$P_n$ is the maximum prime factor of $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2, where $P_1$ = 2. Is there a $P_n =11$? The definition of a sequence ${P_n}$ is as following: $P_1$ = 2, $P_n$ is the maximum prime factor of  $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2. Is there one term equal to 11 in this sequence?
I tried the following
We know that $P_1$ = 2, $P_2$ = 3, and $P_3$ = 7. If there is a $P_n$ = 11, then $1 + P_1P_2P_3…P_{n-1}$ has a maximum prime factor of 11.
Since two consecutive integers will not have common prime factors, $1 + P_1P_2P_3…P_{n-1}$ does not have prime factors of 2, 3, and 7. Therefore, the only possible prime factorization for  $1 + P_1P_2P_3…P_{n-1}$ is $5^a11^b$.How should I start from here? Thanks.
 A: We can continue the line to get the next three elements:
$$
1+2\cdot 3\cdot 7 = 43 \Rightarrow P_4=43 \\
1+2\cdot 3 \cdot 7\cdot 43=1807=13\cdot 139 \Rightarrow P_5 = 139 \\
1+2\cdot 3 \cdot 7\cdot 43\cdot 139=251035=5\cdot 50207 \Rightarrow P_6=50207
$$
A quick search on OEIS shows that this is sequence https://oeis.org/A000946.
There it says that Cox and van der Poorten showed in their paper "On a sequence of prime numbers" that 5, 11, 13, 17, ... are NOT part of this series. I didn't have a look into their proof, but I assume that its not easy.
A: You have verified the claim for $n \leqslant 3$. Now, let $n > 3$ and assume the contrary. As you rightly pointed out, we have:
$$P_1P_2\cdots P_{n-1} + 1 = 5^a \cdot 11^b$$
where $a$ is a non-negative integer and $b$ is a positive integer. Considering the equation modulo $4$, the LHS is $3 \bmod{4}$ since the first term is divisible by $2$ but not $4$. Hence, the RHS must also be $3 \bmod{4}$, which forces $b$ to be an odd integer. Next, consider the equation modulo $3$. The LHS is $1 \bmod{3}$, which forces $a+b$ to be even in the LHS. Since $b$ is odd, $a$ must be odd as well.
Finally, consider the equation modulo $7$. We have:
$$5^a \cdot 11^b \equiv 1 \pmod{7}$$
$$(-2)^a\cdot 4^b \equiv 1 \pmod{7}$$
$$(-1)^a \cdot 2^{a+2b} \equiv 1 \pmod{7}$$
Here, note that since $a$ is odd, we have $2^{a+2b} \equiv -1 \pmod{7}$. However, this has no solutions. This is a contradiction. Hence, $11$ will not appear in the sequence of primes.
