TITLE: Are there infinitely many $N^3$ (especially for prime $N$) that cannot be expressed as a sum of three positive cubes?"
QUESTION [23 upvotes]: 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

REPLY [2 votes]: Don't know about prime $N,$ but otherwise yes: http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html