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

Amritanshu Prasad
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Steven Spallone
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    In case it helps, the generating function for the number of bipartitions apparently satisfies a very nice ADE, similar to the one for partitions https://mathoverflow.net/a/47706/3032: $${{x^2}{{{f(x)}}^2}{{f^{{(iv)}}}(x)}}+{{({8{x^2}{f(x)}{{f'}(x)}}+{5x{{{f(x)}}^2}})}{{f'''}(x)}}-{21{x^2}{f(x)}{{{{f''}(x)}}^2}}+{{({12{x^{2}}{{{{f'}(x)}}^2}}-{15x{f(x)}{{f'}(x)}}+{4{{{f(x)}}^2}})}{{f''}(x)}}+{10x{{{{f'}(x)}}^3}}-{10{f(x)}{{{{f'}(x)}}^2}}=0$$ – Martin Rubey Apr 24 '17 at 17:39
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    I should add that it is clear that it does satisfy an ADE, what may be surprising is that it would be so nice. – Martin Rubey Apr 24 '17 at 17:42
  • After the OP's edit, the generating function is simply: $\prod_{k=1}^{\infty}\frac1{(1-x^k)^2}$. – T. Amdeberhan Apr 25 '17 at 02:51
  • If you want more values etc., see http://oeis.org/A000712 – Amritanshu Prasad Apr 25 '17 at 04:23
  • Just to clarify: I used the definition "square of the generating function for partitions" for the generating function for bipartitions. – Martin Rubey Apr 25 '17 at 05:03
  • Numerical data suggests that $p(a)p(b)>p(n)$ when $a+b=n\geq 10$ and $a,b\geq 2$. I haven't found a nice combinatorial proof though. – Harry Huang Apr 25 '17 at 09:20
  • Numerical data also suggests that, for $n>101$, $p_2(n) > n^2 p(n)$, and for $n>321$, $p_2(n)>n^3 p(n)$, and for $n>701$, $p_2(n)>n^4 p(n)$. Could it be that, for every positive integer $k$, $p_2(n) > n^k p(n)$, provided that $n$ is sufficiently large (how large depends on $k$). – Amritanshu Prasad Apr 25 '17 at 11:07

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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.

Janne Kokkala
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