Are there arbitrarily long prime deserts?

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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 $k$?

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 E. Caicedo, Jan 23 '13 at 01:35
possible duplicate of What is the maximum number of consecutive composite numbers possible?. Also related are http://math.stackexchange.com/q/168103/ and http://math.stackexchange.com/questions/20001/how-many-consecutive-composite-integers-follow-k1 (This comment might be autodeleted if the question is closed, but the links remain on the right.) — Jonas Meyer, Jan 23 '13 at 01:41
And it is expected that every even $n$ is the length of a gap infinitely often. — Andrés E. Caicedo, Jan 23 '13 at 01:42
@hardmath: I don't think you can refer to this as prime gap because there is no assumption that $n$ and $n+k+1$ are prime. Obviously this makes a difference because you can have a prime desert of any length, but this is not true of prime gaps (for example, you can't have an odd prime gap other than $1$). — Michael Albanese, Jan 23 '13 at 01:52
@MichaelAlbanese: Yes, but prime gaps are maximal prime deserts (those which cannot be extended at either end). — hardmath, Jan 23 '13 at 03:12

score 8 · Accepted Answer · answered Jan 23 '13 at 01:37

8

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.

answered Jan 23 '13 at 01:37

Geoff Robinson

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score 5 · Answer 2 · answered Jan 23 '13 at 01:48

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 $\lim_{x\to\infty}\frac{x/(k+1)}{x/\ln(x)}=\lim_{x\to\infty}\frac{\ln(x)}{k+1}=\infty$.

score 3 · Answer 3 · answered Jan 23 '13 at 01:37

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.

Are there arbitrarily long prime deserts?

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