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Can the sum of $2025$ consecutive factorials be a perfect power?

My thoughts:

If the sum of $2025$ consecutive factorials is a perfect power, then the number of trailing zeroes in the sum must be a perfect (number of trailing zeroes)th power.

So this means that if the sum of $2025$ consecutive factorials is a perfect power, then the sum must be a perfect (number of trailing zeores)th residue $\pmod{p}$, where $p$ is a certain prime number.

For example, the sum of $10!+11!+12!+\cdots+2034!$ contains $2$ trailing zeroes as all factorials from $10!$ to $14!$ contains $2$ trailing zeroes, and since the sum contains $2$ trailing zeroes, then it must be a perfect square in order to be a perfect power.

By using Pari GP, I tried to check all factorials from $1!$ to $10^{6}!$, but I did not succeed.

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    $\begingroup$ suggest you investigate five consecutive factorials that may not begin at $1$ Let's see, factor $2! + 3! + 4! + 5! + 6! $ and then factor $ 3! + 4! + 5! + 6! + 7! $ and so on. I can see that things change a little when beginning after 5 itself. The point, though, is that you will have no idea how to deal with 2025 factorials unless you know how to deal with 5 $\endgroup$
    – Will Jagy
    Feb 2 at 4:35
  • $\begingroup$ @willjagy I did already tried that by using Pari GP, but I wasn’t successful. $\endgroup$ Feb 4 at 13:46
  • $\begingroup$ @ThirdyYabata I think Will was suggesting to do a little algebra with pen and paper. For example, what can you do with $5! +6! +7! +8! +9!$. Running programs is good, but sometimes a little algebra can be more illuminating :) $\endgroup$ Feb 5 at 2:16
  • $\begingroup$ One neat thing about the sum is that every subsequent term is a multiple of every preceding term. Perhaps you could use that to help with your question..? $\endgroup$ Feb 5 at 16:47

1 Answer 1

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Suppose $X=n!+(n+1)!+\cdots+(n+2024)!$ is a perfect power. Let $$Y=\frac X{n!}=1+(n+1)+(n+1)(n+2)+\cdots+(n+1)\cdots(n+2024).$$ Note that $$Y<\frac{(n+2025)!}{n!}<(n+2025)^{2025}.$$ For $X$ to be a perfect power, each prime which divides $n!$ exactly once must also divide $Y$. In particular, $$(n/2)^{\pi(n)-\pi(n/2)}\leq \prod_{n/2<p\leq n}p\leq Y\leq (n+2025)^{2025}.$$ This implies that $$\pi(n)-\pi(n/2)\leq 2025\log_{n/2}(n+2025);$$ as long as $n>200$, say, $n+2025<(n/2)^2$, and so we have \begin{equation}\tag{$\star$}\pi(n)-\pi(n/2)\leq 4050.\end{equation} On the other hand, we have explicit bounds on $\pi(n)$ from explicit versions of the prime number theorem: see for example this article of Dusart, as described on Wikipedia, which gives $$\frac{x}{\log x-1}\leq \pi(x)\leq \frac{x}{\log x-1.1}$$ for $x\geq 60184$. This implies, for $n\geq 130000$, \begin{align} \pi(n)-\pi(n/2) &\geq \frac n{\log n-1}-\frac{n/2}{\log(n/2)-1.1}\\ &>\frac n{\log n-1}\left(1-\frac{\log n-1}{2(\log(n/2)-2)}\right)\\ &=\frac n{\log n-1}\cdot\frac{\log n-3}{2\log n-4}. \end{align} As long as $\log n>10$, we have $$\pi(n)-\pi(n/2)\geq \frac n{\log n-1}\cdot\frac 7{16}.$$ Since $n/(\log n-1)$ is an increasing function of $n$, we have that, when $n\geq 130000$, $$\pi(n)-\pi(n/2)\geq \frac 7{16}\frac{130000}{\log(130000)-1}\geq \frac{7}{16}\cdot\frac{130000}{11}\geq 5000.$$ This contradicts ($\star$). We conclude that $X$ can only be a perfect power for $n<130000$. Since you have already checked these cases, we conclude that the sum in question is never a perfect power.

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