# 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.

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 $$a_n = c_n + (2a_{n-1}-a_{n-2})$$.

• Glad this was helpful. Curious what kind of knot theory enumeration brought this up. – Brian Hopkins Apr 3 '20 at 14:59

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.

• great, many thanks! what do you mean by "smaller by a factor $\Pi^2/6n$"? This means that $[z^n]F(z)=\Pi^2/6n a_n$? if yes, how you reach (in a fast way, I assume) such result? – Johnny Cage Apr 2 '20 at 20:31
• if $F(z)=\sum_n c_nz^n$, then $c_n=a_n-2a_{n-1}+a_{n-2}$; inserting the asymptotics for $a_n$ gives $c_n=(\pi^2/6)n^{-1}a_n[1+{\cal O}(n^{-1/2})]$. – Carlo Beenakker Apr 2 '20 at 20:46
• While I do not doubt the final result, it seems to me that the derivation in your comment doesn’t work: if you just plug in the bound from your post then the error term will be bigger than the main term. It can be fixed by a more precise expansion, though (up to $o(1/n)$), and I would be very surprised if it would not follow from the same method. – Aleksei Kulikov Apr 4 '20 at 14:20
• @AlekseiKulikov --- thanks, I have expanded the derivation in the answer box, which I think is now complete (the higher order terms in the expansion of $a_n$ do not seem to contribute to $c_n$ to order $1/\sqrt n$). – Carlo Beenakker Apr 4 '20 at 15:31