Writing p_i for the ith prime, A000040(i); let n_0 = n, and apply the mapping n_i = n_{i1} + p_i (if p_i > n_{i1}) else n_{i1}  p_i. Then a(n) is the least k > 0 for which n_k = 0, or 1 if no such k exists.
In the traversal of n_i for a given n, if it reaches a local minimum after subtracting p_i, it will next reach a local minimum at p_j which will be close to 3p_i.
Conjecture: a(n) > 0 for all n.
For n in { 6 16 20 30 42 50 51 56 70 76 84 85 90 92 }, a(n) is unknown; in each case either a(n) = 1 or a(n) > 2 * 10^12. a(n) is known for all other n <= 100: see the Afile for details.
