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.

  • $\begingroup$ Did you follow up the reference given at the OEIS? $\endgroup$ Jun 14, 2011 at 1:13

1 Answer 1


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)


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.

  • $\begingroup$ Integrality in the Zagier paper follows from the existence of closed forms, all as sums involving products of binomial coefficients. In general (i.e. for longer recurrence equations), closed forms are not known, in which case integrality has only been shown in very specific cases, namely when the solution arises geometrically out of certain Picard-Fuchs equations. $\endgroup$
    – awwalker
    Jul 17, 2011 at 23:08

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