 &101 \\
5 &3 &347 \\
6 &4 &1607 \\
7 &4 &1277 \\
8 &5 &21557  \\
9 &8 &51867197 \\
10 &6 &844427  \\
11 &9 &180078317 \\
12 &10 &1761702947 \\
13 &10 &8776320587 \\
14 &14 &27649987598537 \\
15 &15 &291598227841757 \\
16 &15 &521999251772081\,(?) \\
17 &?? &??\\
\,\vdots\\
21 &21 &234505015943235329417\\
\hline
\end{array}$$
where $\lceil x \rceil$ is the ceiling function. Assuming the prime k-tuples conjecture and Mollin's theorem 2.1 in Prime-Producing Quadratics (1997), this shows that the sequence is defined for $n>0$.
Questions:

Anyone has the resources to compute $A(16),\,A(17)$, etc?
It seems the second column has a comparable rate to the first. By the time it reaches $n=40$ (comparable to Euler's polynomial), what is a ballpark figure for $A$'s number of decimal digits? $40$? $50$?

REPLY [2 votes]: The value $A(16)$ is equal to $521999251772081$:

It is easy to check that it satisfies the given primality conditions.
Minimality can be ascertained by exhaustively testing all smaller candidates; pre-sieving the possible residues modulo small primes reduces the range enough to allow exhaustive search to proceed successfully.

I will let the search run further and update the answer if any further values pop up eventually.