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](https://math.stackexchange.com/questions/576018/finding-the-min-of-a-sequence-related-with-factorials) 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 $2500=10\cdot a\_{10}$.)



Finding the min of a sequence related with factorials
Lemma: Let $n,a$, be positive integers with $r = n \bmod a$, the remainder after
dividing $n$ by $a$. Using Iverson notation [statement] is $1$ if true, $0$ otherwise,
$\lfloor 2n/a \rfloor = 2\lfloor n/a \rfloor + [2r \geq a]$.


From this, one shows that the exponent of a prime $p$ in a prime factorization of
$C\_n={n \choose {\lfloor n/2 \rfloor}}$ is at most $\log\_p n$, and is more likely to
be about half of that if it is nonzero, by looking at the base $p$ expansion of $n$.


Now suppose $n = c + \lfloor N/2 \rfloor$ and $a\_n \gt 1$. Then
$P= n!/(n-c)!$ is such that $P^2$ divides $C\_N$.
For large $N$, $c$ should be bounded by $\min\_{p^2 \leq N} pe\_p/2$ where $e\_p$ is the largest
power of $p$ dividing $C\_n$. An upper bound that should be tight in general is to look at 
$p=2,3$ and $5$ and use the minimum based on those. For other bounds, find a prime
$q$ with $q^2 \gt N$ and $N/2 \bmod q$ close to $q$: the next multiple of $q$ after $N/2$ is
also an upper bound. I do not know how to find $\min(N)$ algebraically, but it is clear
one does not have to look far from $N/2$.


There are similar problems depending on sequences of smooth numbers as well as divisibility by small primes. In addition to one-complexity of an integer, my MathOverflow
favorite can be found at [Factorials in Pascals Triangle](https://mathoverflow.net/questions/17058/factorials-in-pascals-triangle). Perhaps a compilation of these can be solved with an appropriate literature search.


Gerhard "Ask Me About Small Factors" Paseman, 2013.12.04


