On the prime-generating polynomial $m^2+m+234505015943235329417$
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? $40$? $50$?
Solution 1:
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.