Numerical evidence suggests that $p\_2(n) \geq n p(n)$ for large $n$.


Here $p(n)$ is the number of partitions of $n$, and $p\_2(n)$ is the number of bipartitions of $n$, i.e., ordered pairs of partitions $(\alpha,\beta)$ with $|\alpha|+|\beta|=n$. I would be grateful for a proof or a disproof.


Edit: Here is a formula for $p\_2(n)$:


$$p\_2(n)= \sum\_{a+b=n}p(a)p(b),$$
with $0 \leq a,b \leq n$ integers. This gives $p\_2(0)=1, p\_2(1)=2,p\_2(2)=5$.



Counting Bipartitions
Asymptotically
$$
p(n) \sim \frac{1}{4 n \sqrt{3}} \exp\left(\pi \sqrt{\frac{2n}{3}}\right),
$$
which gives (Well, technically this is valid only for even $n$, but odd $n$ wouldn't be much different.)
$$
\frac{p(n/2)^2}{p(n)} \sim \frac{1}{n\sqrt{3} } \exp\left( (2-\sqrt{2}) \pi \sqrt{\frac{n}{3}} \right).
$$


Because $p\_2(n)>p(n/2)^2$, we know that $p\_2(n)/p(n)$ grows faster than any polynomial, as Amritanshu Prasad suggested in a comment.


