In a sequence of numbers, every (intermediate) term is half as large as the arithmetic mean of its neighboring terms. What relationship holds between an arbitrary term, the term with an index two less, and the term with an index two more? - Let the first term of the sequence be 1, and the 9th term be 40545. What is the $25-\mathrm{th}$ term?
I. Let the $n$-th term of the sequence be $a_{n}$. By definition, for $n \geq 2$,

$$
a_{n}=\frac{a_{n-1}+a_{n+1}}{4}
$$

Applying this to the terms in the numerator for $n-1 \geq 2$, i.e., $n \geq 3$,

$$
a_{n}=\frac{1}{4}\left(\frac{a_{n-2}+a_{n}}{4}+\frac{a_{n}+a_{n+2}}{4}\right)=\frac{1}{16}\left(a_{n-2}+2 a_{n}+a_{n+2}\right)
$$

Solving this for $a_{n}$, the desired relationship is:

$$
a_{n}=\frac{1}{14}\left(a_{n-2}+a_{n+2}\right)
$$

II. Similarly, based on (2), for $n-2 \geq 3$, i.e., $n \geq 5$,

$$
\begin{aligned}
a_{n}=\frac{1}{14}\left(\frac{a_{n-4}+a_{n}}{14}+\frac{a_{n}+a_{n+4}}{14}\right), & a_{n} & =\frac{1}{194}\left(a_{n-4}+a_{n+4}\right) \\
a_{n}=\frac{1}{194} \cdot \frac{a_{n-8}+2 a_{n}+a_{n+8}}{194}, & a_{n} & =\frac{a_{n-8}+a_{n+8}}{37634}
\end{aligned}
$$

if $n-4 \geq 5$, i.e., $n \geq 9$. The difference between the indices of the given terms is exactly $9-1=8$, and the difference between the index of the desired term and the indices of the given terms is a multiple of 8, so we can reach our goal without determining the intermediate terms, only $a_{17}$ needs to be calculated. From (4),

$$
\begin{aligned}
a_{n+8} & =37634 a_{n}-a_{n-8}, \text { hence } \\
a_{17} & =37634 \cdot 40545-1=1528870529 \\
a_{25} & =37634 \cdot a_{17}-a_{9}=57424611447841
\end{aligned}
$$

Siklósi István (Budapest, Berzsenyi D. g. III. o. t.)

Remark. Based on (3), then (2), and finally (1), $a_{5}=209, a_{3}=15, a_{2}=4$.