There are n ropes.  Pick two rope ends uniformly at random and tie them together. Repeat until there are no rope ends left. What is the expected number of loops at the end of the process?



When you tied the first pair of ends, you make a loop with probability 1/(2n-1) and are left with a box with (n-1) ropes in it. The expected number of loops from n ropes is 1/(2n-1) plus the expected number from (n-1) ropes. 



In other words expected number of loops is f(n) = 1/(2n-1) + f(n-1) given n ropes. 


a(n) describes the inverse of this, how many ropes we need to start with for the expected number of loops to be >=n.