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Let $(x,q)_k$ denote the $q$-Pochhammer symbol and $\mathrm{Coeff}_n\hspace{0.1em} f(q)$ denote the coefficient of $q^n$ in $f(q)$. What I want to know is the large-$n$ asymptotic of \begin{align} \mathrm{Coeff}_n\hspace{0.1em} \frac{1}{(q;q)_k} \,, \end{align} and also \begin{align} \mathrm{Coeff}_n\hspace{0.1em} \frac{1}{(q;q)_\infty} =\mathrm{Coeff}_n\hspace{0.1em} \bigg( 1 + q + 2q^2 + 3q^3 + 5q^4 + 7q^5 + 11q^6 + 15q^7 + \cdots \bigg) \,. \end{align} How can I find these? I tried to use Mathematica, like this, \begin{align} & \texttt{DiscreteAsymptotic}\Big[ \texttt{SeriesCoefficient}\big[ 1/\texttt{QPochhammer}[q,q,k], \{q,0,n\} \big], n{\to}\infty \Big] \,,\\ & \texttt{DiscreteAsymptotic}\Big[ \texttt{SeriesCoefficient}\big[ 1/\texttt{QPochhammer}[q,q], \{q,0,n\} \big], n{\to}\infty \Big] \,, \end{align} but unfortunately it returned nothing. I would appreciate any help or suggestions.

One of my attempts was to use somehow the following $q$-binomial expression, \begin{align} \frac{1}{(x;q)_N} = \sum_{k=0}^\infty \binom{N+k-1}{k}_{\hspace{-0.15em} q} x^k \,, \end{align} but I am not sure whether if plugging in $q$ to $x$ results in anything useful...

Lightcone
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    Ramanujan and Hardy showed the asymptotic formula $$ \text{Coeff}n \frac{1}{{(q;q)\infty }} = p(n) \sim \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi\sqrt {\frac{2n}{3}} } \right). $$ Here $p(n)$ is the partition function. An exact formula with asymptotic character was given by Rademacher. For more details, see here. – Gary May 24 '22 at 00:15
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    Interesting! Thank you very much. For finite $N$, I think this may provide an answer. – Lightcone May 24 '22 at 02:42

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Too long for a comment.

For the asymptotic, @Gary already gave the answer in comments.

Looking at sequence $A000041$ in $OEIS$, you would find a very good approximation proposed in year $2016$ by Vaclav Kotesovec. It can rewrite $$a_n=\frac{1}{4 \sqrt{3} n}\,\exp\Bigg[\,\pi \sqrt{\frac{2 n}{3}\Bigg[1+\sum_{k=1}^\infty \frac {a_i}{n^{\frac {k+1}2} }\Bigg] }\,\,\, \Bigg]$$ the first coefficients being

$$a_1=\frac{\pi ^2-18}{4 \sqrt{6} \pi ^3}\qquad a_2=\frac{3}{4 \pi ^4}\qquad a_3=\sqrt{\frac{3}{2}}\,\frac{36-5 \pi ^2}{32 \pi ^5}\qquad a_4 =\frac{3348-120 \pi ^2+5 \pi ^4}{1920 \pi ^6}$$

For $n=10^3$, the relative error is $1.31$%

Claude Leibovici
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