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Consider the set $A$ of natural numbers which are of the form $k^3-p$, for $k$ a positive integer and $p$ a positive prime. Does $A$ have a density (of any of the usual kinds for sets of natural numbers) and if so, what is it?

$A$ contains a lot of numbers, and it seems somewhat difficult to me to prove that a particular number is not in $A$, except that most cubes are not in $A$.

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A naive approach would be to say that the chance $n$ is equal to $k^3-p$ for a given $k$ is about $\frac{1}{\ln(k^3-n)}$. Then the chance $n$ is not equal to $k^3-p$ for any $k$ is $$\prod_{k=\sqrt[3]{n}}^\infty {1-\frac{1}{\ln(k^3-n)}}$$ as this goes to zero, "all" numbers should be in A.

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