Algebraic degrees of $j\left(\sqrt{-n}\right)$ and $j\left(\frac{1 + \sqrt{-n}}{2}\right)$ and class numbers of $Q(\sqrt{-n})$

Question

Let $n\in\mathbb N$ be squarefree. Denote by $h(n)$ the class number of $Q(\sqrt{-n})$ and by $d_1(n)$ and $d_2(n)$ the degrees of the algebraic numbers $j\left(\sqrt{-n}\right)$ and $j\left(\frac{1 + \sqrt{-n}}{2}\right)$ respectively. Then it appears that for all $4\le n<500$, we have $$\frac {d_1(n)}{h(n)}= \begin{cases} 3\phantom{\bigl(}\text { if } n\equiv3\pmod8\\ 1\phantom{\bigl(}\text { else} \\ \end{cases}$$ (note that for $n\equiv3\pmod8$, there is a special closed form of $h(n)$, at least for primes), and $$\frac {d_2(n)}{h(n)}= \begin{cases} 1& \text{ for }\ n\phantom{\bigl(} \text {odd}\\ 2& \text{ for }\ n\phantom{\bigl(} \text {even.}\\ \end{cases}$$

Are these identities true for all $n>3$?

If $n$ is not squarefree, both quotients still appear to be integers, but depending in a rather irregular way on the prime factors.
Some patterns related to primes seem to exist however, e.g.

$\dfrac {d_1(4p)}{h(4p)}= \begin{cases} 6\phantom{\bigl(}\text { if } p\equiv3\pmod8\\ 2\phantom{\bigl(}\text { else,} \\ \end{cases}$

$\dfrac {d_1(n)}{h(n)}=\dfrac {d_2(n)}{h(n)}=2\left\lfloor\dfrac{p+1}4\right\rfloor$ for $n=p^2$ with an odd prime $p$ (in fact, for all $n\equiv1 \pmod 4$, whether squarefree or not, $j\left(ni\right)$ and $j\left(\frac{1 + ni}{2}\right)$ appear to have the same minimal polynomial) or

$\dfrac {d_2(n)}{h(n)}=p$ for $n=p^3q$ with odd primes $p,q$.

Is there anything known in this direction?