> I think the answer to your question is lurking in what I said. Note that I considered a second-order differential equation and a second-order recurrence, but the exact same idea works for the third-order one you are interested in. You just need a \$$3 \times 3\$$ matrix instead of a \$$2 \times 2\$$.

I believe we are thinking of the same approach.

In another thread I wrote a few recurrence puzzles. I threw a 9th order one in there. I asked folks to compute \$$10^7\$$ terms in these recurrences.

The only way I know how to compute so many terms uses your trick...