The growth of the number of Fano complete intersection families

Question

I recently calculated the number (possible multidegrees) of Fano complete intersections of dimension $n$ , because I wanted to make the remark that it grows "very rapidly" as $n \rightarrow \infty$ The calculation itself is an "exercise" and in particular it is surely well-known. I would like to know how number theorists/combinatorists would think about the growth of this function.

By the adjunction formula a smooth complete intersection in $\mathbb{P}^k$ of multidegree $(d_{1}, \ldots d_{m})$ is Fano $\iff$ $d_1 + \ldots + d_m \leq k$.

This implies that the number of possible multidegrees of complete intersection Fano $n$-folds is $$\sum_{i=0}^{n-1} P(i) $$, where $P$ is that "partition function" studied by Euler and others.

I am aware that many authors don't consider $\mathbb{P}^n$ as a complete intersection. In which case we can take:

$$\sum_{i=1}^{n-1} P(i) .$$

Question: This function of course grows "very rapidly" as $n \rightarrow \infty$. How would a number theorist express that?

Q1. for example is there a "nicest" function $f(n)$ which approximates this?

Q2. Is there some technical term which describes precisely "how fast" this grows?

Also if there is a proof from first principles which is reasonably accessible for someone from another area then that would be great!

Answer 1

Hardy and Ramanujan obtained the asymptotic $$P(n) \approx \frac{1}{4n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}}$$ which can be summarized as saying that $P(n)$ grows roughly as the exponential of the square root of $n$.

Summing from $1$ to $n$ clearly increases the asymptotic by a factor of at most $n$, so we can still say $P(n)$ grows roughly as the exponential of the square root of $n$, or we can be more precise as follows

$$P (n-c) = \frac{1}{4 (n-c) \sqrt{3}} e^{ \pi \sqrt{\frac{2(n-c)}{3}} } = \frac{1}{4 (n-c) \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} + O\left( \frac{c^2}{ n^{3/2}}\right) } $$

which for $c = o(n^{3/4})$ is $$\approx \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} }$$ and for $c> o (n^{3/4})$ is exponentially smaller than $P(n)$ and can be ignored.

$$ \sum_{i=1}^{n-1} P(i) = \sum_{c=1}^{n-1} p(n-c) \approx \sum_{c=1}^{\infty} \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} }= \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}} \sum_{c=1}^{\infty} e^{ - \frac{ \pi c}{ \sqrt{6 n }} } \approx \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}} \frac{ \sqrt{6n}}{\pi} = \frac{1}{ \sqrt{8n } \pi} e^{ \pi \sqrt{\frac{2n}{3}}} $$