# How do i show that $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$?

I have run some computations for some finitely primes to know the nature of the ratio below (the product of the first few primes over the sum of them), specifically if it is an integer for finitely many positive integers $n$.

My question is: How do i show thiat $\displaystyle\frac{\prod_{k=1}^np_k}{\sum_{k=1}^{n}p_k}$ is an integer for finitely many $n$'s?

Note 1: $p_1<p_2<\cdots$ is the sequence of prime numbers.

Note 2: For $n=3$ the ratio equals $3$.

• For $n=3$ the ratio is $3$, not $7$. Personally, I would think that the ratio is an integer for infinitely many $n$'s, but I don't think we have the technology to prove (or disprove) that. In other words, I think your question is too hard. – GH from MO Jan 24 '17 at 0:34
• Also, it is unfortunate (inappropriate) to phrase a question like "How do I show this or that". This site is about mathematics, not your personal experience or abilities. Mathematics does not know about people. A better way to phrase your question would be: "Is it true that etc." – GH from MO Jan 24 '17 at 0:38
• ok, just i w'd like if there are some advanced theory in number theory help me to show the titled problem, and thanks for your constrictuve comments – zeraoulia rafik Jan 24 '17 at 0:40
• I agree with @GH: there are almost surely infinitely many $n$'s. The problem can be restated as "How often is $S_n = \sum_{k=1}^n p_k$ squarefree with no prime factors larger than $p_n$?" Squarefree numbers are plentiful. Since $p_n > \sqrt{S_n}$, numbers satisfying other condition are plentiful too. One would need an extraordinarily good reason to suspect the number of $n$'s is finite. – user13113 Jan 24 '17 at 1:28
• The $n$ for which this is an integer form OEIS sequence A051838. There are also several related sequences. Unfortunately there is no discussion there of whether the sequence is infinite. – Robert Israel Jan 24 '17 at 3:41

I did experiments, and it looks like the density of those $n$ for which the ratio is an integer approaches a limit, and the limit is approximately $0.2187.$ Of this writing I went up to $1000000$ - (with Mathematica); the convergence seems reasonably rapid, however. Note that the density is not the same as the density of square-free integers - I am not sure if there is any sort of heuristic that predicts what it should be.

However, another experiment shows that the density of square-free numbers amongst sums of the first $n$ primes is about $6/\pi^2,$ as expected. There are two things at play here: the sum of the first $n$ primes is square free, and also its largest prime divisor is no bigger than the $n$-th prime. Now, the sum of the first $n$ primes is of order of $\frac{n^2}{2 \log n},$ and the expected value of the largest prime divisors of numbers smaller than $N$ is $\frac{\pi^2}{12} \frac{n^2}{\log n},$ not sure what the distribution looks like. Assuming the events of square-freeness and "reasonable smoothness" are independent should give a heuristic asymptotic.

UPDATE Now up to $10^7,$ and the probability of integrality is slowly declining (now $0.2174$). So, not sure what to think.

• $6/\pi^2$ times $1-\log 2$ is about $0.186544$. – Lucia Jan 24 '17 at 4:25
• en.wikipedia.org/wiki/Dickman_function – Douglas Zare Jan 24 '17 at 4:27
• @Lucia Why $1-\log 2?$ – Igor Rivin Jan 24 '17 at 4:35
• Dickman function of 2 is $1-\log 2$ --- you're looking for $S= \sum_{j\le n}p_j$ to be essentially $\sqrt{S}$ smooth. There are various $\log \log$'s involved, so I expect the convergence to be slow. – Lucia Jan 24 '17 at 4:48
• @Lucia Ah, I see. Well, the experimental evidence is consistent with this. – Igor Rivin Jan 24 '17 at 4:51