$152 a_{1}, a_{2}, \cdots$ is an integer sequence, satisfying $a_{1} \cdot a_{2}+a_{2} \cdot a_{3}+\cdots+a_{n-1} \cdot a_{n}=$ $\frac{(n-1) \cdot n \cdot(n+1)}{3}(n=2,3, \cdots)$, the number of such sequences is
A. 0 (i.e., does not exist)
B. 2
C. 4
D. infinitely many
152 C. Subtracting the two equations yields
$$
\begin{aligned}
a_{n} \cdot a_{n+1} & =\frac{n(n+1)(n+2)}{3}-\frac{(n-1) n(n+1)}{3} \\
& =n \cdot(n+1) \quad (n=1,2,3, \cdots)
\end{aligned}
$$

Taking $n=1$ in (1) gives $a_{1} \cdot a_{2}=2$. Since $a_{1}$ and $a_{2}$ are both integers, there are four possibilities:
$$
\left\{\begin{array} { l } 
{ a _ { 1 } = 1 } \\
{ a _ { 2 } = 2 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ a _ { 1 } = - 1 } \\
{ a _ { 2 } = - 2 }
\end{array} \text { or } \left\{\begin{array} { l } 
{ a _ { 1 } = 2 } \\
{ a _ { 2 } = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
a_{1}=-2 \\
a_{2}=-1
\end{array}\right.\right.\right.\right.
$$

The entire sequence can be determined from (2) and (1).