Meinardus theorem at use: problems with conditions 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.
 A: Let me first consider the ratio
$$G(z)=\frac{F(z)}{(1-z)^2}=(1-z)^{-2}\prod_{n\geq 1} \frac{1}{(1-z^{2n+1})^2}=\frac{1}{\left(\left(z;z^2\right){}_{\infty }\right){}^2},$$
with $(a;q)_\infty$ the q-Pochhammer symbol.
The series expansion
$$G(z)=\sum_{n=0}^\infty a_n z^n$$
has coefficients $a_n$ listed in OEIS:A022567.    
The large-$n$ asymptotics has been derived (using a variation of the Meinardus method) by V. Kotěšovec in A method of finding the asymptotics of q-series based on the convolution of generating functions, see page 8:
$$a_n=\tfrac{1}{4} 6^{-1/4}n^{-3/4}\,e^{\sqrt{2\pi^2 n/3}}\bigl(1+{\cal O}(n^{-1/2})\bigr).$$
Higher order terms are listed in the OESIS entry, the factor $1+{\cal O}(n^{-1/2})$ expands further into $1+q_1 n^{-1/2}+q_2 n^{-1}+{\cal O}(n^{-3/2})$ with
$$q_1=\frac{\pi}{12\sqrt 6}-\frac{\sqrt{27/2}}{8\pi},\;\;q_2=\frac{\pi^2}{1728}-\frac{45}{256\pi^2}-\frac{5}{64}.$$

The coefficients in the series expansion of $F(z)=\sum_{n=0}^\infty c_n z^n$ follow by equating $c_n=a_n-2a_{n-1}+a_{n-2}$ and expanding $n^{3/4}e^{-\sqrt{2\pi^2 n/3}}c_n$ in powers of $1/\sqrt n$. After some algebra I find
$$c_n= \frac{\pi^2}{6n}a_n\left(1+{\cal O}(n^{-1/2})\right).$$ 
 The terms $q_1$, $q_2$ do not contribute to this order.
A: 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.
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 $a_n = c_n + (2a_{n-1}-a_{n-2})$.
