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Oct 12, 2017 at 12:20 answer added Rosie F timeline score: 0
Apr 1, 2015 at 3:58 history edited Tito Piezas III
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Apr 1, 2015 at 3:55 comment added Tito Piezas III Note that the next step with 16 consecutive cubes $1^3,2^3,3^3,\dots, 16^3$ yields the equi-partition, $$1^k + 4^k + 6^k + 7^k + 10^k + 11^k + 13^k + 16^k = 2^k + 3^k + 5^k + 8^k + 9^k + 12^k + 14^k + 15^k$$ which is good for $k=0,1,2,3$. See also this post.
Mar 24, 2015 at 10:12 answer added barak manos timeline score: 2
Mar 24, 2015 at 9:52 history edited A is for Ambition CC BY-SA 3.0
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Mar 24, 2015 at 7:40 answer added David timeline score: 3
Mar 24, 2015 at 7:36 history asked A is for Ambition CC BY-SA 3.0