The greedy alternating Egyptian fraction for e-2 must begin with 1/1 as the (-1)^0 term. - Greg Huber, May 17 2018

a(n) >= a(n-1)^2 + a(n-1) for n >= 1, so the ratio log(a(n))/2^n is strictly increasing. But does it approach a limit? Conjecture: lim_{n->infinity} log(a(n))/2^n = 0.9748... - Jon E. Schoenfield, Jun 22 2018