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!



The growth of the number of Fano complete intersection families
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.


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


