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 $n = 0\,\text{to}\,39$.
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 not diagonals in the Ulam spiral.