Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
What is the maximum number of consecutive composite numbers possible?

Define a prime desert of length $k$ to be a sequence of numbers $n + 1, n + 2, ..., n + k $ such that $n + i$ is composite for $1 \le i \le k$. So my question is given a positive integer $k$ is there a prime desert of length $k$?

share|cite|improve this question

marked as duplicate by Jonas Meyer, Rahul, hardmath, Henry T. Horton, rschwieb Jan 23 '13 at 2:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes. The usual approach is to look at the numbers $(k+1)!+2,(k+1)!+3,\dots,(k+1)!+k,(k+1)!+(k+1)$. – Andrés Caicedo Jan 23 '13 at 1:35
A common name for this is prime gap. – hardmath Jan 23 '13 at 1:37
Related: – Jonas Meyer Jan 23 '13 at 1:38
possible duplicate of What is the maximum number of consecutive composite numbers possible?. Also related are and… (This comment might be autodeleted if the question is closed, but the links remain on the right.) – Jonas Meyer Jan 23 '13 at 1:41
And it is expected that every even $n$ is the length of a gap infinitely often. – Andrés Caicedo Jan 23 '13 at 1:42

3 Answers 3

up vote 5 down vote accepted

Yes there is always such a prime desert. Consider $(k+1)! +1, (k+1)! +2,\ldots (k+1)! +(k+1).$ Then for $2 \leq i \leq k+1,$ we see that $(k+1)! +i$ is divisible by $i,$ but is strictly greater than $i,$ so can't be prime.

share|cite|improve this answer

Here's an alternative approach that uses some heavier machinery.

We know that, if $\pi(x)$ denotes the number of primes $\leq x$, then $\pi(x)$ grows at roughly the same rate as $\frac{x}{\ln x}$, in the sense that $\lim_{x\to\infty} \frac{\pi(x)}{x/\ln (x)}=1$. (This statement is called The Prime Number Theorem.) Now, if the largest prime desert were of size $k$, then $\pi(x)\geq\frac{x}{k+1}$. But $\lim_{x\to\infty}\frac{x/(k+1)}{x/\ln(x)}=\lim_{x\to\infty}\frac{\ln(x)}{k+1}=\infty$.

share|cite|improve this answer

Yes. It is a nice exercise to show that the numbers $k!+2,\dots, k!+k$ are all composite and thus this is a prime desert of length $k-1$ (so we can get arbitrarily long ones by letting $k$ be large enough). We can of course do better for how large the numbers are by for example starting with $k!-2$ and going down to $k! - k$. I do not know how small we can make the smallest number in order to get a desert of length $k$ though.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.