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)](http://turnbull.mcs.st-and.ac.uk/~john/Zagier/Problems.html)). That's not a very satisfying answer though. Is there a good reason for this behavior?



Simple recurrence that fails to be integer for the first time at the 44th term
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](https://oeis.org/A292996) gives indices of first noninteger terms in similar sequences.


Some spin-off questions:


* [Is there any positive integer sequence $c\_{n+1}=\frac{c\_n(c\_n+n+d)}n$?](https://mathoverflow.net/q/282159/7076)
* [Can these sequences stay integer-valued as many times as we want and then fail?](https://mathoverflow.net/q/282174/7076)


