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In the sequence (A052530 - OEIS) $2, 8, 30, 112, ...$ where $a(n)= 4a(n-1) - a(n-2)$, the prime factors demonstrate periodic behavior for the first $28$ elements of the sequence. For example, $3$ and $5$ are factors of only the elements whose index is a multiple of $3$ (period $3$ within the sequence). On the other hand, $7$ is a prime factor for only the elements whose index is a multiple of $4$ (period $4$). $11$ and $19$ have period $5$, $13$ has period $6$, $41$ and $71$ both have period $7, 97$ has period $8, 17$ has period $9$ and so on.

It really surprised me that the first few primes behaved like this, even for the short span of the sequence that I was able to check. My question is, do all of the primes have periodic behavior within this sequence? Has periodicity of prime factors been observed before in recursive sequences (references)?

Here is my conjecture: If $n$ is the index of the sequence and $k$ is a natural number, then there is a prime of the form $2kn +1$ or $2kn-1$ that is a factor of the $n$th element with a period $n$ within the sequence. Take $41$ and $71$, which first show up as factors of the $7$th element $5822$. They are also factors of the $14$th, $21$st and $28$th elements. $41 = 2(3)(7)-1$ and $71 = 2(5)(7)+1.$ This general observation seems to hold for at least the primes up to $23$, albeit for a small sample of the sequence.

If the conjecture is true, then a natural outcome is an algorithm for finding large primes, ie. take a prime index of the sequence, and then find prime factors of this element by successively dividing $2kn \pm 1$. Repeat using the highest prime output as an index again, and so on.

Bob Tivnan

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    Did you try reducing $a(n)=4a(n−1)−a(n−2)$ modulo $p$ ? here $a(n)-4a(n-1)+a(n-2)= 0$ is inversible modulo $p$, that is knowing $a(n),a(n -1) \bmod p$ you can find $a(n-2) \bmod p$, and so $a(n) \bmod p$ is periodic since the beginning. In general, with such a recurrence $a(n) = f(a(n-1),\ldots,a(n-k))$ it only ends up being periodic $\bmod p$, but not necessarily from the beginning. – 2017-01-29
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    I think I understand you. Any term a(n) has a prime factor p iff a(n) mod p = 0. Rearranging in mod p, a(n-2) = 4a(n-1) - a(n) mod p, which is simply the involution of the original recursive function. So, it turns out that the forward progression of terms mod p can be shown to have opposites mod p working backwards ...-a(n+2)mod p, -a(n+1)mod p, a(n)mod p, a(n+1)mod p, a(n+2)mod p...If any forward term a(n+k) mod p = 0, then there must be another term -a(n+k) mod p = 0. This will continue ad infinitum in the forward direction, but has a limit backwards since the terms are positive. – 2017-01-30

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