TITLE: Simple recurrence that fails to be integer for the first time at the 44th term
QUESTION [16 upvotes]: The sequence defined by $a_0=a_1 =1$ and 
$$
a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1
$$
fails to be integer for the first time at $a_{44}$. Why??
You can verify the statement by computing the sequence mod 43 (see more commentary here (day 5, problem 3)). That's not a very satisfying answer though. Is there a good reason for this behavior?

REPLY [24 votes]: Copying my explanation from https://mathoverflow.net/a/217894/25028
The recurrence formula can be rewritten as
$$a_2=2,\qquad a_{n+1}=\frac{a_n\cdot (a_n+n-1)}n,\quad n\geq 2,$$
which somewhat justifies why $a_n$ remains integer for quite a while. It shows that $a_n$ accumulates most of the factors of the previous terms and gains some new ones. The division by $n$ happens to hit the existing factors up until $n=43$.
Another example of this kind is given by
$$b_2=2,\qquad b_{n+1}=\frac{b_n\cdot (b_n+n+5)}n,\quad n\geq 2,$$
which remains integer for up to $n=59$.
ADDED. OEIS A292996 gives indices of first noninteger terms in similar sequences.
Some spin-off questions:

Is there any positive integer sequence $c_{n+1}=\frac{c_n(c_n+n+d)}n$?
Can these sequences stay integer-valued as many times as we want and then fail?