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This might be a very stupid question, so apologies in advance)

Yesterday I was trying to fall asleep, and noticed a strange pattern with prime numbers:

Let $p_{1}$ be prime, and $p_{2} = p_{1} + 2$ also be prime, and $p_{i} = p_{i - 1} + 2 \cdot (i - 1) $. Then, all the next numbers until $p_{1}^2$ will also all be prime.

For example: $$3, 5, 9 = 3^{2}$$

$$5, 7, 11, 17, 25 = 5^2$$

$$ 11, 13, 17, 23, 31, 41, 53, 67, 83, 101, 121 = 11^2$$

But this does not always work, for example $$ 29, 31, 35 = 5 \cdot 7$$

Is there an obvious reason for this, and can it be generalized, or is it just a coincidence?

Legolas131
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  • Very closely related: https://mathworld.wolfram.com/PrimeTriplet.html and https://mathworld.wolfram.com/PrimeConstellation.html – Adam Rubinson Jan 25 '24 at 11:41
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    The sequence starting with $17$ is the values of the polynomial $n^2+n+17$. It's known that this is prime for $n=0,1,2,\dotsc,15$ (but obviously not for $n=16$) because $4\times17-1=67$ and the ring of integers of ${\bf Q}(\sqrt{-67})$ is a unique factorization domain. Even better is starting with $41$, which gives values of $n^2+n+41$, prime for $0\le n\le39$, because ${\bf Q}(\sqrt{-163})$ is a unique factorization domain. But that's the last such example. This is all very well-known. – Gerry Myerson Jan 25 '24 at 12:17
  • See, e.g., https://mathworld.wolfram.com/Prime-GeneratingPolynomial.html or any site that comes up when you type prime-producing polynomial into the internet. – Gerry Myerson Jan 25 '24 at 12:20
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    These are the initial prime values of the famous Euler prime producing polynomials $,n^2+n+k,$ given by the Theorem in the linked dupe, since for $,k = 3,5,11,17,41,$ the ring of integers of $,\Bbb Q(\sqrt{1-4k}),$ is a UFD. See the Monthly paper linked there for an elementary exposition. – Bill Dubuque Jan 25 '24 at 12:43

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