5. Given integers $a_{1}, a_{2}, \cdots, a_{5}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{5}=15 \text {, }
$$

and for any $i=2,3, 4$ there is $a_{i}^{2} \leqslant a_{i-1} a_{i+1}+1$. Then the number of $\left(a_{1}, a_{2}, \cdots, a_{5}\right)$ that satisfy the conditions is $\qquad$ groups.
5.3.

When $i=2$,
$$
\begin{array}{l}
a_{2} \geqslant a_{1}+1=2, \\
a_{2}^{2} \leqslant a_{3}+1<15+1=16 \Rightarrow a_{2}<4 .
\end{array}
$$

Thus, $a_{2}=2$ or 3.
(1) When $a_{2}=3$, $a_{3} \geqslant a_{2}^{2}-1=8$,
$$
a_{4} \geqslant \frac{a_{3}^{2}+1}{3} \geqslant 22 \text {, }
$$

which is a contradiction.
(2) When $a_{2}=2$, $a_{3} \geqslant a_{2}^{2}-1=3$.
By $a_{3}^{2} \leqslant 2 a_{4}+1<30+1=31$
$\Rightarrow 3 \leqslant a_{3} \leqslant 5$.
Also, $a_{4}^{2} \leqslant 15 a_{3}+1, 2 a_{4}+1 \geqslant a_{3}^{2}$, thus,
when $a_{3}=3$, $4 \leqslant a_{4} \leqslant 6$;
when $a_{3}=4$, $a_{4} \leqslant 7$ and $a_{4} \geqslant 8$, which is a contradiction;
when $a_{3}=5$, $a_{4} \leqslant 8$ and $a_{4} \geqslant 12$, which is a contradiction;
In summary, $\left(a_{2}, a_{3}, a_{4}\right)=(2,3,4),(2,3,5)$.