Is it true that $6n^2+p$ gives primes for $n=0,1,2,\dots,p-1$ iff $Q(\sqrt{-6p})$ has class number $4$?

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Let $p$ be a prime number, are the following statements true?

1.Quadratics of the form $6n^2+p$ gives primes for $n=0,1,2,\dots,p-1$ iff $Q(\sqrt{-6p})$ has class number $4$. And all such primes are $p\in\{5,7,13,17\}.$

2.Quadratics of the form $10n^2+p$ gives primes for $n=0,1,2,\dots,p-1$ iff $Q(\sqrt{-10p})$ has class number $4$. And all such primes are $p\in\{3,7,13,19\}.$

I know that quadratics of the form $2n^2+p$ gives primes for $n=0,1,2,\dots,p-1$ iff $Q(\sqrt{-2p})$ has class number $2$. And all such primes are $p\in\{3,5,11,29\}.$ See here.

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You can verify or disprove the "if" part of your claim easily with a computer because the only quadratic imaginary fields with class number $4$ are given by $\mathbb{Q}(\sqrt{-d})$ where $d$ is one in the list

14,17,21,30,33,34,39,42,46,55, 57,70,73,78,82,85,93,97,102,130, 133,142,155,177,190,193,195,203,219,253, 259,291,323,355,435,483,555,595,627,667, 715,723,763,795,955,1003,1027,1227,1243,1387,1411,1435,1507,1555.

Here is a list of imaginary quadratic fields with small class numbers.