I am working on an enumerative problem related to knot theory, and I have found the following generating function


$$F(z)=\prod\_{n\geq 1} \frac{1}{(1-z^{2n+1})^2}.$$
I am interested on getting asymptotic estimates for this GF. I am not an expert on partition function theory, but it seems that trying to apply Meinardus Theorem on this context should be a good idea (in this case, the sequence of coefficients to be studied is $a\_n=2$ for $n=2k+1$ and $\geq 3,$ and $a\_n=0$ in the rest of the cases.


Once trying to apply Meinardus Theorem (as it is stated in the reference book of George Andrews: The theory of partitions) I have seriuos troubles with condition III, namely that


$$f(t,y)=Re(g(e^{-t-2\pi i y}))-g(e^{-t})\leq -C t^{-\varepsilon}$$


for $g(z)=\sum\_{n\geq 1} a\_n z^n$, $|y|\leq 1/2$, $t$ small enough and convenient choices of positive $C$ and $\varepsilon$. Essentially, the problem I am finding is that when taking $y=1/2$ and $t$ small (or in a neighbourhood), the value I get for $f(t,y)$ is $-2t+o(t)$, which cannot fit with the condition $\leq -C t^{-\varepsilon}$.


So: which are the alternatives that one can use to get the assymptotics in such a situation (without having to make the computation of the contour integral computation?)


I was thinking also to try to get results exploiting the asymptotics of the partition function, but I do not see a direct way.



Meinardus theorem at use: problems with conditions
This is a postscript to Carlo Beenakker's answer, a combinatorial explanation for the relationship between $a\_n$ and $c\_n$.


Your $F(z)=\sum\_n c\_nz^n$ is the generating function for the number of partitions of $n$ into odd parts 3 or greater where there are two kinds of each part. Carlo's $G(z) = \sum\_n a\_nz^n$ allows two kinds of 1's also, matching one of the descriptions given for [A022567](https://oeis.org/A022567).


Rewrite the equation in Carlo's comment as $a\_n = c\_n + 2a\_{n-1} - a\_{n-2}$. To see this, condition the partitions counted by $a\_n$ by whether they contain (either kind of) 1 as a part. Those with no 1's are given by your $c\_n$. To build these partitions of $n$ with 1's from smaller partitions, take the partitions of $n-1$ counted by $a\_{n-1}$ and include a $1\_1$. Repeat and include a $1\_2$ to complete the $2a\_{n-1}$ term. Some partitions of $n$ arise twice, though---precisely those that include both a $1\_1$ and a $1\_2$; the number of those is $a\_{n-2}$ by adding $1\_1$ and $1\_2$ to each partition of $n-2$. So $a\_n = c\_n + (2a\_{n-1}-a\_{n-2})$.


