29° Olimpiada Matemática del Cono Sur

In order to do this, we will show that if the sequence is alagoana, then $a_n$ cannot have any prime factor. Consider a prime $p$. For every positive integer $n$, let $\alpha (n)$ be the exponent of $p$ in the factorization of $a_n$. We will prove that $\alpha (n)=0$ for every $n$. By the second condition on the sequence, we know that $\alpha (n)$ is a multiple of $n$; that is, there exists a non-negative integer $k_n$ such that $\alpha (n)=n\cdot k_n$. From the first condition, we have that $a_{n!}=a_1\cdot a_2\cdot \cdots \cdot a_n$. This implies that $\alpha (n!)=\alpha (1)+\alpha (2)+\cdots +\alpha (n)$. In particular, $\alpha (n!)=\alpha (n)+\alpha ((n-1)!)$, so $\alpha (n)=\alpha (n!)-\alpha ((n-1)!)$, that is, $n\cdot k_n=n!\cdot k_{n!}-(n-1)!\cdot k_{(n-1)!}$. Defining $\beta (n)=(n-1)!$, we have $$k_n=\frac{\beta (n)}{n}(n\cdot k_{n!}-k_{\beta (n)}).(1)$$ For $n\ge 4$, we will show, recursively, that ${\beta }^j(n)$ divides $\alpha (n)$ for every positive integer $j$, where ${\beta }^j(n)$ is the function $\beta$ applied $j$ times. To this end, we will apply formula (1) recursively. We first establish some facts we will use. Let ${\gamma }_m=\frac{\beta (m!)}{m!}$ be the factor appearing in formula (1) when applied to $n=m!$. Note that, for $m\ge 3$, ${\gamma }_m$ is an integer (since $m!-1>m$) and divides $k_{m!}$. In addition, ${\gamma }_m$ divides ${\gamma }_{m+1}$, since $\frac{((m+1)!-1)!}{(m+1)!}=\frac{1}{m+1}\cdot ((m+1)!-1)((m+1)!-2)\cdots m!\cdot \frac{(m!-1)!}{m!}$, and $(m+1)!-(m+1)=(m+1)(m!-1)>m!$; then, ${\gamma }_m$ divides ${\gamma }_n$ for every $n>m\ge 3$. Now, we will prove recursively that, for every positive integer $j$, we have that $$k_n=\frac{{\beta }^j(n)}{n}\cdot N_j.$$ where $N_j$ is an integer that can be expressed as a sum of $2^j$ multiples of integers of the form $k_{m!}$, where the smallest subindex $m!$ that appears is ${\beta }^j(n)$. For $j=1$, the statement is clear from identity (1), since $\beta (n)=(n-1)!$. Assume the result holds for $j\ge 1$. Since every term in the expression of $N_j$ is a multiple of $k_{m!}$ for an integer $m$, and the smallest of the integers $m$ involved is ${\beta }^{j-1}(n)-1$ (since, by definition, ${\beta }^j(n):=({\beta }^{j-1}(n)-1)!$), it follows that each term is a multiple of ${\gamma }_{{\beta }^{j-1}(n)-1}=\frac{\beta (({\beta }^{j-1}(n)-1)!)}{({\beta }^{j-1}(n)-1)!}=\frac{{\beta }^{j+1}(n)}{{\beta }^j(n)}$. Moreover, as a consequence of identity (1) applied to $m!$, the quotient in the division of $k_{m!}$ by ${\gamma }_{{\beta }^{j-1}(n)-1}$ can be written as a sum of a multiple of $k_{(m!)!}$ and a multiple of $k_{\beta (m!)}$. Then, $N_j$ is a multiple of $\frac{{\beta }^{j+1}(n)}{{\beta }^j(n)}$, and the quotient $N_{j+1}$ can be expressed as a sum of $2^{j+1}$ terms that are multiples of integers of the form $k_{m!}$, where the smallest subindex that appears is $\beta ({\beta }^j(n))={\beta }^{j+1}(n)$ (note that, if $m_1>m_2$, then $m_1!>\beta (m_1)=(m_1-1)!\ge m_2!>(m_2-1)!=\beta (m_2)$). We conclude that $$k_n=\frac{{\beta }^j(n)}{n}\cdot N_j=\frac{{\beta }^j(n){\beta }^{j+1}(n)}{n{\beta }^j(n)}\cdot N_{j+1}=\frac{{\beta }^{j+1}(n)}{n}\cdot N_{j+1},$$ as we wanted to prove. Thus, if $n\ge 4$, for every positive integer $j$, we can write $\alpha (n)=n\cdot k_n={\beta }^j(n)\cdot N_j$ for a non-negative integer $N_j$. Finally, to conclude that $\alpha (n)=0$ for all $n\ge 4$, we notice that ${\beta }^j(n)>2^j$ for every $j$; for $j=1$, $\beta (n)=(n-1)!\ge 6>2$, since $n\ge 4$, and for $j\ge 1$, $\frac{{\beta }^{j+1}(n)}{{\beta }^j(n)}=\frac{(m-1)!}{m}\ge 2$, since $(m-1)!\ge 2m$ for every integer $m\ge 4$. The remaining cases follow now easily from the identity $\alpha (1)+\alpha (2)+\alpha (3)=\alpha (3!)=\alpha (6)=0$, which implies that $\alpha (1)=\alpha (2)=\alpha (3)=0$.