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** :

$$\min(2)=2,\min(3)=2,\min(4)=3,\min(5)=3,\min(28)=16,\min(2008)=1005.$$

**Remark** : This question has been asked previously on math.SE without receiving any answers.

**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 $\min(N)$ in general. 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 $k=14$ 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 $2500=10\cdot a_{10}$.)