, 2022

Solution:

Say $n$ is $p$-good if $\sum_{k=1}^{n}(-1)^{\nu_{p}(k!)}<0$, where $p$ is an odd prime.

Claim. $n$ is $p$-good iff
$$n+1=\sum_{i=0}^{k} a_{i} p^{2 i+1}$$
where $a_{i}$ is an even integer with $\left|a_{i}\right|<p$.

The proof of this claim will be deferred to the end of the solution as it is rather technical, and we believe that it would be more illuminating for the reader to graph the function $n \mapsto \sum_{k=1}^{n}(-1)^{\nu_{p}(k!)}$ and examine its properties, instead of focusing on the formal proof.

A consequence of the claim is that if $n$ is $p$-good then $p$ divides $n+1$, and $p^{2 k-1}<n+1<p^{2 k}$ for some $k \in \mathbb{Z}^{+}$.

Now suppose $n$ is $p$-good and $q$-good for distinct odd primes $p<q$. Then $n+1 \geq p q>p^{2}$, so we must have $n+1>p^{3}$.

Checking $p=3$, the smallest potential $n+1$ 's are
- $2 \cdot 3^{3}-2 \cdot 3=48$, which does not have a prime factor $q>3$.
- $2 \cdot 3^{3}=54$, which does not have a prime factor $q>3$.
- $2 \cdot 3^{3}+2 \cdot 3=60$, which does not work because 60 is the wrong size for $q=5$.
The next value $2 \cdot 3^{5}-2 \cdot 3^{3}-2 \cdot 3$ is already bigger than 230.

Checking $p=5$, the smallest potential $n+1$ 's are
- $2 \cdot 5^{3}-4 \cdot 5=230$, which works for $q=23$.

For $p \geq 7, n+1 \geq p^{3}>230$, so 229 is the smallest value of $n$.

It suffices to prove the claim. We argue via a series of lemmas. We introduce the notation of $S(a, b)=\sum_{k=a}^{b-1}(-1)^{\nu_{p}(k!)}$. Note that $n$ is $p$-good if and only if $S(0, n+1) \leq 0$.

Lemma 1. If $n \leq p^{2}, S(0, n)$ is the distance to the nearest even multiple of $p$.

Proof. This follows straightforwardly from the fact that $\nu_{p}(k!)=\lfloor k / p\rfloor$ for $n \leq p^{2}$.

Lemma 2. If $a$ and $b$ are positive integers so that $b \leq p^{\nu_{p}(a)}$, then $S(a, a+b)=(-1)^{\nu_{p}(a!)} S(0, b)$.

Proof. Note that for $0<k<b, \nu_{p}(a+k)=\nu_{p}(k)$, so it follows that $\nu_{p}((a+k)!)=\nu_{p}(a!)+\nu_{p}(k!)$. The result follows.

Lemma 3. For any nonnegative integer $a, \nu_{p}\left(\left(a p^{2}\right)!\right)$ is the same parity as $\nu_{p}(a!)$.

Proof. Note that $\nu_{p}\left(\left(a p^{2}\right)!\right)-\nu_{p}(a!)=a p+a=a(p+1)$, which is even as $p$ is odd.

Lemma 4. For a nonnegative integer $a, S\left(a p^{2},(a+1) p^{2}\right)=p(-1)^{\nu_{p}(a!)}$.

Proof. Combine Lemmas 1, 2, and 3.

Lemma 5. For a nonnegative integer $a, S\left(0, a p^{2}\right)=p S(0, a)$.

Proof. Apply Lemma 4 and sum.

Lemma 6. If $a, b$ are nonnegative integers with $b<p^{2}$, then $S\left(0, a p^{2}+b\right)=p S(0, a)+(-1)^{\nu_{p}(a!)} S(0, b)$.

Proof. Combine Lemmas 2, 3, and 5.

We are now ready to prove the claim. Call a nonnegative integer neat if it can be written as $\sum_{i=0}^{k} a_{i} p^{2 i+1}$ for integers $a_{i}$ with $\left|a_{i}\right|<p$. For a nonnegative integer $n$, let $P(n)$ be the following statements:
- $S(0, n) \geq 0$.
- $S(0, n)=0$ if and only if $n$ is neat. In this case, $\nu_{p}(n!)$ is even.
- $S(0, n)=1$ if and only if $n+1$ is neat or $n-1$ is neat. If $n+1$ is neat, then $\nu_{p}(n!)$ is odd. If $n-1$ is neat, then $\nu_{p}(n!)$ is even.

It suffices to show $P(n)$ for all $n$, which we will prove by induction on $n$. The base case of $n=0$ is obvious.

Now take some $n>0$ and suppose $n=a p^{2}+b$ for $0 \leq b<p^{2}$. Lemma 6 tells us that $S(0, n)=p S(0, a)+(-1)^{\nu_{p}(a)} S(0, b)$. Since $0 \leq S(0, b) \leq p$ (by Lemma 1), the only way for $S(0, n)$ to be less than 0 is if $S(0, a)=0$ and $(-1)^{\nu_{p}(a)}=-1$, which is impossible since $P(a)$ holds.

There are two ways for $S(0, n)=0$ to be true. The first case is that $S(0, a)=S(0, b)=0$, which implies by $P(a)$ and Lemma 1 that $a$ is neat and $b$ is a multiple of $2 p$. This captures the neat numbers with $a_{0} \geq 0$. Note that in this case $\nu_{p}(n!)=\nu_{p}\left(\left(a p^{2}\right)!\right)+\nu_{p}(b!)$ (by the same logic as Lemma 2), which is even as $\nu_{p}\left(\left(a p^{2}\right)!\right)$ is even by $P(a)$ and Lemma 3 and $\nu_{p}(b!)=b / p$ which is even.

The second way for $S(0, n)$ to be 0 is if $S(0, a)=1, S(0, b)=p$, and $(-1)^{\nu_{p}(a!)}=-1$. By the inductive hypothesis, this is equivalent to $a+1$ being neat and $b$ being an odd multiple of $p$. This captures exactly the neat numbers with $a_{0}<0$. Also, $\nu_{p}(n!)=\nu_{p}\left(\left(a p^{2}\right)!\right)+\nu_{p}(b!)$, which is even as $\nu_{p}(a!)$ is odd and $\nu_{p}(b!)$ is odd.

Analyzing the possibilities where $S(0, n)=1$ is almost exactly the same as the above, so we will omit it here. We encourage the reader to fill in the details.