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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:

  1. Anyone has the resources to compute $A(16),\,A(17)$, etc?
  2. 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? $40$? $50$?
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  • 1
    $\begingroup$ The discriminant $d$ of the 8th polynomial, $m^2+m+21557$, is $d = -23^2\cdot 163$. $\endgroup$ – Tito Piezas III Sep 14 '15 at 6:53
  • $\begingroup$ $A(16)\leq 521999251772081$ based on the Mersenneforum post. $\endgroup$ – Peter Košinár Feb 18 '16 at 8:46
  • $\begingroup$ @PeterKošinár: This is a good find! Thanks. Can you convert this to an answer? $\endgroup$ – Tito Piezas III Feb 18 '16 at 15:05
  • $\begingroup$ Ok, my computer-powered verification confirmed that $A(16)$ is indeed equal to $521999251772081$. $\endgroup$ – Peter Košinár Feb 29 '16 at 18:47
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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.

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  • $\begingroup$ By the way, for $n>1$, all so far have $\lceil log_{10}A\rceil\leq n$, or the number of $A$'s decimal digits is $\leq n$. Is there a heuristic to explain this? $\endgroup$ – Tito Piezas III Feb 29 '16 at 21:37

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