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The Gessel sequence is known for Ira Gessel's Lattice Path Conjecture of $2001$, which has been proved by Kauers, Koutschan and Zeilberger in $2009$ with the aid of a computer. Later, other proofs were found ("human proofs"), e.g., by using Weierstrass elliptic functions (see here).

I was wondering whether or not it is true that the integers $a_n$ of the Gessel sequence are never prime numbers for $n\ge 3$. There is an easy recursion for this sequence, namely $a(0):=1$ and $$ a_{n+1}=\frac{4(6n+5)(2n+1)}{(3n+5)(n+2)}a_n $$ for all $n$. Is there some (obvious) reason that all $a_n$ are composite ?

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Yes. Towards a contradiction, suppose $n \geq 2$ is such that $a_{n+1}$ is prime. Say, $a_{n+1}=p$. Then, $(3n+5)(n+2)p=4(6n+5)(2n+1)a_n$. Evidently, $p$ does not divide $a_n$ since $a_{n+1} > a_n$. Thus, $p$ divides $6n+5$ or $2n+1$. This implies that $6n+5 \geq p=a_{n+1}$, which is a contradiction for all $n \geq 2$.

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    $\begingroup$ What a coincidence a_1 and a_2 to be prime. $\endgroup$
    – joro
    Commented Apr 15, 2016 at 16:05
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It is easy to see from the recurrence that every prime factor of $a_{n}$ is less than $6n$. But the sequence grows much faster than that.

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