TITLE: Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$?
QUESTION [5 upvotes]: Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$?
The recurrences $\mod p$ can be different, possibly depending on $p$ and the recurrence need not be linear, any recurrence will do.

REPLY [4 votes]: YES, define $a_k$ by $a_0=0$, $a_k=a_{-k}$ for $k<0$ and for $k>0$ $a_k=m!$ where $m\ge 1$ is as large as possible subject to $k$ being a multiple of $m!$. Then $$a_{k+m!} \equiv a_k \mod m.$$ However we have $a_j=a_{m\cdot m!+j}$ for $1 \le j\le m!-1$ but not for $j=m!$ when $a_{m!} \ne a_{(m+1)!}$ so the sequence can't satisfy a recurrence of finite order.
By request: Here it is  from $a_{-9}$ to $a_{33}:$ 
$${\small \cdots 1,2,1,6,1,2,1,2,1,}\mathbf{0}\small{,\overline{1,2,1,2,1,6,1,2,1,2,1,6,1,2,1,2,1,6,1,2,1,2,1,24},1,2,1,2,1,6,1,\cdots}$$
$a_0=\mathbf{0}$ and all other terms are positive. The length 24 sequence with the overline keeps repeating except the $4!=24$ is $5!=120$ in positions 120,240,360,480,600 (but not 720) 840,960
Three notes: 

One could use the least common multiple of $\lbrace1,2,3\cdots m\rbrace$ in place of $m!$ 
Since the question asked only about primes one could make it have period #$p$ (p primorial) $\mod p$
If $a_k$ satisfies a recurrence of order $n$ mod $m$ then it is periodic $\mod m$ with a period $P=P_m$ which is no greater than $m^n$. Hence it is enough to ask: " If $<a_k>$ is an integer sequence which is periodic mod $m$ (with a period $P_m$ depending on $m$) for every $m$, must it satisfy a finite recurrence?