TITLE: A congruence involving binomial coefficients
QUESTION [19 upvotes]: The following open problem was shown to me by Maxim Kontsevich. I
state it in a different but equivalent form. Let $a(n)$ be the sequence
at http://oeis.org/A131868, that is,
  $$ a(n) =\frac{1}{2n^2}\sum_{d|n}(-1)^{n+d}\mu(n/d){2d\choose d}. $$
Is it true that $6a(n)/n$ is always an integer? The sequence begins
(starting at $n=1$) $(1,1,1,2,5,13,35,100,300,925,2915,9386,\dots)$.
If $n$ is prime then the result is true. It is equivalent to
the well-known fact that ${2p\choose p}\equiv 2\,(\mathrm{mod}\,p^3)$
for a prime $p>3$. See for instance Enumerative Combinatorics, vol.
1, second ed., Exercise 1.14(d).

REPLY [2 votes]: I'd like to draw your attention to a recent paper On p-adic approximation of sums of binomial coefficients where we briefly discuss (in Section 4) the divisibility of this kind. Namely, we claim that for any integers $a\geq b>0$ and $m>0$, 
$$m^3\mid M(a,b)\cdot\sum_{d\mid m}\mu\left(\frac md\right)\binom{ad}{bd}$$
and
$$m^3\mid M'(a,b)\cdot\sum_{d\mid m}(-1)^{m+d}\mu\left(\frac md\right)\binom{ad}{bd},$$
where $M(a,b) = \frac{12}{\gcd(12,ab(a-b))}$ and 
$M'(a,b) = \frac{3}{\gcd(3,ab(a-b))}\cdot 2^\delta$, where 
$$\delta = \begin{cases}
\min\{1,\nu_2(b)\}, & \text{if}\ \nu_2(a-b)=\nu_2(b),\\
2, & \text{otherwise}.
\end{cases}
$$
In particular, for the case $(a,b)=(2,1)$ (questioned here), we have $M(2,1)=6$ and $M'(2,1)=3$.
We don't give a proof in the paper (as this is only loosely related to the main subject), but we can share it if anyone is interested.
In fact, the paper is mainly devoted to generalization of the congruence ${2p\choose p}\equiv 2\pmod{p^3}$ ($p>3$) to higher powers of $p$.