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Define a sequence $a_n$ as such. $a_0=1$, $a_n=a_{n-1}+a\lfloor\frac{n}{3}\rfloor, \forall n\ge1$ Find all primes $p$ such that p divides infinitely many values of $a_i$.

Edit: This is extension to p=2,3..13. Suppose $a_i$ divides p. Then $a_{3i-1}$ to $a_{3i+2}$ is congruent mod p (if they are divisible by 13, we are done), and thus $a_{9i-4}$ to $a_{9i+8}$ forms a complete residue set mod p (as $p\le13$, and $a_{3i-1}$ to $a_{3i+2}$ is not divisible by 13) Can this method be extended to larger primes? By going one more layer down to $a_{27i-13}$ to $a_{27i+26}$, or going a few more layers?

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What have you tried? For a start, compute a bunch of terms and stare at them. Is $[\frac n3]$ the floor? We have \lfloor and \rfloor for that. –  Ross Millikan Apr 19 '13 at 15:27
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It doesn't hurt to mention that you already know that $p=2, 3, 5, 7, 11, 13$ work. This sequence is also the number of partitions of $3n$ into powers of $3$, which can be found here. –  Ivan Loh Apr 19 '13 at 15:29

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