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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 $k$?

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

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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.

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

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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.

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