Can the sum of $n$ consecutive primes be $n$ times a prime? I recently saw a "coffin problem" asking to prove that the sum of two consecutive primes is not twice a prime. This got me wondering if three consecutive primes can sum to be three times a prime, and they can. I figured out that it is possible if and only if the smallest and largest of the three primes sum to twice the middle prime (some examples are 3, 5, 7 and 47, 53, 59). I tried to figure out whether or not 4 consecutive primes can sum to be 4 times a prime, but I have made no progress.
So, I am wondering if it is known or possible to deduce for which values of $n$ can $n$ consecutive primes sum to be $n$ times a prime.
 A: According to the Hardy-Littlewood $k$-tuples conjecture (the natural generalization of the twin prime conjecture) there should be infinitely many prime quadruples of the form $(p,p+6,p+12,p+30)$ and hence infinitely many solutions to the original problem.
While this remains a (most likely extremely hard) conjecture, it seems an interesting question whether we can still prove that infinitely often the sum of four consecutive primes is four times a prime (since this is strictly weaker than even this special case of the $k$-tuples conjecture).
In general, it should be relatively easy to find such a "linear parametrization" when generalizing to $n \ge 5$ so that we are done by the $k$-tuples conjecture. However, again it might be interesting to ask whether we can prove unconditionally that such primes always exist.
A: The Green-Tao theorem (https://en.wikipedia.org/wiki/Primes_in_arithmetic_progression) tells us that there are arbitrarily long arithmetic progressions of primes. It does not allow us to control whether the primes are consecutive, so I don't think we know that there are arbitrarily long arithmetic progressions of consecutive primes. But for arbitrary arithmetic progressions the following can be found.
Since the sum of an odd length arithmetic progression is divisible by the number of terms, the answer is yes for $n$ odd, and the total will be $n$ times the middle prime.
For even lengths, if the first prime is $p$ and the common difference is $d=2r$ (the prime 2 is clearly not involved, so the difference will always be even), the sum of the progression will be $$np+(n-1)(n-2)r$$
Now $n=2m$, so $n$ has the common factor $2$ with $(n-1)(n-2)$, and you will need $m|r$ or equivalently $n|d$.
So to make this work for $4$ you need to find an arithmetic progression of length $4$ with common difference $4$ or $8$ or $12 \dots$ eg $17, 29, 41, 53$
We can use the Green-Tao theorem again. For example we know that there must be an arithmetic progression of primes of length $33$. If the difference is $d$, then $p, p+4d, p+8d, p+12d, p+16d, p+20d, p+24d, p+28d, p+32d$ is a sequence whose common difference is divisible by $8$. So we can pick out a subsequence of a long progression to get the common difference we need. If you investigate you will discover that such long sequences of primes in arithmetic progression can get very large and be hard to find - but we do know they are there.
I think you should be able to show that this works for any $n$ you choose.
