Continued fraction expansion of the square root of n: part II

In an earlier blog post, $\delta(n)$ is defined as the smallest term in the periodic part of the continued fraction of $\sqrt{n}$ and I showed that if $r$ is even then $\delta((\frac{rm}{2})^2+m) = r$ for all $m\geq 1$. and if $r$ is odd, then $\delta((rm)^2+2m) = r$ for all $m\geq 1$. Note that $\delta(n)$ is only defined if $n$ is not a perfect square.

If you look at the first few numbers $n$ that satisfy $\delta(n) = r$, it would appear that they all follow the quadratric equations above. However, not all integers $n$ such that $\delta(n) = r$ are of the forms above. In particular, if $r  > 0$ is even, then $\sqrt{\frac{r^4}{4} + r^3 + 2r^2 + 3r + 2} = \sqrt{\frac{(r^2-2)^2}{4}+(r+1)^3}$ has continued fraction expansion $\left[\frac{(r+1)^2+1}{2};\overline{r+1,r,r+1,(r+1)^2+1}\right]$ and thus $\delta\left(\frac{r^4}{4} + r^3 + 2r^2 + 3r + 2\right) = r$ and it is not of the forms above.

Similarly, if $r$ is odd, then $\sqrt{r^4 + r^3 + \frac{5(r+1)^2}{4}}$ has continued fraction expansion $\left[\frac{(r+1)(2r-1)+2}{2};\overline{r,2r-1,r,(r+1)(2r-1)+2}\right]$ and thus $\delta\left(r^4 + r^3 + \frac{5(r+1)^2}{4}\right) = r$ and it is not of the forms above either.