# The largest number $y$ such that $(x!)^{x+y}|(x^2)!$

Since the multiplication of $$n$$ consecutive integers is divided by $$n!$$, then $$(n!)^n|(n^2)!$$ with $$n$$ is a positive integer.

Are there any formula of the function $$y=f(x)$$ that shows the largest number $$y$$ in which $$(x!)^{x+y}|(x^2)!$$ with $$x>1$$. If so, for a positive integer $$y$$, are there infinitely many $$x$$ such that $$(x!)^{x+y}|((x^2)!)$$ (with $$x, y, n$$ are all positive integers) ?

If not, what are the conditions of $$m$$ so that there are infinitely many $$x$$ such that $$(x!)^{x+m}|((x^2)!)$$ ? (with $$m$$ is a positive integer and $$m \leq y$$) (For example, with $$m=2015$$, are there infinitely many $$x$$ such that $$(x!)^{x+2015}|((x^2)!)$$ ?

• You could generate the first 50 terms or so with a computer, subtract one from each and use oeis.org – Martin Rubey Oct 21 '18 at 10:27
• (sorry, subtracting one was nonsense) – Martin Rubey Oct 21 '18 at 12:19
• "Computationally" the progression is oeis.org/A187279 – Xarles Oct 21 '18 at 16:22

$$\newcommand{\eps}{\varepsilon}$$It seems that for any $$y$$ the number of such $$x$$ is infinite.

First of all, let's fix a prime $$p$$ and compute $$v_p(\frac{(x^2)!}{(x!)^x})$$ -- the exponent of $$p$$ in the prime factorization of this ratio. For any natural number $$n$$ we have $$v_p(n!)=\frac{n-S_p(n)}{p-1}$$ where $$S_p(n)$$ is the sum of digits in the base p expansion of $$n$$. So, we get $$v_p(\frac{(x^2)!}{(x!)^x})=\frac{x^2-S_p(x^2)}{p-1}-x\frac{x-S_p(x)}{p-1}=\frac{xS_p(x)-S_p(x^2)}{p-1}$$

Thus, a number $$y$$ satisfies $$(x!)^{x+y}|(x^2)!$$ if and only if for every prime $$p$$ th inequality $$y\frac{x-S_p(x)}{p-1}\leq \frac{xS_p(x)-S_p(x^2)}{p-1}$$ holds. As was obvious from the beginning, $$x!$$ is only divisible by primes not greater than $$x$$ so let's assume $$p\leq x$$. We then get that the maximal $$y$$ such that $$(x!)^{x+y}|(x^2)!$$ is given by $$\min\limits_{p\leq x}\frac{xS_p(x)-S_p(x^2)}{x-S_p(x)}$$. The function $$S_p(x)$$ is hard to compute explicitely but, for sure, there is an estimate $$S_p(x)\leq (p-1)(\log_p(x)+1)$$. So, for a fixed $$p$$ as $$x$$ tends to $$\infty$$ the expression $$\frac{xS_p(x)-S_p(x^2)}{x-S_p(x)}$$ is asymptotically equivalent to $$S_p(x)$$. However, here as $$x$$ gets bigger we should take into account bigger and bigger primes.

Anyway, let's prove that $$f_p(x):=\frac{xS_p(x)-S_p(x^2)}{x-S_p(x)}$$ is not much smaller than $$S_p(x)$$ for every $$p$$ and then, using the prime number theorem we'll find infinitely many $$x$$ such that $$S_p(x)$$ is bigger than a fixed number $$y$$ for every $$p$$.

We have $$f_p(x)=S_p(x)+\frac{S_p(x)^2-S_p(x^2)}{x-S_p(x)}$$. Let's separately consider cases $$p> \sqrt{x}$$ and $$p\leq\sqrt{x}$$. In the first case we have $$\frac{S_p(x)^2-S_p(x^2)}{x-S_p(x)}>-\frac{4(p-1)}{p-1}=-4$$ because $$x^2$$ has at most $$4$$ digits. So $$f_p(x)>S_p(x)-4$$. If $$p\leq \sqrt{x}$$ then $$x\geq p^2$$. Pick number $$k$$ such that $$p^k\leq x. We have $$\frac{f_p(x)}{S_p(x)}=1+\frac{S_p(x)-\frac{S_p(x^2)}{S_p(x)}}{x-S_p(x)}\geq 1+\frac{-(p-1)(2k+2)}{p^k-1}=1-\frac{2k+2}{p^{k-1}+..+p+1}\geq 1-\frac{2k+2}{p(k-1)}\geq 1-\frac{7}{p}$$ because $$k\geq 2$$.

Thus, there exists a positive constant $$\eps$$ such that for big enough $$x$$ we have $$f_p(x)>\eps S_p(x)$$ for every prime $$p\leq x$$(strictly speaking, the inequality above shows this only for $$p>7$$ but any finite set of primes can be easily covered by the above remark about the asymptotics of $$S_p(x)$$).

Let's now fix number $$y$$ and denote by $$\mathcal{B}_y(N)$$ the number of integers in $$[1, N]$$ which have $$S_p(-)$$ less than $$y$$ for some number $$p\leq N$$. For a given prime $$\#\{x|S_p(x) is smaller than $$\binom{\log_pN+1+y}{y}y^y$$ -- because such number $$x$$ has at most $$\log_pN+1$$ digits, at most $$y$$ of them are non-zero, and every non-zero digit is at most $$y$$(I've added $$y$$ to $$\log_pN+1$$ to account for the case when $$log_p+1 -- this is a very rough estimate, of course). By the prime number theorem, for big enough $$N$$ the number of primes smaller than $$N$$ is smaller than $$\frac{3}{2}\frac{N}{\ln N}$$ (of course, $$\frac{3}{2}$$ can be replaced by any number bigger than $$1$$). Combining these two observations get $$\mathcal{B}_y(N)<\sum\limits_{p<\sqrt{N}} \binom{\log_2 N+y+1}{y}y^y+\sum\limits_{N\geq p\geq \sqrt{N}}\binom{\log_{\sqrt{N}}N+y+1}{y}y^y<\frac{3}{2}\frac{\sqrt{N}}{\ln\sqrt{N}}\binom{\log_2 N+y+1}{y}y^y+\frac{3}{2}\frac{N}{\ln N}\binom{y+3}{3}y^y where $$C(y)$$ is a constant depending only on $$y$$.

Thus, $$\lim \frac{\mathcal{B}_y(N)}{N}=0$$ so, for a given $$y$$ the set of $$x$$ satisfying your condition even has density $$1$$.