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Here's the reason why keeping primes with multiplicity makes the answer "no." If $p_n$ denotes the product of all the numbers you have so far, where $p_1$ is the product of the primes you start with, then $p_n = p_1 ... p_{n-1} + 1$. But we can rewrite this as $p_n = p_{n-1}(p_{n-1} - 1) + 1 = f(p_{n-1})$ where $f(x) = x^2 - x + 1$, and it is well-known that any prime divisor of $f(n)$ for an integer $n$ must be $2, 3$, or congruent to $1 \bmod 3$, i.e. the primes $5, 11, 19$ 17, ...$will never appear (unless they divide$p_1$to start with.) (Sketch: if$q | x^2 - x + 1$then$q | x^3 + 1$, hence$x$has order$6 \bmod q$or$q = 2, 3$. Since the multiplicative group$\bmod q$has order$q - 1$, this is possible if and only if$6 | q-1$.) 1 Here's the reason why keeping primes with multiplicity makes the answer "no." If$p_n$denotes the product of all the numbers you have so far, where$p_1$is the product of the primes you start with, then$p_n = p_1 ... p_{n-1} + 1$. But we can rewrite this as$p_n = p_{n-1}(p_{n-1} - 1) + 1 = f(p_{n-1})$where$f(x) = x^2 - x + 1$, and it is well-known that any prime divisor of$f(n)$for an integer$n$must be$2, 3$, or congruent to$1 \bmod 3$, i.e. the primes$5, 11, 19$will never appear (unless they divide$p_1$to start with.) (Sketch: if$q | x^2 - x + 1$then$q | x^3 + 1$, hence$x$has order$6 \bmod q$or$q = 2, 3$. Since the multiplicative group$\bmod q$has order$q - 1$, this is possible if and only if$6 | q-1\$.)