How many solutions are there to $\sum_{i=1}^3 x_i^2+x_iy_i+y_i^2=k$? Let $k$ be a positive integer. Let
$$Q=
\begin{pmatrix}
 1 &1/2&   &   &   &   \\
1/2& 1 &   &   &   &   \\
   &   & 1 &1/2&   &   \\
   &   &1/2& 1 &   &   \\
   &   &   &   & 1 &1/2\\
   &   &   &   &1/2& 1 
\end{pmatrix}.
$$

How many solution $x\in\mathbb Z^6$ are there to $\quad  x^tQx=k$?

This is equivalent to: 

How many solution $x\in\mathbb Z^6$ are there to 
  $$x_1^2+x_1x_2+x_2^2+ x_3^2+x_3x_4+x_4^2+ x_5^2+x_5x_6+x_6^2=k?$$

or to

How many solution $x\in \mathbb Z\left[\omega \right]^3$ are there to $\quad x^* I_3 x=k$?

where $I_3$ is the $3\times3$-identity matrix and $\omega=\frac{1+\sqrt{-3}}{2}$.
I know that there is a formula for this number (there is only one class in its genus), but I don't know it.
This question is related to


*

*which integers take the form x^2 + xy + y^2 ?

*https://math.stackexchange.com/questions/44139/how-many-solutions-are-there-to-fn-m-n2nmm2-q/
but they don't answer my question. 
 A: This is a supplement to Noam Elkies' nice answer. The coefficients $s(k)$ can be expressed as
$$ s(k)=27\sum_{d\mid k}\chi(k/d)d^2-9\sum_{d\mid k}\chi(d)d^2, $$
hence the function $\varphi$ is a linear combination of
$$E_1:=\sum_{k=1}^\infty\sum_{d\mid k}\chi(k/d)d^2q^k
\quad\text{and}\quad
E_2:=1-9\sum_{k=1}^\infty\sum_{d\mid k}\chi(d)d^2q^k.$$
The latter functions are proportional to the standard Eisenstein series
$$ E_1':=\sum'_{m,n\in\mathbb{Z}}\chi(m)(mz+n)^{-3}
\quad\text{and}\quad
E_2':=\sum'_{m,n\in\mathbb{Z}}\chi(n)(mz+n)^{-3},$$
which form a basis of the space of modular forms $M_3(\Gamma_0(3),\chi)$, hence indeed $\varphi$ lies in this space. For more details see Section 7.1 in Miyake: Modular Forms, especially Lemma 7.1.1 and Theorem 7.1.3.
A: Let's recall that the number $R(k)$ of representations of $k$ as $x^2+y^2$ can be written as follows: write $k=2^\alpha bc$ where $b$ is composed entirely of primes congruent to 1 (mod 4) and $c$ is composed entirely of primes congruent to 3 (mod 4). Then $R(k)=0$ unless $c$ is a square, in which case $R(k) = 4\tau(b)$, where $\tau(b)$ is the number of divisors of $b$.
A very similar proof would surely address the number $S(k)$ of representations of $k$ as $x^2+xy+y^2$: write $k=3^\alpha bc$ where $b$ is composed entirely of primes congruent to 1 (mod 3) and $c$ is composed entirely of primes congruent to 2 (mod 3). Then I believe that $S(k) = 0$ unless $c$ is a square, in which case $S(k) = 4\tau(b)$. (Or maybe it's $6\tau(b)$.) I guess we should also mention $S(0)=1$.
In your original question, the number of representations of $k = x^t Qx$ where $x\in{\mathbb Z}^6$ will be exactly the triple convolution $\sum_{m=0}^k \sum_{n=0}^{k-m} S(m)S(n)S(k-m-n)$. ($Q$ is positive definite so we needn't worry about negative integers.) This probably leads to a rather different-looking formula than one would get from modular forms.
A: The formula that emiliocba seeks seems to be as follows.
Let $\chi$ be the Dirichlet character mod $3$.  For $k>0$
write $k = 3^e n$ with $n \equiv \pm 1 \bmod 3$.  Then
the number of representations of $k$ by this quadratic form $A_2^3$ is
$$
s(k) :=
9 (3^{2e+1}-\chi(n)) \phantom. \sum_{d|n} \phantom. \chi(n/d)\phantom. d^2.
$$
I append gp code that verifies that this holds for $k \leq 432$.
To prove it in general it will be enough to check that
$$
\varphi := 1 + \sum_{k=1}^\infty \phantom. s(k) q^k
$$
is a modular form of weight $3$ and character $\chi$ for $\Gamma_0(3)$,
and to match a few coefficients with the theta function $\theta_{A_2^3}$.
In principle, it is enough to match only the $q^0$ coefficient:
the dual of $A_2^3$ is isomorphic with the scaling of $A_2^3$ by $1/3$,
so by Poisson summation $\theta_{A_2^3}$ is modular also for
the normalizer $\Gamma_0^+(3)$ of $\Gamma_0(3)$ (generated by $\Gamma_0(3)$ and
the involution $w_3 : \tau \longleftrightarrow -1/3\tau\phantom.$);
and $\Gamma_0^+(3)$ has only one cusp,
and no cusp forms of weight less than $6$ (the weight of
$\eta(\tau)^6 \eta(3\tau)^6$), so the normalized Eisenstein series $\varphi$
is the only candidate for $\theta_{A_2^3}$.
H = 24
A2 = sum(m=-H,H,sum(n=-H,H,q^(m^2+m*n+n^2))) + O(q^(3*H^2/4+1));
L = A2^3;

chi3(m) = kronecker(m,3)
{
s(k, e,n) =
  e = valuation(k,3);
  n = k / 3^e;
  9 * (3^(2*e+1)-chi3(n)) * sumdiv(n, d, chi3(n/d)*d^2)
}

L == 1 + sum(k=1,3*H^2/4,s(k)*q^k)

