Is there a function that grows faster than exponentially but slower than a factorial?

Hint For exponential you have the growth $$\frac{a_{n+1}}{a_n}=\mbox{constant}$$

For the factorial you have the growth $$\frac{b_{n+1}}{b_n}=n$$

Take any function $g(n)$ which grows to infinity slower than $n$ and set $$\frac{c_{n+1}}{c_n}=g(n)$$

For example, $g(n)=\sqrt{n}$ gives the example $\sqrt{n!}$ given by AntonioVargas.

Another interesting example is $g(n)=\ln(n)$ which gives $$d_n =\prod_{k=2}^n \ln(k)$$

Given any two positive functions $f$ and $g$ such that $\frac{f(x)}{g(x)}$ tends to zero, let $j(x) = \sqrt{f(x)g(x)}$ (this is the geometric mean of $f$ and $g$).

Then $\frac{f}{ j} = \frac{j}{ g} = \sqrt{\frac{f}{g}}$ which must also tend to zero, so $j$ is an intermediate complexity class.

For variety, here are two striking examples of such a $j$ that demonstrate just how narrow the big-theta complexity classes are when applied to such rapidly growing functions:

• $j(n) = 3^n$
• $j(n) = (n-1)!$

The first demonstrates something you may have misunderstood: exponential growth is a much wider complexity class than merely $O(2^n)$.