Problem 69 - J

I decided to skip problem 68, at least for a little bit - it involves magic and polygons and all sorts of weird stuff. So, instead I decided to solve problem 69. This problem is rather simple in J, where a reasonably efficient implementation of Euler's totient function is part of the language: (5 p: y) is the totient of y. So, with that, writing a solution was not hard at all. Now, I am sure there are better solutions, as mine runs in about 20 seconds, but what works works.

f =: 3 : 'y % (5 p: y)'
lst =: f i.1000000
lst i.{.\:~lst