Coincidence in dimensions of Hilbert modular cusp forms I'm wondering if someone can shed some light onto the following, which most likely is just a silly coincidence, but I would be interested to know if there is more to it.
I found myself needing to compute some dimensions of spaces of Hilbert modular cusp forms over $\mathbb{Q}(\sqrt{5})$ of parallel weight $[2k,2k]$ and level $\Gamma_0(\mathfrak{p}_{11})$  where $\mathfrak{p}_{11} $ is a prime ideal dividing $11$. Now I started by doing the naive thing of simply going to MAGMA and asking for Dimension(HilbertModularCuspForms(...)). Now this is not a good idea since it will become very slow very quickly. But still, if you wait a bit you find that the sequence [Dimension, weight] is $[ 1, 4 ],[ 5, 6 ]
,[ 9, 8 ]
,[ 17, 10 ]
,[ 25, 12 ]
,[ 33, 14 ]
,[ 45, 16 ]
,[ 57, 18 ]
,[ 73, 20 ]
,[ 89, 22 ]
,[ 105, 24 ]
,[ 125, 26 ]
,[ 145, 28 ]
,[ 169, 30 ]$ Now I looked at the sequence of dimensions and wondered if there was a pattern, so I did the usual trick of typing OEIS into google and then putting in the dimensions. It turns out the dimensions match up (as far as I've computed) with the number of (integer) solutions of $w+2x+5y=0$ for $x,y,w \in [-k \dots k]$.
So my question is: Is this nothing more than a coincidence? or is there any connection here?  The dimension formulas for Hilbert modular forms seem to usually be a bit messy so I can't really see how the two things are connected (but this likely just because I haven't looked at them carefully yet). My hope was to eventually have a quick way of computing these dimensions and having such a description would make computations much easier. 
Full disclosure: The sequence on OEIS actually starts $1,1,5,9,17,$.. and the first "1" would correspond to the dimension in weight $[2,2]$, which is actually zero, so the sequences don't match up in the first term. But if you look at the corresponding space on the quarternionic side (via the Jacquet-Langlands correspondence), then in weight $[2,2]$ one needs to quotient out the space by the stuff factoring through the reduced norm map, which in this case is 1-dimensional, so this might be the where the "1" is.
Thank you.
 A: This answer is very late (and D. Loeffler's formula in the comments is certainly correct), but I want to point out that the dimension formula for parallel-weight modular forms over real-quadratic fields is not all that messy after all.
By Shimizu (in particular, the formulation of Theorem 2.15 in Thomas and Vasquez, Rings of Hilbert modular forms, Compos. Math. 48(2) 139-165), for every $k \ge 2$ and every congruence subgroup $\Gamma \le \Gamma_K := \mathrm{PSL}_2(\mathcal{O}_K)$,
$$\mathrm{dim}\, S_{2k,2k}(\Gamma) = 2k(k-1) \zeta_K(-1) [\Gamma_K : \Gamma] + \chi - a_3 \delta_K s_D - \varepsilon_k a_5 /5,$$ where:
$\zeta_K$ is the Dedekind zeta function;
$\chi$ is the arithmetic genus (which is $1 + \mathrm{dim}\, S_{2,2}(\Gamma)$; in particular, this formula is off by one at $k=1$);
$a_n$ counts the elliptic points with stabilizer of size $n$;
$\delta_k$ and $\varepsilon_k$ are defined by $$\delta_k = \begin{cases} 1: & k \equiv 2\, (\text{mod} \, 3); \\ 0: & \text{otherwise}; \end{cases} \quad \varepsilon_k = \begin{cases} 2 : & k \equiv 2,4 \, (\text{mod}\, 5); \\ 1: & k \equiv 3 \, (\text{mod}\, 5); \\ 0: & \text{otherwise}; \end{cases}$$
and where $s_D$ depends on the discriminant $D$ of $K$: $$s_D = \begin{cases} 1/6: & D \not \equiv 0 \, (\text{mod}\, 3); \\ 4/15: & D > 12 \; \text{and} \; D \equiv 3 \, (\text{mod} \, 9); \\ 1/3: & D = 12 \; \text{or} \; D \equiv 6 \, (\text{mod}\, 9). \end{cases}$$
There are at most three ``unknowns" in this formula, $\chi$, $a_3$ and $a_5$, and if all else fails you can simply read these off the dimensions of $S_{2k,2k}(\Gamma)$ for $k=1,2,3$. (And $a_5 = 0$ unless $K = \mathbb{Q}(\sqrt{5})$. Unfortunately you are in that exceptional case.)
In the case you looked at, we find $\zeta_K(-1) = 1/30$; $[\Gamma_K: \Gamma] = 12$; $\chi = 1$; $a_3 = 0$; $a_5 = 4$ and therefore $$\mathrm{dim}\, S_{2k,2k}(\Gamma) = \frac{4k(k-1)}{5} + 1 - \frac{4}{5}\varepsilon_k, \; \; k \ge 2.$$
