TITLE: Is there a closed form or approximation to $\sum_{i=0}^n\binom{\binom{n}{i}}{i}$
QUESTION [12 upvotes]: I tried to calculate the sum
$$
\sum_{i=0}^n\binom{\binom{n}{i}}{i}
$$
but it seems that all my known methods are poor for this.
Not to mention the intimate recursion, that is
$$
\sum_{i=0}^n\binom{\binom{{\binom{n}{i}}}{i}}{i}
$$
Any ideas?

REPLY [7 votes]: Lets try to find the maximum of
$$f(i)=\binom {\binom ni}i$$
Consider $i\in \{n/4,3n/4\}$. In this range a good approximation (from the central limit theorem) is
$$\binom ni\simeq\frac{2^n}{\sqrt{\frac 12n\pi}}e^{-\frac{(i-n/2)^2}{n/2}}$$
This is far larger than $i$, so we have
$$f(i)=\binom {\binom ni}i\simeq\frac{\binom ni^i}{i!}\simeq\frac{2^{ni}}{i!(\frac 12n\pi)^{i/2}}e^{-\frac{i(i-n/2)^2}{n/2}}$$
Taking logarithms and using the Stirling approximation
$$\log f(i)\simeq ni\log 2 -\frac i2\log(\frac 12n\pi)-i\log i+i-\frac 12\log(2\pi i)-\frac{i(i-n/2)^2}{n/2}$$
These terms are all negligable compared to the first and last, so
$$\log f(i)\simeq ni\log 2 -\frac{i(i-n/2)^2}{n/2}$$
$$=-\frac 2ni^3+2i^2+(\log2-\frac{1}2)ni$$
Therefore
$$\frac{d\log f(i)}{di}\simeq-\frac 6ni^2+4i+(\log 2-\frac 12)n$$
With roots
$$i=\frac{-4\pm\sqrt{16+24(\log 2-\frac 12)}}{-12/n}\simeq-0.0452n\text{ and }0.712n$$
The first is a minimum (and outside the sensible range) but the second is a maximum. Let $\alpha\simeq0.712n$ be this maximum.
So using our above approximation
$$\log f(\alpha)\simeq [-2\times 0.712^3+2\times 0.712^2+(\log2-\frac{1}2)\times 0.712]n^2\simeq0.430n^2$$
So bounding $\sum_i f(i)$ below by its largest term gives an approximation of about $e^{0.430n^2}$, which is in line with Claude Leibovici's empirical results.
This isn't a rigorous lower bound. The main problem is that $\alpha$ isn't an integer, and so $f(i)$ might not actually attain this maximum. Since $\alpha$ is within $\frac 12$ of an integer you can fix this by evaluating the second derivative of $\log f$ at $\alpha$ an use this to approximate $f(\alpha\pm\frac 12)$.