Possibility of having exactly 5 primes in a sequence of $10$ consecutive positive integers.

• Does a sequence of positive integers $a_n$ such that the sequence $a_n, \space a_n+1,\space a_n+2,\space \cdots, \space a_n+9$ contains exactly $5$ primes exist?
• Is such sequence finite or infinite?
• Does it have a general term? If it does, then kindly prove it and provide its limit as $n \to + \infty$.

$a_0 = 2$ because the sequence $2, \space 3,\space \cdots, \space 11$ contains exactly $5$ primes $2,\space 3,\space 5,\space 7,\space 11$

I found a finite list of prime and I couldn't find any other integer and I was wondering if such integers except from 2 exist.

• What about a sequence for exactly $4$ primes?
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Of 10 consecutive numbers, exactly five are even, one or two are multiples of six and three or four ar emultiples of three. Thus at least one of the multiples of three is odd. This leaves at most four numbers that are neither divisible by two nor three. Therefore, unless $2$ and $3$ appear themselves among the primes, no five primes can occur.
I think that's it for 5 primes. A sequence of 10 consecutive positive integers has 5 even numbers. It also contains 3 multiples of 3, at least one of which is odd, leaving you with 4 odd numbers not divisible by 3. For $a_0$ through $a_0+9$, that's enough since it actually has primes divisible by 2 and 3, namely 2 and 3 themselves.