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**quasi** 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 $\displaystyle \log _3\left(\frac{n}{2}\right)+1$. 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: $\displaystyle \frac{1}{2}-\frac{1}{3}+\frac{1}{5}-\frac{1}{11}+\frac{1}{17}-\frac{1}{41}+\frac{1}{83}-\frac{1}{137}+\ldots $, it *appears* to approach $\displaystyle \frac{1}{\pi }$, but finding more terms in the sequence is currently an inefficient process, so I cannot test this hypothesis any further.)