2016 is absolutely awesome 2

Note that it is equivalent to saying that for $n\ge 4$ , $a_n$ is the least common multiple of $\{a_1a_{n-1}, a_2a_{n-2}, \ldots,a_{n-1}a_1\}$ .

This formulation makes it clear that an has no prime factors other than $2$ and $3$ , for any $n$ . We also notice that we can consider these two prime factors independently, since the LCM of the above set will be the highest power of two present times the highest power of three present.

In particular, let $b_n$ denote the sequence defined as above, except $b_0 = b_1 = 1$ and $b_2 = b_3 = 2$ .

Similarly, let $c_n$ denote the sequence defined as above, except $c_0 = c_1 = c_2 = 1$ and $c_3 = 3$ .

It is clear that $a_n = b_nc_n$ for all $n$ .

We claim that $b_n = 2^{\left\lfloor n/2\right\rfloor}$ . The proof is by induction.

If $n = 2k$ , then $b_ib_{2k-i} = 2^{\left\lfloor i/2\right\rfloor}.2^{\left\lfloor (2k-i)/2\right\rfloor}\le 2^{i/2}.2^{k-i/2}= 2^k$ , with equality when $i \equiv 0 \pmod 2$ .

And, if $n = 2k + 1$ , then $b_ib_{2k+1-i} = 2^{\left\lfloor i/2\right\rfloor}.2^{\left\lfloor (2k+1-i)/2\right\rfloor}=2^{i/2}.2^{(2k+1-i)/2}.2^{-1/2}= 2^k$ , because exactly one of $i$ and $2k+1-i$ is odd.

A similar argument, with 3 cases modulo $3$ , works to show $c_n = 3^{\left\lfloor n/3\right\rfloor}$ .

Hence, in general we have $a_n = 2^{\left\lfloor n/2\right\rfloor}3^{\left\lfloor n/3\right\rfloor}$ .

So we have $\sqrt[2016]{a_{2016}}=\sqrt[2016]{2^{1008}.3^{672}}=\sqrt2.\sqrt[3]{3}\approx\boxed{2.040}$ .