## Problem of the Day #51: Recurrence mod 53
*May 9, 2011*

*Posted by Mitchell in : potd , trackback*

Suppose that the sequence $a_0, a_1, a_2, \cdots$ of elements of $\mathbb{Z}^{107}$ satisfies the following properties:

- $a_0 = (1, 0, 0, \cdots, 0)$.
- For $n \geq 0$, if $a_n = (x_0, x_1, \cdots, x_{106})$ and $a_{n+1} = (y_0, y_1, \cdots, y_{106})$, where $x_0, x_1, \cdots, x_{106}, y_0, y_1, \cdots, y_{106}$ are integers, then $y_k = x_{k-1} + 2 x_{k-2} + 3 x_{k-3} + \cdots + 106 x_{k+1}$ for all $k$ (where indices are considered cyclically, so $x_{k} = x_{k+107}$ for all $k$).

Find the remainder when the second entry of $a_{53}$ is divided by $53$.

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