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.

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 $p_2(0)=1, p_2(1)=2,p_2(2)=5$.