It's Trivial?-2

It's easy to see that for all $n \in \mathbb{N}$ , $a_n$ is of the form $\prod_{k=1}^{6} a_k^{b_{k,n}}$ , where $b_{k,n} \in \mathbb{Q}$ .

For $1 \le k \le 6$ , let $c_k = \lim_{n \to \infty} b_{k,n}$ . (I won't bother proving that these limits exist because it's a little tedious). Hence, $L = \lim_{n \to \infty} a_n = \prod_{k=1}^{6} a_k^{c_k}$ Note that this formula holds regardless of what the initial conditions are. Hence, if we were to ignore the first number in the sequence and use the next $6$ numbers as the initial conditions, the same formula will work (and they will have the same limit because they are parts of the same sequence). Therefore, $\prod_{k=1}^{6} a_k^{c_k} = L = \prod_{k=1}^{6} a_{k+1}^{c_k}$ Now $a_7 = \sqrt[6]{\prod_{k=1}^{6} a_k}$ . Therefore, $\prod_{k=1}^{6} a_k^{c_k} = L = a_{1}^{c_6/6} a_2^{c_1+c_6/6} a_3^{c_2+c_6/6} a_4^{c_3+c_6/6} a_5^{c_4+c_6/6} a_6^{c_5+c_6/6}$ Matching up the exponents gives $c_1 = \frac{c_6}{6}, c_2 = c_1 + \frac{c_6}{6}$ , and so on. Simplifying a little, we get that $c_k = k c_1$ for $1 \le k \le 6$ . Therefore, $L = \left(\prod_{k=1}^{6} a_k^k\right)^{c_1}$

So now we need to find $c_1$ . Suppose, for a moment, that all the initial conditions are equal to the same number $a$ . Then logically, $L = a$ because the geometric mean of $6$ equal numbers always gives the same number. Hence, plugging in these numbers gives us: $a = \prod_{k=1}^{6} a^{c_k} = a^{\sum_{i=1}^{6} c_i}$ Hence, $\sum_{k=1}^{6} c_k = 1$ $\sum_{k=1}^{6} k c_1 = 1$ $21c_1 = 1$ Thus $c_1 = \frac{1}{21}$ . Therefore, we finally get the elegant formula: $L = \sqrt[21]{\prod_{k=1}^{6}a_k^k} = 1^1 2^2 3^3 4^4 5^5 6^6 = 4031078400000$ This number has $\boxed{5}$ trailing zeroes.