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Define $F(n,i)=\prod_{j=1}^nj^{j^i}$. $F(n,0)=n!$ is just the standard factorial, whereas $F(n,1)$ is the so-called hyperfactorial.

Is there a term for $F(n,i)$?

How fast do these grow?

Is the growth rate given by $2^{\frac{c_in^{i+1}\log_2n}{i+1}}$ with some constant $c_i>0$ (based on comments below)?

Is there a relation to some special function for every $i$ (just like we have for the factorial and the hyperfactorial)?

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  • $\begingroup$ $\log F(n,i)=\sum_{j=1}^n j^{i+\log\log j/\log j}$. $\endgroup$
    – András Salamon
    Commented Nov 16, 2015 at 12:30
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    $\begingroup$ It's not reasonable to guess $\Theta$-level precision of rapidly growing functions. Try expressing $\Theta(2^{n!})$ in a nontrivial fashion. Stirling's formula $n! \sim \sqrt{2\pi n} (n/e)^n$ does not narrow down $n!$ enough to approximate $2^{n!}$ up to a constant or anywhere close. $\endgroup$
    – Douglas Zare
    Commented Nov 16, 2015 at 14:57
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    $\begingroup$ $n^{n^{i+1}}$ is too much, it should be rather something like $\exp((1+o(1))n^{i+1}(\log n)/(i+1))$. Well, the $\Theta(n^{n^{i+1}})$ bound is already contradicted by the $i=0$ case. $\endgroup$
    – Emil Jeřábek
    Commented Nov 16, 2015 at 15:24
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    $\begingroup$ The log of this product is just $\sum_{j=1}^n j^i\log j$ and this sum is easily approximable by all the usual methods (e.g., Euler-Maclaurin); in particular, the lead term is $\Theta(n^{i+1}\log n)$ and the constant is easily found. $\endgroup$
    – Steven Stadnicki
    Commented Nov 16, 2015 at 19:34
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    $\begingroup$ (Also, pedantry requires me to belatedly point out that by mathematical standards these functions aren't particularly 'fast-growing' at all - they're dominated by functions at level 3 of the usual fast-growing hierarchies. :-) See en.wikipedia.org/wiki/Fast-growing_hierarchy for details) $\endgroup$
    – Steven Stadnicki
    Commented Apr 16, 2016 at 0:06

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Expanding on my comments as requested, bounds similar to what you ask for are $2^{n^{i+1}/(i+1)} \le F(n,i) \le 2^{n^{i+1+1/(e\ln 2)}}$.

Let $S_i(n) = \sum_{k=1}^n k^i$. To prove the claim, first observe that $\log F(n,i) = \sum_{k=1}^n k^{i+\log\log k/\log k} \ge S_i(n)$, for logarithms of any fixed positive base. Moreover, $S_i(n) \ge n^{i+1}/(i+1)$ by Bernoulli's formula (it is true for small $i$ and the sum of the less significant terms is positive for large $i$). For the other side, $S_i(n) \le \sum_{k=1}^n n^i = n^{i+1}$. The maximum of $\log_b\log_b x/\log_b x$ is at $x=b^e$ and is $1/(e\ln b)$. Hence $\log_2 F(n,i) \le S_{i+1/(e\ln 2)}(n) \le n^{i+1+1/(e\ln 2)}$.

See also Steven Stadnicki's suggestion to directly use Euler-Maclaurin.

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