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?

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    $\begingroup$ Lower than what? $\endgroup$ Dec 3, 2014 at 14:28
  • $\begingroup$ @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? $\endgroup$
    – Matt Groff
    Dec 3, 2014 at 14:30
  • $\begingroup$ Your LCM is $\lambda(p_n\#)$. May be you get lower bounds from Carmichael/primorial formulas. $\endgroup$ Dec 3, 2014 at 14:32
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    $\begingroup$ This is sequence A058254 in OEIS. Unfortunately nothing is said about its asymptotic behavior. $\endgroup$ Dec 3, 2014 at 15:07
  • $\begingroup$ @gammatester: I don't think this would do very much; the bounds on $\lambda$ are very weak. $\lambda(48665323350093056511370687590824766511200)=2520,$ for example. $\endgroup$
    – Charles
    Dec 3, 2014 at 16:35

2 Answers 2


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


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