I cannot seem to find a precise definition of what "super-exponential" is supposed to refer to when one's talking about an algorithm's time complexity.

For instance, if an algorithm runs for $nC(n-1)$ steps, where $C(\cdot)$ is the Catalan number, is this algorithm super-exponential in $n$?

  • $\begingroup$ We discourage posts that simply state a question out of context, and expect the community to solve it. Can you provide a few urls or books where you have seen this "super-exponential"? Or do you want to initiate the usage of that term? $\endgroup$
    – John L.
    Nov 12, 2018 at 14:03
  • $\begingroup$ Here is a definition of super-exponential growth: services.math.duke.edu/education/postcalc/growth/growth4_1.html $\endgroup$
    – Ahti Ahde
    Aug 19, 2020 at 16:46

2 Answers 2


"Super-exponential" just means more than exponential, so a function is super-exponential if it grows faster than any exponential function. More formally, this means that it is $\omega(c^n)$ for every constant $c$, i.e., if $\lim_{n\to\infty} f(n)/c^n=\infty$ for all constants $c$.

Conversely, a function is "sub-exponential" if it is $o(c^n)$ for every constant $c>1$, i.e., $\lim_{n\to\infty} f(n)/c^n=0$ for all constants $c>1$.

Asymptotically, the $n$th Catalan number is $\Theta(4^n\, n^{-3/2})$. This is $o(4^n)$, so the Catalan numbers are not super-exponential; it is $\omega(2^n)$, so they're not subexponential either. The Catalan numbers are just exponential.

An exception to the above definitions is that, in some contexts, functions of the form $b^{n^k}$ for constants $b,k>1$ are considered to be exponential, even though $$ \lim_{n\to\infty}\frac{b^{n^k}}{c^n}=\lim_{n\to\infty}b^{n^k-n\log_b c}=\infty\,.$$ For example, the complexity class EXP is defined as the class of languages decided by Turing machines running in time $O(2^{n^k})$ for any $k$. Thanks to Yuval Filmus for pointing this out.

  • $\begingroup$ Can you provide some references? If there is no significant reference or usage, then it is rather void of practical importance to define a term just for the sake of that term, even though there might be a natural meaning for it. It might be better to leave its definition for better occasions. $\endgroup$
    – John L.
    Nov 12, 2018 at 13:35
  • $\begingroup$ References for which part? "Super-exponential" follows, e.g., "super-polynomial"; "sub-exponential" follows, e.g., "sub-linear", and the definition of EXP is in any complexity theory textbook. $\endgroup$ Nov 12, 2018 at 13:47
  • $\begingroup$ References to existing articles, textbook, papers that define and use "super-exponential", the term in the question. In fact, let me ask the OP. $\endgroup$
    – John L.
    Nov 12, 2018 at 13:59

Using David's definition; a function is super-exponential if it grows faster than any exponential function. More formally, this means that it is $\omega(c^n)$ for every constant $c$, i.e., if $\lim_{n\to\infty} f(n)/c(n)=\infty$ for all constants $c$.

The $n$-th formula of Catalan Numbers is given by Wikipedia as;

$$ C_n \approx \frac{4^n}{n^{3/2}\sqrt \pi} $$

Than we have;

$$4^n > \frac{4^n}{n^{3/2}\sqrt \pi} > 2^n$$

By the definition of super-exponential time, $4^n$ not super-exponential time, this implies $C_n$ also not super-exponential time. Since we have;

$$\lim_{n \rightarrow \infty} \frac{f(n)}{4^n} = \infty,$$ with super-exponential growing function $f$.

This implies that Catalan numbers are exponential, not super-exponential.

    1. Note: Answer updated after David's comments.
    1. Note: In, Analytic Combinatorics, Sedgewicks says that Catalan Numbers is roughly comparable to an exponential, $4^n$, modulated by a subexponential factor. here $1/\sqrt{π n^3}.$

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