I just answered an old post Relatives of Heegner numbers? about prime-generating polynomials of form,
$$F(n)=n^2+n-p^2$$
for $p = (2,3,5,7,13)$. It turns out that since the discriminant of $F(n)$ is $d=1+4p^2$, then they were real quadratic fields with class number 1 and related continued fraction of period 3. (In my answer, I also included period 1).
Which made me think about $d$ with continued fraction of period 2. Checking Mathworld's Prime-Generating Polynomials, we find
$$F(n) = 7n^2 - 371n + 4871,\quad d = 1253\quad$$
which is prime for $n = 0\text{ to }23$. It has composite discriminant $d=7\times179$ and a check with OEIS shows that, surprisingly, it is the largest $d$ for period 2 (A050951). We can transform this to a more aesthetic form,
$$F(n) = 7n^2+49n+41,\quad d = 1253$$
But we can make it consistent with the form for period 1 which is $F(n) = an^2+abn-1$. Choosing those with highest ranges $R=1\text{ to } m$ where $m=(8,10,12,20,24)$, we have in order,
\begin{align} F(n) &=\; 3n^2 + 15n - 1,\quad\; d = 237\\ F(n) &= 19n^2 + 19n - 1,\quad d = 437\\ F(n) &=\; 3n^2 + 21n - 1,\quad\; d = 453\\ F(n) &= 11n^2 + 33n - 1,\quad d = 1133\\ F(n) &=\; 7n^2 + 35n - 1,\quad d = 1253 \end{align}
which are the five largest $d$ of A050951.
Question: The fourteen $d$ of A050951 are,
$$d= (3, 6, 11, 21, 38, 77, 83, 93, \color{blue}{227}, 237, 437, 453, 1133, 1253)$$
However, I'm having trouble with $d=\color{blue}{227}$ since it is prime. Anyone can suggest a nice prime-generating polynomial with it as the discriminant? (Nice = relatively small coefficients.)
Part of the problem is you do not actually have quadratic polynomials with integer coefficients whose discriminants are $\equiv3\bmod 4$. If you look at the integers of $\mathbb{Q}[\sqrt{m}]$ for $m\in\{2,3\}\bmod 4$ ($m$ squarefree), you discover that they solve quadratic equations with discriminant $4m$ not $m$.
So in your case, aim for a discriminant of $4×227=908$. The polynomial $2n^2+34n+31$ is prime for $n\in\{0,1,2,...,13\}$, with no prime factors below $29$ for any $n$ (the latter property is inherent to the chosen discriminant $908$). The translated polynomial $2n^2+30n-1$ matches the form of the quadratic polynomials given for the larger radicands in the question and is prime for $n\in\{1,2,3,...,14\}$.