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Mar 12, 2015 at 9:52 history edited Jose Arnaldo Bebita CC BY-SA 3.0
streamlined the previous answer with respect to Ochem's improved lower bound for $I(n)$
Mar 12, 2015 at 9:46 history edited Jose Arnaldo Bebita CC BY-SA 3.0
streamlined the previous answer with respect to Ochem's improved lower bound for $I(n)$
Jun 12, 2011 at 21:35 vote accept Jose Arnaldo Bebita
Mar 3, 2011 at 5:31 comment added Jose Arnaldo Bebita As you can see, this method can be likened to a form of descent, coupled with exploiting symmetry via the divisibility constraint $\gcd(q, n) = 1$. More on this soon.
Mar 3, 2011 at 5:30 history edited Jose Arnaldo Bebita CC BY-SA 2.5
fixed an error in an inequality
Mar 3, 2011 at 5:14 history answered Jose Arnaldo Bebita CC BY-SA 2.5