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Let $(a_n)$ be the sequence defined by $$a_{n+1}=2na_n-n^2a_{n-1}$$ and $a_0=0$ and $a_1=1$. I would like to prove that there exist infinitely many primes $p$ such that $p$ does not divide $a_p$. Any help or reference will be welcome.

Thanks in advance

edit: I made a mistake for the first terms: $a_0=0$ and $a_1=1$.

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    $\begingroup$ This is OEIS sequence 9940, see oeis.org/A009940 . What's the significance of primes $p$ such that $p \nmid a_p$? $\endgroup$
    – Noam D. Elkies
    Jun 1, 2015 at 3:27
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    $\begingroup$ Following Noam's comment and the explicit expression $a_n=n!L_n(1)$, where $L_n$ is the $n$-th Laguerre polynomial, it follows that $a_p$ is congruent to $-1$ modulo $p$ for any prime number $p$. $\endgroup$
    – Leonardo
    Jun 1, 2015 at 5:54
  • $\begingroup$ My bad. I made a mistake about the two first terms: $a_0=0$ and $a_1=1$ $\endgroup$
    – joaopa
    Jun 1, 2015 at 8:54
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    $\begingroup$ With the new definition, $a_{n} = -(n+1)! L_{n+1}(1)$ and it follows that $a_{p}$ is a multiple of $p$ if $p$ is an prime. $\endgroup$
    – Jeremy Rouse
    Jun 1, 2015 at 12:25
  • $\begingroup$ Jeremy. I am not convonced by your answer. The first 20 terms of the sequences are 0, 1, 2, 4, 6, -16, -310, -3144, -28826, -260000, -2345094, -20901880, -176084986, -1216168944, -1862029910, 186232275544, 6005924996070, 144514137334976, 3177768345524954, 67577079942366120, 1420754665075404166. They don't occur as terms of your sequence $\endgroup$
    – joaopa
    Jun 1, 2015 at 16:52

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