Asymptotics of $\operatorname{lcm} ((2-1), (3-1), (5-1), (7-1), (11-1), \dotsc, p_n-1 )$

Question

$\DeclareMathOperator\lcm{lcm}$Let $p_k$ be the $k$th prime number. Set $$L(n) = \lcm(p_1-1, p_2-1, \dotsc, p_n-1). $$

What can we say about the growth of $L(n)$? Trivially, one has that $L(n) < p_1p_2 \dotsb p_n$. One can do better than that since after $2$, every prime is odd. Thus, if $n \geq 3$, one has $$L(n) \leq \frac{p_1p_2 \cdots p_n}{2^{n-1}}.$$ And one can continue with larger primes, using explicit versions of Dirichlet's theorem on arithmetic progressions, to put in powers of 3, and then 5, and so on in the denominator in the same way.

In the other direction, one can use Linnik's theorem to get lower bounds on $L(n)$ but these are weak.

$L(n)$ is A058254 in the OEIS, but that doesn't give any non-trivial bounds, even heuristically.

Question: Can we get an asymptotic to $\log L(n)$, ideally with explicit bounds? Closely related: is it true that $\log L(n) = o(p_n)$? This is a natural bound to ask about because $\log \lcm (p_1, p_2 \cdots p_n ) = \sum_{p \leq p_n} \log p \sim p_n$ by the Prime Number Theorem.

For the application I'm interested in, I'd like to then use this to get explicit bounds on $\lcm (p_{n}-1, p_{n+1}-1, p_{n+2}-1, \dotsc, p_{n+m}-1)$ where $m$ is at least about $n^2$, and I would like that $\log \lcm (p_{n}-1, p_{n+1}-1, p_{n+2}-1, \dotsc, p_{n+m}-1)$ grows slower than $\sum_{p_n \leq p \leq p_{n+m}} \log p \sim p_{n+m}$. So it may be possible to get a useful bound even without an asymptotic for $L(n)$.

Answer 1

It is true that $\ln L(n)=o(p_n)$ for $n\to+\infty$. To prove it, let us get a bound for contribution of large primes into $\ln L(n)$ and then estimate the contribution of the rest trivially. Choose a parameter $R<\sqrt p_n$. Let $M(n)=\mathrm{lcm}[1,2,\ldots,p_n]$. Then we have $\ln M(n)=\psi(p_n)\sim p_n$, where $\psi$ is Chebyshev's function. Now, $$ \sum_{p^k\mid L(n), p<p_n/R}\ln p\leq \sum_{p^k\mid M(n),p<p_n/R}\ln p=O(p_n/R). $$ Next, if $p\geq p_n/R$ contributes to $\ln L(n)$, then there is a natural number $k\leq R$ and a prime $q\leq p_n$ such that $q-1=kp$. We will evaluate the terms that come from different $k$. As $q\leq p_n$, for a given $k$ we should have $p<p_n/k$. The number of primes $p<x$ such that $kp+1$ is prime is known to be bounded by $$ C\frac{k}{\varphi(k)}\frac{x}{\ln^2 x} $$ (see, for example, H. Iwaniec, "Sieve methods", Section 2.6). Applying this to our case, we get the bound of the form $$ O\left(\frac{1}{\varphi(k)}\frac{p_n}{\ln^2 p_n}\right) $$ for the number of primes we are interested in, and their contribution is, therefore, $$ O\left(\frac{1}{\varphi(k)}\frac{p_n}{\ln p_n}\right) $$ for any $k\leq R$. Summing over all $k$ and using the fact that $$ \sum_{k\leq R}\frac{1}{\varphi(k)}=C_1\ln R+O(1), $$ we obtain $$ \sum_{p^k\mid L(n), p\geq p_n/R}\ln p=O\left(\frac{p_n\ln R}{\ln p_n}\right). $$ Combining the first and the second estimate and choosing $R=\ln p_n$, we get $$ \ln L(n)=O\left(\frac{p_n\ln\ln p_n}{\ln p_n}\right). $$ The results used in the proof do not rely on anything particularly non-explicit, so the constants can be made completely explicit. Also, the bound for the number of primes of the form $kp+1$ is conjectured to be sharp, so the bound should probably also be asymptotically sharp, unless I am missing something.

Answer 2

Here is $\log(L(n))/p_n$ for $10^2\leq n\leq 10^4$: