Since Euclid, we have known that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new (possible) prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \$13\$ and \$139\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

###Challenge Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

###Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5 

Since Euclid, we have known that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new (possible) prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \$13\$ and \$139\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

Challenge

Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5 

Since Euclid, we know that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new (possible) prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \$13\$ and \$139\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

###Challenge Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

###Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5 

Since Euclid, we have known that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new (possible) prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \$13\$ and \$139\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

###Challenge Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

###Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5 

Since Euclid we know that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, lets say \$p_1,p_2,...,p_n\$. Then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so it's prime factorization must yield a new prime that was not in the list yet. So the assumption that only finitely primes exist, is false.

Now lets assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$ so both \$13\$ and \$139\$ are new primes, etc. In the case we get a composite number, we just take the least new prime. This results in A000945.

###Challenge Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

###Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5 

Since Euclid, we know that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \$2\$ is the only prime. The method from above yields \$2+1=3\$ as a new (possible) prime. Applying the method again yields \$2\cdot 3+1=7\$, and then \$2\cdot 3\cdot 7+1=43\$, then \$2\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \$13\$ and \$139\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

###Challenge Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

###Examples

For \$p_1 = 2\$:

1 2 2 3 3 7 4 43 5 13 6 53 7 5 8 6221671 9 38709183810571 

For \$p_1 = 5\$ (A051308):

1 5 2 2 3 11 4 3 5 331 6 19 7 199 8 53 9 21888927391 

For \$p_1 = 97\$ (A051330)

1 97 2 2 3 3 4 11 5 19 6 7 7 461 8 719 9 5