The first term of a number sequence is 2, the second term is 3, and the subsequent terms are formed such that each term is 1 less than the product of its two neighbors. What is the sum of the first 1095 terms of the sequence?
Let the $n$-th term of the sequence be denoted by $a_{n}$, thus according to the problem,

\[
\begin{aligned}
a_{1} & =2, & & a_{2}=3 \\
a_{n} & =a_{n-1} a_{n+2}-1 & & (n=2,3, \ldots)
\end{aligned}
\]

Accordingly, if $a_{n-1} \neq 0$, then $a_{n-1}$ and $a_{n}$ uniquely determine the value of $a_{n+1}$:

\[
a_{n+1}=\frac{1+a_{n}}{a_{n-1}}
\]

Notably, if $a_{n-1}$ and $a_{n}$ are positive, then $a_{n+1}$ is also positive. Therefore, if $a_{1}$ and $a_{2}$ are positive, then the values of $a_{1}$ and $a_{2}$, along with the condition (1), uniquely determine the entire sequence (and every term of this sequence is positive). This is also the case for the sequence in the problem. By calculating the initial terms of the sequence according to (2), we obtain the following values:

\[
a_{3}=2, \quad a_{4}=1, \quad a_{5}=1, \quad a_{6}=2, \quad a_{7}=3
\]

We observe that $a_{6}=a_{1}$ and $a_{7}=a_{2}$, and thus $a_{8}=a_{3}, a_{9}=a_{4}, a_{10}=a_{5}$. The terms of the sequence repeat every five terms, since the calculation of each new term only uses the two immediately preceding terms.

The sum of the first 5 terms is 9, and the number of periods is $1095:5 = 219$, so the desired sum is $219 \cdot 9 = 1971$.

György Csapó (Debrecen, KLTE Gyak. Ált. Isk., 8. o. t.)

Remark. The observed repetition does not depend on the values of the first two terms, but rather on the formation rule (1). If the first two terms are $a_{1}$ and $a_{2}$, then

\[
\begin{aligned}
& a_{3}=\frac{1+a_{2}}{a_{1}}, \quad a_{4}=\frac{1+a_{3}}{a_{2}}=\frac{a_{1}+1+a_{2}}{a_{1} a_{2}}, \quad \text { if } a_{1}, a_{2} \neq 0 \\
& a_{5}=\frac{1+a_{4}}{a_{3}}=\frac{\frac{\left(1+a_{1}\right)\left(1+a_{2}\right)}{a_{1} a_{2}}}{\frac{1+a_{2}}{a_{1}}}=\frac{1+a_{1}}{a_{2}}, \quad \text { if } 1+a_{2} \neq 0 \\
& a_{6}=\frac{1+a_{5}}{a_{4}}=\frac{\left(a_{2}+1+a_{1}\right) a_{1} a_{2}}{a_{2}\left(a_{1}+1+a_{2}\right)}=a_{1}, \quad \text { if } 1+a_{1}+a_{2} \neq 0 \\
& a_{7}=\frac{1+a_{6}}{a_{5}}=a_{2}
\end{aligned}
\]

Accordingly, the sequence can only not be formed if any of the equalities $a_{1}=0, a_{2}=0, a_{2}=-1, a_{1}+a_{2}=-1$ hold.