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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.

Gary
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owen88
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  • 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. – hardmath Jul 05 '11 at 16:14
  • @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. – owen88 Jul 05 '11 at 17:18
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    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. – hardmath Jul 05 '11 at 17:34

2 Answers2

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

Charles
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The asymptotic formula is an approximation of the leading term within an asymptotic expansion formulated by Ramanujan and Hardy: $$ p(n) = \frac{2\sqrt 3 }{24n - 1}\sum_{m = 1}^{\lfloor c \sqrt{n} \rfloor} \frac{A_m (n)}{\sqrt m } \left( 1 - \frac{6m}{\pi \sqrt {24n - 1} } \right)\exp \left( \frac{\pi }{6m}\sqrt {24n - 1} \right) + \mathcal{O}(n^{ - 1/4} ) $$ as $n\to+\infty$. Here, $c$ denotes an arbitrary fixed positive number, $$ A_m(n) = \sum_{0 \le k < m, \, (k, m) = 1} \mathrm{e}^{ \pi \mathrm{i} \left( s(k, m) - 2 nk/m \right) } $$ and $s(k,m)$ is the Dedekind sum. Later, Rademacher found an exact, convergent expansion for $p(n)$: $$ p(n) = \frac{{4\sqrt 3 }}{{24n - 1}}\sum_{m = 1}^\infty {\frac{{A_m (n)}}{{\sqrt m }}}\! \left( {\cosh \left( {\frac{\pi }{{6m}}\sqrt {24n - 1} } \right) - \frac{{6m}}{{\pi \sqrt {24n - 1} }}\sinh \left( {\frac{\pi }{{6m}}\sqrt {24n - 1} } \right)} \right). $$ In this more recent paper, the following Poincaré-type asymptotic expansion for $p(n)$ was obtained: $$ p(n) \sim \frac{\exp \left( \pi \sqrt {\frac{2n}{3}} \right)}{4\sqrt 3 n}\sum_{m = 0}^\infty\frac{\omega_m }{n^{m/2}}=\frac{\exp \left( \pi \sqrt {\frac{2n}{3}} \right)}{4\sqrt 3 n} \left( {1 - \frac{{\pi ^2 + 72}}{{24\sqrt 6 \pi }}\frac{1}{{n^{1/2} }} + \frac{{\pi ^2 + 432}}{{6912}}\frac{1}{n} - \ldots } \right), $$ with an explicit expression for the coefficients $\omega_m$ given by $$ \omega_m = \frac{(-1)^m}{(4\sqrt 6)^m }\sum_{k=0}^{\lfloor{(m + 1)/2}\rfloor}\binom{m+1}{k}\frac{m+1-k}{(m+1-2k)!}\left(\frac{\pi}{6}\right)^{m-2k} . $$

Gary
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