\begin{aligned} & [\text { Recurrence relations (other) }] \\ & {[\text { Periodicity and non-periodicity }]}\end{aligned}

The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13.

Find the 2013th term of the sequence.
Let's compute the first few terms of the sequence: $a_{2}=16 \cdot 13=208, a_{3}=10 \cdot 13=130, a_{4}=4 \cdot 13=52, a_{5}$ $=7 \cdot 13=91, a_{6}=10 \cdot 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a period of 3. The number 2013 is divisible by 3, so $a_{2013}=a_{3}=130$.

## Answer

130.