# 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?

• Lower than what? Dec 3, 2014 at 14:28
• @barakmanos: Sorry, I should explain. I'm searching for an asymptotic function/expansion that the LCM is greater than or equal to. Does this help? Dec 3, 2014 at 14:30
• Your LCM is $\lambda(p_n\#)$. May be you get lower bounds from Carmichael/primorial formulas. Dec 3, 2014 at 14:32
• This is sequence A058254 in OEIS. Unfortunately nothing is said about its asymptotic behavior. Dec 3, 2014 at 15:07
• @gammatester: I don't think this would do very much; the bounds on $\lambda$ are very weak. $\lambda(48665323350093056511370687590824766511200)=2520,$ for example. Dec 3, 2014 at 16:35

## 2 Answers

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 $$e^{(\frac{1}{2} + o(1))p_n}$$.

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