Consider the recurrence given by
$(n+1)^2 a_{n+1} = (9n^2+9n+3)a_n-27n^2 a_{n-1}$
$a_0 = 1, a_1 = 3$.
Clearly, $a_n$ is rational, but unexpectedly, the recurrence seems to output only integral values. Can you prove that this is indeed so? For more terms, see A006077 in the OEIS.
This looks a lot like Apery's recurrence $$(n+1)^2a_{n+1}=(11n^2+11n+3)a_n+n^2a_{n-1}$$ Many papers have been written on generalizations of Apery's recurrence. I don't know whether any of them have studied the particular recurrence of this question, but it might be worth having a look at
Frits Beukers, On Dwork's accessory parameter problem, Math. Z. 241 (2002), no. 2, 425–444, MR1935494 (2003i:12013)
and
Don Zagier, Integral solutions of Apéry-like recurrence equations, Groups and symmetries, 349–366, CRM Proc. Lecture Notes, 47, Amer. Math. Soc., Providence, RI, 2009, MR2500571 (2010h:11069).
EDIT: Beukers' paper is available online; he writes that Zagier did a computer search for equations of this type returning only integral values, gives a table of what Zagier found, and the current equation is in the table. I can't find the Zagier paper online, but I'm guessing it will have the detailed proof of integrality.