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Jan 18, 2014 at 8:35 comment added Felix Marin $k^{q - 1}$ instead of $k^{q}$.
Oct 23, 2013 at 14:52 review Low quality posts
Oct 23, 2013 at 14:53
Oct 23, 2013 at 14:34 history answered Brendan McKay CC BY-SA 3.0