In big-O notation the complexity class $O(2^n)$ is named "exponential". The complexity class $O(n!)$ is named "factorial".
I believe that $f(n) = O(2^n)$ and $g(n) = O(n!)$ means that $\dfrac{f(n)}{g(n)}$ goes to zero in the limit as $n$ goes to infinity.
Is there any known function between the factorial and exponential complexity classes?
In other words is there any known function $j(n)$ that dominates every function in $O(2^n)$, such that:
$$ (j(n) \neq O(2^n)) \wedge (j(n) = O(n!)) \wedge (n! \neq O(j(n)))
$$
or, informally, $j(n)$ grows asymptotically strictly faster than $2^n$ but not as fast as $n!$?
Or perhaps it has been proven that no such function can exist?
Note: this may seem like a Computer Science question, but in fact I am attempting to prove that any periodic, convergent power series must have coefficients whose inverses grow asymptotically as fast as $n!$ but not faster. I think I can show they most grow faster than $O(2^n)$, but that does not prove they are in $\Theta(n!)$ unless there is no complexity class between $O(2^n)$ and $O(n!)$.
Best Answer
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 $$d_n =\prod_{k=2}^n \ln(k)$$