Smallest Number $k \in \mathbb{N}$ Such That $(2k-9)! \equiv 0\pmod{k!^2}$

Question

Find the Smallest Number $k \in \mathbb{N}$ Such That $(2k-9)! \equiv 0\pmod{k!^2}$

My Attempt

We want a natrual number k such that $\frac{(2k-9)!}{k!^2}$ is a whole number. so:

$\frac{(2k-9)!}{k!^2}$ = $\frac{(k+1)(k+2)...(2k-9)}{k!}$

We can see that some terms of $k!$ might cancel out, like $k+2 $ and $\frac{k}{2} +1$ if $k$ is even. Although, I haven't been able to generalize such behaviour. What can we do now?

P.S. I ran a relatively efficient python script, and the number (I don't know what it is) is bigger than $10^5$. Also, I have defined the sequence $a_n = $ the smallest number $k \in \mathbb{N}$ such that $(2k-n)! \equiv 0\pmod{k!^2}$. This is how I thought of this problem. From the python script: $a_1 = 1,a_2 = 1,a_3 = 210,a_4 = 210,a_5 = 3478,a_6 = 3478,a_7 = 8178,a_8 = 8178, a_9 = ?$

Is there any info on this?

Answer 1

My computations say that $k=252970$ verifies this and it is the smallest one, in your notation $a_9=252970$. Also $a_{10}=a_9$.

I explain how I did it. First, I use the well-known formula $$\nu_p(m!)=\sum_{i=1}^{\infty} \left\lfloor \frac{m}{p^i} \right\rfloor $$ (where $\nu_p(m)$ means the $p$-adic valuation of the number $m$, so the maximum $n$ such that $p^n$ divides $m$).

Now, your condition is equivalent to $$ \nu_p(2k-9) \ge 2\nu_p(k)$$ for all primes $\le k$. That's what I computed.