The number of primitive reduced forms with discriminant $D< 0$ is called the class number of discriminant $D$. We denote it by $h_f(D)$.
Write down a Delta symbol
$$ \Delta(D)=\begin{cases} \operatorname{sqfr}(D)&\operatorname{sqfr}(D)\equiv1\pmod{4}\\ 4\operatorname{sqfr}(D)&\operatorname{sqfr}(D)\not\equiv1\pmod{4}\\ \end{cases}$$
where the square-free part of the integer $n$ is represented by $\operatorname{sqfr}(n)$.
The class number of the imaginary quadratic field $K=\mathbb{Q}(\sqrt{D})$ is denoted as $h(D)$, where $D$ is a negative integer, then
$$h(D)=h(\Delta(D))=h_f(\Delta(D))\qquad\,D<0$$
The class number of the imaginary quadratic field $K=\mathbb{Q}(\sqrt{d_K})$ is a denoted as $h(d_K)$, where $d_K$ is fundamental discriminant, and its definition refers to the link [1], then
$$h(d_K)=h(\Delta(d_K))=h_f(\Delta(d_K))=h_f(d_K)\qquad\,d_K\in\mathrm{FD}_K$$
I want to ask, is this not missing some situations? In other words, when the negative integer $D$ is not a fundamental discriminant, is there a strict inequality in class numbers?
This fact seems to explain why there are more quadratic forms with class number equal to one than quadratic fields with class number equal to one.
\begin{align*} h_f(d) = 1 &\Leftrightarrow d = −3, −4, −7, −8, −11, −12, −16,\\ &\qquad\quad\>\>−19, −27, −28, −43, −67, −163;\\ h(d_K) = 1 &\Leftrightarrow d_K = −3, −4, −7, −8, −11,\\ &\qquad\qquad−19, −43, −67, −163.\\ \end{align*}