# The quality of specific triples based on the first $2n$ prime numbers

I'm interested in the abc conjecture. Sometimes I do tasks as examples, with the purpose to understand more about this conjecture.

I am thinking about the definition of quality that provide us this Wikipedia as background to understand the formulation ABC Conjecture III.

Let $p_k$ the $kth$ prime number. And $\vartheta(x)$ the the first Chebyshev function.

Few minutes ago I've considered the following coprime triples $$a(n)=2^{n!}3^{n!}\ldots p_n^{n!},$$ $$b(n)=p_{n+1}^{n!}p_{n+2}^{n!}\ldots p_{2n}^{n!}$$ with $c(n):=a(n)+b(n)$. Thus the quality should be $$q(a(n), b(n),c(n))=\frac{\log(c(n))}{\vartheta(p_{2n})+\log\operatorname{rad}(c(n))}.\tag{1}$$

Question. I hope that previous simple calculation is right. Is feasible to continue the study of the quality of our triples for $n\geq 1$? Specifically I am asking if is it possible to deduce some aspect of the asymptotic behaviour of $q(a(n), b(n),c(n))$ as $n\to \infty$? Thanks in advance.

Of course I presume that from the literature we know very well upper and lower bounds for the first Chebyshev function, and the asymptotic behaviour of the first Chebyshev function. But I don't know what can I say about $c(n)$.

• Well $\theta(p_{2n}) \sim p_{2n} \sim 2n\log(2n)$ by the prime number theorem. This doesn't matter at all compared to $\log(c(n))$ since this is very stupidly bounded below by $\log(p_{2n}^{n!}) = n!\log(p_{2n}) \sim n!\log(2n\log(2n)) \sim n!\log(2n)$. So the quantity you are interested in is basically $\frac{\log(c(n))}{\log rad(c(n))}$ which is always greater than $1$. I'm not sure if it blows up to infinity... – mathworker21 Jul 9 '17 at 18:32
• Many thanks for your contribution.Feel free to add your calculations and reasonings as an answer @mathworker21 – user243301 Jul 9 '17 at 19:28

## 1 Answer

Note $\theta(p_{2n}) \sim p_{2n} \sim 2n\log(2n)$ by the prime number theorem. This doesn't matter at all compared to $\log(c(n))$ since this is very stupidly bounded blow by $\log(p_{2n}^{n!}) = n!\log(p_{2n}) \sim n!\log(2n\log(2n)) \sim n!\log(2n)$. So the quantity you are interested in is basically $\frac{\log(c(n))}{\log rad(c(n))}$ which is always greater than $1$. I'm not sure how this fraction behaves.

Note $p_i \not \mid c(n)$ for $1 \le i \le 2n$, so $c(n)$ just has large prime factors, but the deviation of $c(n)$ from $rad(c(n))$ of course concerns how many high powers of primes divides $c(n)$. I see no reason why $c(n)$ can't just be a product of distinct primes or just a large power of a prime, so my best guess is the fraction is unbounded (as $n$ increases) but still gets arbitrarily close to $1$ sometimes.

• Many thanks for your answer. – user243301 Jul 9 '17 at 19:53