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The approximation is usually written as $$ p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}} $$

But every graph I've seen makes it look like this is an asymptotic upper bound for $p(n)$. Is that true?

Kidplaydough
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2 Answers2

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It might be interesting to note, after having the answer for all $n\ge n_0$ in the comment, that we can easily prove a weaker bound for all $n\ge 1$. In fact, in Tom Apostol's book on analytic number theory, section $14.7$ formula $(11)$, the following upper bound for all $n\ge 1$ for $p(n)$ is proved in a completely elementary way: \begin{equation*} p(n)<\frac{\pi}{\sqrt{6n}}e^{\alpha\sqrt{n}} \quad \text{for all} \;\; n\ge 1 \end{equation*} Here $\alpha=\pi\sqrt{2/3}$.

Edit: The asymptotic formula can be written as follows:

$$ p(n)=e^{\pi\sqrt{\frac{2n}{3}}} \left( \frac{1}{4n\sqrt3} -\frac{72+\pi^2}{288\pi n\sqrt{2n}} +\frac{432+\pi^2}{27648n^2\sqrt3} +O\left(\frac{1}{n^2\sqrt n}\right) \right) $$

Dietrich Burde
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    Possible typo: should $n$ (under the square root) be deleted in the expression for the constant $\alpha$? Anyway, your bounding function is (much) bigger than the OP's, so it doesn't say anything about whether the OP's function is an upper bound. – John Bentin Dec 15 '19 at 13:02
  • @JohnBentin It doesn't say this, yes, but this was my point. Since the question is answered by the comment, it might be interesting to note that we can show this weaker estimate easily, which then holds for all $n\ge 1$. – Dietrich Burde Dec 17 '19 at 15:25
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This is true. In fact, $$ \frac{{{\rm e}^{\pi \sqrt {2n/3} } }}{{4n\sqrt 3 }}\left(1-\frac{1}{2\sqrt n}\right)< p(n) < \frac{{{\rm e}^{\pi \sqrt {2n/3} } }}{{4n\sqrt 3 }}\left(1-\frac{1}{3\sqrt n}\right) $$ for any $n\ge 1$. This is Theorem $1.3$ of this paper.

Gary
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