Given the first $n$ primes, find the least common multiple of $p_1 - 1$, $p_2 - 1$, ..., $p_n - 1$ Given the first $n$ primes, we can label the $k$th prime as $p_k$.  So, what is the least common multiple(LCM) of {$p_1 - 1$, $p_2 - 1$, $p_3 - 1$, ..., $p_n-1$}?  In other words, if we subtract $1$ from each of the first $n$ primes, and wish to find the LCM of these new values, can we find a lower bounds for this LCM?
 A: By the most recent bound on Linnik's Theorem, there is an absolute constant $c$ such that for every prime $q < cp_n^{1/5}$, there is a prime $p < p_n$ such that $p \equiv 1 \pmod{q}$. Your least common multiple is therefore divisible by all primes below $cp_n^{1/5}$. The prime number theorem implies that the product of all primes below $cp_n^{1/5}$ is $e^{(c + o(1))p_n^{1/5}}$, and it follows that this is a lower bound on your lcm as well. 
Conjecturally there is a prime $p < p_n$ such that $p \equiv 1 \pmod{q}$ for every $q < cp_n^{1-\epsilon}$, and this would provide a lower bound of $e^{(c + o(1))p_n^{1-\epsilon}}$. On the other hand, your lcm is not divisible by any prime larger than $\frac{1}{2}p_n$ so, again using the prime number theorem, a straightforward upper bound is $e^{(\frac{1}{2} + o(1))p_n}$.
A: I believe there is a recursive formula for this:
Let L(k) = LCM({$p_1 - 1$, $p_2 - 1$, $p_3 - 1$, ..., $p_k$}
Then L(k+1) = L(k) ($p_{k+1}-1$) / GCD(L(k),$p_{k+1}-1$)
where GCD = greatest common divisor
