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Well few days ago i asked a question on perfect numbers and Tito Piezas III answered the question in a very intriguing way which has helped me to get a lead on it.But his answer and perfect numbers made us land on a very interesting question about cubes.

The sequence A023042 on the OEIS website shows that a large percentage of $N^3$ are a sum of three positive cubes. OEIS lists only N<1770, but we can extend that:

$$\begin{array}{|c|c|} N&\text{%}\\ 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\\ \end{array}$$

This means that 94.2% of all N<10000 have a solution to $a^3+b^3+c^3=N^3$ in positive integers. Thus, if we pick a random N in the high end of that range, there is a very good chance that there is an a,b,c.

Now my question is that:Are there infinitely many $N^3$ (especially for prime $N$) that cannot be expressed as a sum of three positive cubes?

For example, there are no positive integers,

$$a^3+b^3+c^3=999959^3$$

even though the percentage of N<1000000 with solutions should be close to 99%.

Please help me in this question or provide me with any lead. The OEIS website link is: oeis.org/A023042

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  • $\begingroup$ I'd get an approximate formula for the number f (N) of triples a ≤ b ≤ c with a^3 + b^3 + c^3 ≤ N^3, calculate f' (N) and see if that grows to infinity. $\endgroup$ – gnasher729 Jan 26 '15 at 10:29
  • $\begingroup$ @Sayan: I fixed the table. $\endgroup$ – Tito Piezas III Jan 27 '15 at 5:54
  • $\begingroup$ @gnasher729: Actually, it is $0<a<b<c$. It is known that $x^3+2y^3=z^3$ has no rational solution. $\endgroup$ – Tito Piezas III Jan 27 '15 at 5:59
  • $\begingroup$ Thanks Tito :) well i am getting some leads on the perfect numbers $\endgroup$ – user210387 Jan 27 '15 at 10:53
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Don't know about prime $N,$ but otherwise yes: http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html

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