Interesting.. There's some info here, I don't know all the details, but there is a list up to the 1002nd term!2, 3, 5, 11, 17, 41, 83, 137, 257, 641, 1097, 2329, 4369, 10537, 17477, 35209, 65537, ...
I have been unable to find any function, f(n), that gives the nth term in this sequence.
For those that are curious to know how I came across it:
Euler's phi(n) function finds how many natural numbers less than n there are that are coprime to n (treating 1 as coprime to every number). If you recursively apply phi() enough times, the number eventually reaches 1. I wrote a function (in the programming sense of the word) that finds the least number of times you must apply phi(n) to get it down to 1, and I plotted the results. The most striking feature of the plot is that the points, while scattered about, seem to follow a logarithm. Specifically, the lower bound of the region in which they reside is . However, I have failed to find an upper bound. That is what the sequence is for: some points on the upper bound are (2,1),(3,2),(5,3),(11,4),(17,5),...
So you see, I want a function y=f(x) for the upper bound. Can anyone help me? Notice that they are not all prime, and the Euler phi of every 4th term, beginning with the first, is a power of 2. If you can solve this, you will have my eternal thanks.
(Interestingly, if you take the sum: , it appears to approach , but finding more terms in the sequence is currently an inefficient process, so I cannot test this hypothesis any further.)