Mediterranean Mathematical Competition

Consider the Adriatic sequence $\langle a_1, \dots, a_n \rangle$ starting with $a_1 = 1$. Construct a new sequence
$$\langle a_2 - 1, a_3 - a_2, \dots, a_n - a_{n-1}, 10^6 \rangle$$
from it. Note that the new sequence is Tyrrhenian, as
$$(a_2 - 1) + (a_3 - a_2) + \dots + (a_n - a_{n-1}) = a_{k-1} - 1 < a_k - a_{k-1}$$
holds for $k = 2, \dots, n-1$ and as
$$(a_2 - 1) + (a_3 - a_2) + \dots + (a_n - a_{n-1}) = a_n - 1 < 10^6.$$
Next, consider the Thyrrenian sequence $\langle t_1, t_2, \dots, t_n \rangle$ with $t_n = 10^6$. Construct a new sequence
$$\langle 1, 1 + t_1, 1 + t_1 + t_2, \dots, 1 + t_1 + \dots + t_{n-1} \rangle$$
from it. Note that this new sequence is Adriatic, since
$$2(1 + t_1 + t_2 + \dots + t_{k-1}) \le 1 + t_1 + \dots + t_k$$
is equivalent to the Thyrrenian property $t_1 + \dots + t_{k-1} < t_k$.

These two constructions yield two injections and demonstrate that the number of Adriatic sequences equals the number of Thyrrenian sequences. (In fact the second injection is the inverse of the first injection, so that we have a clean bijection between the two sets.) $\Box$