proof $\lfloor\log n\rfloor !$ is an exponential function?

Question

Can anyone prove that $\lfloor\log n\rfloor !$ is an exponential function?

I've tried a lot but i didn't find anything relating to the solution except e number that i guess it can help.

Answer 1

$$\log(\lfloor \log n\rfloor!)=\sum_{k=2}^{\lfloor\log n\rfloor}\log k<\int_2^{\lfloor\log n\rfloor}\log(x+1)\,dx\\ =(\lfloor\log n\rfloor+1)(\log(\lfloor\log n\rfloor+1)-1)-3\log3+3.$$

Taking the antilogarithm, you get an $O(n^{\log\log n})$ expression.

Answer 2

Let $f(n)= [\log n] !$ for $n\geq 1.$ We have $f(n) \leq (1+\log n)!=(\log ne)! .$

From Stirling's Formula $x!=(x/e)^x\sqrt {2\pi x}\;(1+d(x))$ where $\lim_{x\to \infty}d(x)=0$ we have $$f(n)\leq ((\log ne)/e)^{\log ne}\sqrt {2 \pi \log ne}\;(1+d(\log ne)).$$ Taking logs we have $$\log f(n)\leq (\log ne)^2 -\log ne +\log \sqrt {2 \pi}+ (1/2)\log \log ne+\log (1+d(\log ne)).$$ For any $k>0$ we have therefore, $$\lim_{n\to \infty} (\log f(n))/\log e^{kn}=(\log f(n))/k\log n=0$$ because $(\log ne)^2/n\to 0$ as $n\to \infty.$

So as $n\to \infty,$ the function $f(n)$ grows more slowly than any increasing exponential function in $n $.

Footnotes: (i). FYI, for $n\geq 1$ we have $0<d(n)<1+1/6n.$ (ii). It is not hard to show there exists $L>0$ such that $L=\lim_{n\to \infty} (\log n!)-( (n+1/2) \log n -n),$ which suffices for this Q . Proving that $L=\log \sqrt {2\pi}$ is more difficult.

Answer 3

An elementary disproof

The family of exponential functions is characterized by the following functional equation

$$f(x+y)=f(x)f(y).$$

For our case, this means that we should be able to show that

$$\lfloor\log(n+m)\rfloor !=\lfloor\log(n)\rfloor !\lfloor\log(m)\rfloor !.$$

Assume for the sake simplicity that the base of our $\log$ is $10$.

For the integers $N<M$ let $n=10^N, m=10^M$. Then

$$\lfloor\log(10^{N}+10^M)\rfloor!<\lfloor\log(2\times 10^{M})\rfloor!=\lfloor\log(2)+ M\rfloor!= M!$$ and

$$\lfloor\log(10^{N})\rfloor!\lfloor\log(10^{M})\rfloor!=(N!)(M!)>M!.$$

So, our function is not exponential.