Finding the min of a sequence related with factorials Let $N,n$ be natural numbers.
Let us define $a_n=m$ when $N!$ can be divided by $(n!)^m$ and it cannot be divided by $(n!)^{m+1}$.
For a given $N(\ge 2)$, let $\min(N)$ be the min of $na_n\ (2\le n\le N)$.
Then, here is my question.

Question : What is $\min(N)$?

Example : 
$$\min(2)=2,\min(3)=2,\min(4)=3,\min(5)=3,\min(14)=16,\min(2008)=1005.$$
Motivation : I've known a question to find $\min(2008)$. Then, I got interested in its generalization. However, I cannot find any good way to find $\min(N)$. I'm afraid that this question might be solved only by brute-force computer search.
Note that it is not true that $\min(2k)=k+1$. See the above $k=7$ case. 
By the way, we can lead $na_n\approx N$, which shows the meaningfulness to treat $na_n$.
The exponent of a prime $p$ of $N!$ can be represented as
$$\sum_{k=1}^{\infty} \left\lfloor{\frac{N}{p^k}}\right\rfloor\approx \sum_{k=1}^{\infty} \frac{N}{p^k}=\frac{N/p}{1-1/p}=\frac{N}{p-1}.$$
On the other hand, the exponet of a prime $p$ of $n!$ can be represented as
$$\sum_{k=1}^{\infty} \left\lfloor{\frac{n}{p^k}}\right\rfloor\approx \cdots =\frac{n}{p-1}.$$
Hence, by considering these ratio, we get $a_n\approx\frac{N}{n}$, namely, $na_n\approx N$.
The followings are the examples of the $N=2008$ case.
$$8\cdot a_8=2280, 9\cdot a_9=2250, 10\cdot a_{10}=2500, 11\cdot a_{11}=2189,12\cdot a_{12}=2388,$$
$$13\cdot a_{13}=2145, 14\cdot a_{14}=2310, 15\cdot a_{15}=2475, 16\cdot a_{16}=2128,17\cdot a_{17}=2108,$$
$$18\cdot a_{18}=2232, 19\cdot a_{19}=2090, 20\cdot a_{20}=2200, 21\cdot a_{21}=2310,22\cdot a_{22}=2178,$$
$$23\cdot a_{23}=2070, 24\cdot a_{24}=2160.$$
(By the way, we can prove that the max of $na_n\ (8\le n\le 2008)$ for the $N=2008$ case is $2500=10\cdot a_{10}$.)
Update : I crossposted to MO.
 A: I've just found that we can easily get $\min(N)$ in the following way, but a more direct answer is welcome. Thanks to The Masked Avenger's comment on MO, I've just got the following :
$\min(N)=M$ where $M$ is the minimum number which satisfies 
$$a_M=1,\ \ \left\lfloor\frac{N}{2}\right\rfloor +1\le M\le N^{\prime}.$$
Here, $N^{\prime}$ is the minimum number in the form of $p^k\ (k\ge 0\in\mathbb Z)$ where $p$ is a prime number.
Proof : Let $h(n)=na_n$. We can see if $\{(kn)!\}^{a_{kn}}$ can divide $N!$, then $(n!)^{ka_{kn}}$ can also divide $N!$ because for $k\in\mathbb N$, 
$$\{(kn)!\}^{a_{kn}}=\left(\binom{kn}{n}\cdot\binom{(k-1)n}{n}\cdots\binom{n}{n}\right)^{a_{kn}}\times (n!)^{ka_{kn}}$$
holds. Hence, we get $a_n\ge ka_{kn}.$ Multiplying the both sides by $n$ gives us $h(n)\ge h(kn).$ 
Thus, letting $M$ be the minimum number such that $a_M=1$, we get the followings : 
$(1)$ For $n$ such that $n\le M-1$, since we know that for each $n$ there exists $k\in\mathbb N$ such that $M\le kn\lt N$, we get $h(n)\ge h(kn)=kna_{kn}\ge kn\ge M$.
$(2)$ For $n=M$, $h(n)=h(M)=Ma_M=M\times 1=M.$ (Note that $a_n$ is a monotone decreasing sequence.)
$(3)$ For $n$ such that $M\lt n\le N$, since $a_n=1$, we get $h(n)=na_n=n\gt M.$
Hence, we lead that $\min(N)=M$.
In addition to this, noting that
$$\frac{(2m)!}{m!m!}\in\mathbb N, \frac{(2m-1)!}{(m-1)!(m-1)!}\in\mathbb N,$$
we can see that $M\ge m+1$ for $N=2m$, and that $M\ge m$ for $N=2m-1$. These can be represented as $\left\lfloor\frac{N}{2}\right\rfloor +1\le M.$ On the other hand, $M\le N^{\prime}$ is obvious where $N^{\prime}$ is the minimum number in the form of $p^k\ (k\ge 0\in\mathbb Z)$ where $p$ is a prime number. Q.E.D.
As I mentioned at the top of this answer, a more direct answer is welcome; especially I would like to see a better pair $(a,b)$ to estimate $M$ such that $a\le M\le b$.
A: I'm posting an answer just to inform that Gerhard Paseman gave some upper bounds for $\min(N)$ on MO.
