How to compute the smallest integer which is sum of cubes in 13 ways? I would like to know the smallest integer which is sum of three positive cubes in 13 ways, such that $1^3+2^3+3^3=2^3+1^3+3^3$ are same ways of representation? What kind of theory there is behind that? I saw the question Sum of squares in at least 64 ways? but I don't know if that can be generalized as I'm not sure how to generalize Gaussian integers.
 A: This is not the smallest solution, but it gives you a simple method to find an integer that is the sum of three positive cubes in an arbitrary number of ways. All you have to do is use Lehmer's Identity,
$$(-9 p^3q + q^4)^3 + (-9p^4 + 3p q^3)^3 + (3p^2)^6 = q^{12}$$
If $q = 1$, then you get near-misses to the eqn $a^3+b^3 = c^3$.  
However, you can also choose a large enough constant $q$ such that the first two addends are positive for $p = 1,2,3,\dots,n$. For $n=13$, I find $q=28$ is enough, so,
$$\begin{aligned}
614404^3+65847^3+3^6 &= 28^{12}\\
612640^3+131568^3+12^6 &=28^{12}\\
607852^3+196839^3+27^6 &=28^{12}\\
598528^3+261120^3+48^6 &=28^{12}\\
\vdots\end{aligned}$$
and so on for 13 $p$ before the first addend becomes negative. Of course, if you choose larger $q$, then the addends will be positive for an even longer range of $p$.
A: To simplify the notation used in the Wikipedia article for generalized taxicab numbers, let $T(k,j,n)$ be the smallest (positive) integer that can be written as the sum of the $k$th powers of $j$ positive integers in $n$ different ways (up to rearrangements of summands).
The present Question asks for $T(3,3,13)$.  At first this seems a daunting computation, given that for $k=3$, $j=2$, ordinary taxicab numbers $T(3,2,n)$ are known only for $n=1,2,3,4,5,6$. In particular:
$$ T(3,2,6) = 24153319581254312065344 $$
according to the NMBRTHRY ListServ announcement by Uwe Hollerbach (2008).
However $T(3,3,n)$ could well be easier to find than $T(3,2,n)$ despite involving the determination of an extra summand.  The main reason for this is that there are far more numbers expressible as sums of three positive cubes than as sums of two positive cubes.
OEIS A024981 lists the numbers "that are the sum of 3 positive cubes, including repetitions." By downloading the first 2000 such integers into a database table, we can find by a SQL query that:
$$ T(3,3,2) = 251 $$
$$ T(3,3,3) = 5104 $$
$$ T(3,3,4) = 13896 $$
Comparison with the known "ordinary" taxicab number $T(3,2,n)$ suggest that these are not only smaller but growing much less rapidly with $n$.
On the other hand techniques for finding "ordinary" taxicab numbers outlined by David W. Wilson in The Fifth Taxicab Number are adaptable to $n$-way sums of three positive cubes.  One of these Wilson calls combination, to produce an $n+1$-way sum from a suitable pair of $n$-way sums, and it is a slight generalization of the technique I sketched in a Comment above.
Let $s$ and $s'$ be $n$-way expressible sums of three positive cubes that have equal cube-free part $e$:
$$ s = ed^3 = a_1^3 + b_1^3 + c_1^3 = \ldots = a_n^3 + b_n^3 + c_n^3 $$
$$ s' = ed'^3 = a_1'^3 + b_1'^3 + c_1'^3 = \ldots = a_n'^3 + b_n'^3 + c_n'^3 $$
We further assume that $s,s'$ are distinct primitive sums in the sense that the $a_i,b_i,c_i$ (respectively $a_i',b_i',c_i'$) taken together have greatest common factor $1$.
Then consider:
$$ \begin{align*}
   s'' &= e(dd')^3
   \\    &= sd'^3 = (a_1 d')^3 + (b_1 d')^3 + (c_1 d')^3 = \ldots
 = (a_n d')^3 + (b_n d')^3 + (c_n d')^3 
   \\    &= s'd^3 = (a_1' d)^3 + (b_1' d)^3 + (c_1' d)^3 = \ldots
 = (a_n' d)^3 + (b_n' d)^3 + (c_n' d)^3 \end{align*} $$
This provides $2n$ representations of $s''$ as sums of three positive cubes, at least $n$ of which are distinct since one set of $n$ is an exact multiple of the distinct representations of $s$.  If exactly $n$ are distinct, then one can divide out common factors to arrive at the same primitive $n$-way sum for $s$ and $s'$, contradicting their distinctness.  So the conclusion is this produces at least $n+1$ representations as sums of three positive cubes.
The construction sketched earlier was the special case $e=1$.
A: I guess it is 
119095488
= 24^3+204^3+480^3
= 48^3+ 85^3+491^3
= 72^3+384^3+396^3
= 113^3+264^3+463^3
= 114^3+360^3+414^3
= 149^3+336^3+427^3
= 176^3+204^3+472^3
= 190^3+279^3+449^3
= 207^3+297^3+438^3
= 226^3+332^3+414^3
= 243^3+358^3+389^3
= 246^3+328^3+410^3
= 281^3+322^3+399^3

The solution was given by Matti K. Sinisalo in http://keskustelu.suomi24.fi/node/12866739
