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Given the *Ulam spiral* with center $c = 41$ and the numbers in a clockwise direction, we have,

$$\begin{array}{cccccc}

\color{red}{61}&62&63&64&\to\\

60&\color{red}{47}&48&49&50\\

59&46&\color{red}{\small{c=\,}41}&42&51\\

58&45&44&\color{red}{43}&52\\

57&56&55&54&\color{red}{53}&\downarrow\\

\end{array}$$

The main diagonal is defined by *Euler’s polynomial* $F(n) = n^2+n+41$, and yields *distinct* primes for 40 consecutive $n = 0\,\text{to}\,39$.

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If we let $c = 3527$ as in this old sci.math post, we get,

$$\begin{array}{cccccccc}

\color{blue}{3569}&3570&3571&3572&3573&3574&\to\\

3568&\color{red}{3547}&3548&3549&3550&3551&3552\\

3567&3546&\color{red}{3533}&3534&3535&3536&3553\\

3566&3545&3532&\color{red}{\small{c=\,}3527}&3528&3537&3554\\

3565&3544&3531&3530&\color{red}{3529}&3538&3555\\

3564&3543&3542&3541&3540&\color{red}{3539}&3556\\

3563&3562&3561&3560&3559&3558&\color{red}{3557}&\downarrow\\

\end{array}$$

The polynomial is $G(n) = 4n^2-2n+3527$ and is prime for 23 consecutive $n = -2\,\text{to}\,20$. Its square-free discriminant is $d = -14107$ and has *class number* $h(d) = 11$. This is the 3rd largest (in absolute value) with that $h(d)$. The blue number, $G(-3)=3569$ is not prime.

** Question**: For $F(n) = n^2+n+p$, the record is held by Euler’s polynomial. For the form $G(n) = 4n^2\pm 2n+p$, is there a better one?

** P.S.** Other polynomials such as $F(n) = 6n^2+6n+31$ are prime for $n=0\,\text{to}\,28$, but are

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(*Updated*.) Knowing more *Mathematica* coding, I re-visited this old question. Given,

$$P(n) = 4n^2\pm2n+p\tag1$$

I searched the first $million$ primes $p$. The *complete* table of record $P(n)$ with at least $14$ consecutive $n$ in the range $n = -5 \to 60\,$ yielding primes are in the table below.

$$\begin{array}{|c|c|c|c|}

\hline

\text{#} & P(n)=an^2+bn+c & d = b^2-4ac & h(d) & Prime\; range\; n &Total\,(T)\\

1& 4n^2-2n+41 &\color{red}{-163} & 1&-19 \to 20& 40\\

2& 4n^2-2n+3527 &\color{blue}{-14107} & 11&-2 \to 20& 23\\

3& 4n^2+2n+21377 &\color{red}{-85507} & 22&47 \to 64& 18\\

4& 4n^2-2n+9281 &\color{red}{-37123} & 17& 0 \to 16& 17\\

5& 4n^2-2n+17 &\color{blue}{-67} & 1&-7 \to 8&16\\

6& 4n^2+2n+41201&\color{red}{-164803} & 32& 52 \to 66& 15\\

7& 4n^2+2n+12821&-51283& 21& 8 \to 21& 14\\

8& 4n^2-2n+3461 &\color{red}{-13843} & 10&34 \to 47&14\\

9& 4n^2-2n+1277 &\color{blue}{-5107} & 7&19 \to 32&14\\

\hline

\end{array}$$

If the discriminant $d$ in red, then it the *largest* $|d|$ of a class number $h(d)$. If it is in blue, then is one of the $3$ largest $|d|$ of that class.

For example, using #3, if you have an Ulam spiral with center $c=21377$, then in its main diagonal there are $18$ primes in a row. It involves $d = -85507$ which is the largest $|d|$ with $h(d) = 22$.

**P.S.** Considering that the millionth prime is $p(10^6) = 15485863$,

it seems odd there are no large $p$ found within the search radius. However, I only searched $n = -5\to 60$, so another choice of range might yield other $P(n)$.

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