Factors of central binomial coefficient The central binomial coefficient $\binom{2n}{n}$ is divisible by $n+1$, as seen from the
identity
$$\binom{2n}{n} = (n+1)\binom{2n}{n} -(n+1)\binom{2n}{n+1}.$$
In fact the Catalan numbers
$$C_n=\frac{1}{n+1}\binom{2n}{n}$$
have a very large (over 200 according to R. Stanley's book) combinatorial interpretations. The identity
$$ \binom{2n}{n} =2(2n-1)C_{n-1}$$
shows that $\binom{2n}{n}$ is also divisible by $2n-1$. So it is natural to ask when
$\binom{2n}{n}$  is divisible by the product $(n+1)(2n-1)$. If $n\equiv 0$ or $1$ $\pmod{3}$, then $\gcd(n+1,2n-1)=1$, and so $(n+1)(2n-1)$ will be a factor of $\binom{2n}{n}$ in this case. But if $n\equiv 2 \pmod{3}$, then $\gcd(n+1,2n-1)=3$ so $(n+1)(2n-1)$ will always be a factor of $3\binom{2n}{n}$, but it may or may not be a factor of $\binom{2n}{n}$. According to (limited) computer calculations it seems that it is more likely that it is a factor. So I would be interested in understanding the set
$$\left\{n: (n+1)(2n-1) \mbox{ is not a factor of } \binom{2n}{n}\right\}.$$
According to computer calculations the first few elements of the set are
$$2,5,8,11,14,26,29,32,35,38,41,80,83,86,89,92,95,107,110,113,116,119,122,242,245,248,251,254,257,269,272,275,278,281,284.$$
So looking at the first 300 integers, about 35% of those that are congruent to $2\pmod{3}$ will be such that $(n+1)(2n-1)$ is not a factor of $\binom{2n}{n}$.
 A: Here is a different proof of the nice result found by Dark Malthorp.
Let
$$K_n=\frac{3}{(2n-1)(n+1)}\binom{2n}{n}$$
We will use the easily verified identities:
$$ K_n=\frac{6}{n(n+1)}\binom{2(n-1)}{n-1}\hspace{10ex} (1) $$
and
$$K_n=\frac{3}{2(2n-1)(2n+1)}\binom{2(n+1)}{n+1} \hspace{10ex} (2)$$
We will also make use of Legendre's formula for the $p$-adic valuation of the factorial. If
$p$ is a prime number, let $\nu_p(n)=\max\{k\in{\mathbb Z}^+: p^k|n\}$ (so in the notation of the previous answer,
$\ell_m=\nu_3(K_{3m+2})$). We also denote by $s_p(n)$ the sum of the digits of $n$ in its base $p$ expansion. Legendre's formula is
$$
\nu_p(n!)=\frac{n-s_p(n)}{p-1}
$$
The following lemma follows immediately from the observation that the presence of a $2$ in the base $3$ expansion of
$n$ would necessarily result in a carry when $n$ is multiplied by $2$.
Lemma
For each positive integer $n$, we have
$$ 
s_3(2n)\leq 2s_3(n)
$$
and equality occurs if and only if there are no $2$'s in the base $3$ expansion of $n$.
Using Legendre's formula on (1) and (2), we find
$$\nu_3(K_n)=s_3(n-1)-\frac{1}{2}s_3(2(n-1))-\nu_3(n) -\nu_3(n+1)+1 \hspace{10ex} (3)$$
$$\nu_3(K_n)=s_3(n+1)-\frac{1}{2}s_3(2(n+1))-\nu_3(2n-1)-\nu_3(2n+1) +1
\hspace{10ex} (4)$$
Suppose now $n\equiv 2 \pmod{3}$. Then $\nu_3(n)=\nu_3(2n+1)=0$.
If $n-1$ has no $2$'s in its base $3$ expansion, since $\nu_3(n-1)=0$,
the base $3$ expansion of $n-1$ ends in $d1$, where $d\in\{0,1\}$. This means that $n+1$ ends in
$(d+1)0$, and since $d+1\leq 2$, $\nu_3(n+1)=1$.
It then follows from (3) and the Lemma that $\nu_3(K_n)=0$.
If $n+1$ has no $2$'s in its base $3$ expansion, then since $\nu_3(n+1) >0$,
$n+1$ (in base $3$) ends in $0$, so $n$ ends in $2$, $2n$ ends in $21$, and $2n-1$ ends in $20$.
So $\nu_3(2n-1)=1$, and it follows from (4) and the Lemma that $\nu_3(K_n)=0$. This proves that if
$n\equiv 2 \pmod{3}$ and either $n-1$ or $n+1$ have no $2$'s in their base $3$ expansion, then $K_n$ is an integer.
For the converse, suppose that $\nu_3(K_n)=0$. Then (3) and (4) give us
$$s_3(n-1)=\frac{1}{2}s_3(2(n-1))+\nu_3(n+1)-1 \hspace{10ex} (5)$$
$$s_3(n+1)=\frac{1}{2}s_3(2(n+1))+\nu_3(2n-1)-1 \hspace{10ex} (6)$$
If $n-1$ has some $2$'s in its base $3$ expansion, then by the Lemma and (5) $\nu_3(n+1)\geq 2$.
So  $n+1$ (in base 3) ends in $00$, $n$ ends in $22$, $2n$ ends in $21$, and $2n-1$ ends in $20$. Hence
$\nu_3(2n-1)=1$ and from (6) and the Lemma, $n+1$ has no $2$'s in its base $3$ expansion.
