Why is $a_{214}$ nonintegral where $a_0 = 1, a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}?$ Define the sequence $a_n$ by the following. 

$$a_0 = 1,$$
$$a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}$$

$a_{214}$ is nonintegral.
（https://oeis.org/A108394）
Please tell me the proof.
(By computing the sequence modulo 251 it is easy to show that $a_{251}$ is not integral.)
 A: In general, we can take the recursion $$a_0=1,\qquad a_{n+1}={1+a_0^k+a_1^k+\cdots+a_n^k\over n+1}$$ and rewrite it as $$(n+1)a_{n+1}=a_n(a_n^{k-1}+n)$$ Then choose some prime $p$ and start calculating the $a_i\bmod p$, and if $a_{p-1}(a_{p-1}^{k-1}+p-1)$ isn't a multiple of $p$, you win –– $a_p$ isn't an integer. 
It seems that this works for $k=2$ with $p=43$, for $k=3$ with $p=89$, and for $k=4$ with $p=97$. 
The trouble with $k=5$ is that 214 isn't a prime, so the calculations don't get off the ground. 
I wonder whether 214 isn't a typo, and maybe it's supposed to be some prime like 241 or 251. It's probably worth having a look at the Ibstedt paper to which I linked in my comment on the question. 
A: EDIT: This isn't fully correct and contains some imprecise calculations. The core is using modulo 215. To misquote the OP: "By computing the sequence modulo 215 it is easy to show that $a_{215}$ is not integral."
ORIGINAL:
Doing the calculations modulo 215 gives the following:
$$a_0=1$$
$$a_1=\frac{1+1^5}{1}=2$$
$$a_2=\frac{1+1^5+2^5}{2}=17$$
$$a_3=\frac{1+1^5+2^5+17^5}{3}=473297\equiv82\pmod{215}$$
$$a_4\equiv\frac{1+1^5+2^5+17^5+82^5}{4}\equiv42\pmod{215}$$
$$a_5\equiv\frac{1+1^5+2^5+17^5+82^5+42^5}{5}\equiv171\pmod{215}$$
$$a_6\equiv\frac{1+1^5+2^5+17^5+82^5+42^5+171^5}{6}\equiv171\pmod{215}$$
All consecutive terms are 171 module 215 which leads to your conclusion.
