Sequence Challenge: Find $a_{2015}$

[magneto1]

Let $(a_i)_{i\in \Bbb{N}}$ be a sequence of nonnegative integers such that $a_2 = 5$, $a_{2014} = 2015$, and $a_n=a_{a_{n-1}}$ for all other positive $n$. Find all possible values of $a_{2015}$.

 

[jvicens]

Hello! This is an interesting problem. Let's first try to understand the sequence given in the problem. We are given that $a_2 = 5$, which means that $a_5 = a_{a_4} = a_{a_{a_3}} = a_{a_{a_{a_2}}} = a_{a_{a_{a_{a_1}}}} = a_{a_{a_{a_{a_0}}}}$. Since $a_n = a_{a_{n-1}}$ for all positive $n$, we can see that this sequence is recursive in nature.

Now, let's try to find a pattern in the sequence. We can see that $a_3 = a_{a_2} = a_5 = 2015$. Similarly, $a_4 = a_{a_3} = a_{2015}$. Continuing this pattern, we can see that $a_{2015} = a_{a_{a_{a_{...a_5}}}}$, where $a_5$ is repeated $2012$ times. This means that $a_{2015}$ is the $2013$th term in the sequence, which is equal to $2015$.

Therefore, the only possible value of $a_{2015}$ is $2015$. This can also be verified by plugging in different values for $a_2$ and $a_{2014}$.

I hope this helps! If you have any further questions or need clarification, please feel free to ask.