I am needing to use the asymptotic formula for the partition number, $p(n)$ (see here for details about partitions).

The asymptotic formula always seems to be written as,

$ p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi \sqrt{\frac{2n}{3}}}, $

however I need to know the order of the omitted terms, (i.e. I need whatever the little-o of this expression is). Does anybody know what this is, and a reference for it? I haven't been able to find it online, and don't have access to a copy of Andrews 'Theory of Integer Partitions'.

Thank you.

  • $\begingroup$ I believe the asymptotic nature of the Hardy-Ramanujan formula, notwithstanding its use to get exact values of $p(n)$, means that a "little-o" notation for omitted terms would be misplaced. $\endgroup$ – hardmath Jul 5 '11 at 16:14
  • $\begingroup$ @Hardmath, I'm about to answer my own question (!), but also justify that there is a little-o representation, since in fact if f is asymptotically equivalent to g, then f = (1 + o(1))g... So as pointed out, I've now answered my original question... silly me. $\endgroup$ – owen88 Jul 5 '11 at 17:18
  • 3
    $\begingroup$ Okay, I thought perhaps the exact convergent series given by Rademacher (1937) that refines the Hardy-Ramanujan formula (which forms the first term of the series) and its order of convergence might be of interest. G. Andrews has a chapter about this in his book Theory of Integer Partitions. $\endgroup$ – hardmath Jul 5 '11 at 17:34

The original paper addresses this issue on p. 83:

$$ p(n)=\frac{1}{2\pi\sqrt2}\frac{d}{dn}\left(\frac{e^{C\lambda_n}}{\lambda_n}\right) + \frac{(-1)^n}{2\pi}\frac{d}{dn}\left(\frac{e^{C\lambda_n/2}}{\lambda_n}\right) + O\left(e^{(C/3+\varepsilon)\sqrt n}\right) $$ with $$ C=\frac{2\pi}{\sqrt6},\ \lambda_n=\sqrt{n-1/24},\ \varepsilon>0. $$

If I compute correctly, this gives $$ e^{\pi\sqrt{\frac{2n}{3}}} \left( \frac{1}{4n\sqrt3} -\frac{72+\pi^2}{288\pi n\sqrt{2n}} +\frac{432+\pi^2}{27648n^2\sqrt3} +O\left(\frac{1}{n^2\sqrt n}\right) \right) $$


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