Stay positive

Nothing novel in the solution; let $x = a_{1}$ , express the first give or take 10 items with $x$ and evaluate the condition for them to be positive.

You could also write a small code if you wanted to be clear. $1241, ... , 1250$ for $a_{1}$ will give $l = 7, 9, 9, 9, 9, 9, 11, 10, 10, 8$ , showing $\fbox{1247}$ is the answer.

However, there is a shorter way. To keep the sequence going as long as possible, we want to minimize the fluctuation of the ratio $\alpha$ between two adjacent items (larger item over the other); i.e., if $a_{n} > a_{n+1}$ then $a_{n} = \alpha \times a_{n+1}$ . Now we have $\alpha(\alpha - 1) = 1$ , which is the famous equation for the golden ratio $\alpha = {1 + \sqrt{5} \over 2}$ . One can therefore shortcut to the answer by $a_{1} = {a_{0} \over \alpha} = {2018 \over {1 + \sqrt{5} \over 2} } = 1247.192589 \simeq \fbox{1247}$ (and checking cases 1246 and 1248 to ensure the answer).