Comment by possiblywrong

Following is the closed form solution linked in the article (from a since-deleted Reddit comment):

    from functools import reduce

    def recurrence(ops, a0, n):
        def transform(x, op):
            return eval(repr(x) + op)
        ops = ops.split()
        m = reduce(transform, [op for op in ops if op[0] in ('*', '/')], 1)
        b = reduce(transform, ops, 0)
        for k in range(n + 1):
            print('Term 0: ', a0 * m ** k + b * (m ** k - 1) / (m - 1))

> This is really only interesting if a particular (potentially really large) term of the sequence is desired, as opposed to all terms up to a point. The key observation is that any sequence of the given set of operations reduces to a linear recurrence which has the given solution.