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The numerator counts the number of different adjacency matrices. I think Sterlings approximation helps to anwser my question but I fail to derive the answer. So, is there a polynomial function $g(x)$ such that $$f(x) \leq g(x)$$

John D.
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  • Apply $\log$ and use that $\log (n!)\ge \frac12 n \log n$. See http://math.stackexchange.com/a/543644/589. – lhf Nov 01 '13 at 13:18

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It's not polynomially bounded. For $n \geqslant 3$, we have $\frac12(n^2-n) \geqslant \frac13 n^2$, so

$$\frac{2^{\frac12(n^2-n)}}{n!} > \frac{2^{\frac13 n^2}}{n^n} = \left(\frac{2^{n/3}}{n}\right)^n.$$

$2^{n/3}$ has exponential growth, so altogether we have superexponential growth.

Daniel Fischer
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