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I noticed that for a Fibonacci sequence starting with seeds $(2,1)$, there is an awful amount of primes in the first $20$ elements of the sequence ($11$ primes), far more than $(0,1)$'s prime density. For $100$ elements, the density is much less than $\frac12 (18)$, but still surprisingly more than the prime density of the first $100$ 'normal-Fibonacci' integers.

Seeing this, I got curious about other seeds that could potentially give better prime density results. I don't know where to start from just guessing, though, and still don't know why the seed $(2,1)$ has a higher prime density. Is it just a coincidence? Can anyone help me out?

Work Until Now

Proving if $L_a$ is prime, $a$ must be a power of $2$, prime, or $0$:

In my proof, I show why every Lucas prime's index (n-value) must be $0$, a power of $2$, or a prime.

A Lucas number, by definition, is a number $L_n$, such that $L_n = A^n+B^n$, where $A$ and $B$ are the roots of $x^2-x-1$.

We want to prove that for any odd $m$, $L_{nm}$ can be divided. By acknowledging this fact, we can show that for $L_a$ to be prime, $a$ must not have odd factors other than $1$. If $a$ does not have any odd factors, it either is a power of $2$, prime, or $0$.

If $m$ is odd, then $X^m+Y^m = (x+y)(x^{m−1}−x^{m−2}y+\dots)$ Plugging in $A^n$ and $B^n$ for $X$ and $Y$, we get, $A^{nm}+B^{nm}=(A^n+B^n)\times S$, where $S$ is the rest of the binomial equation. This means $L_{nm}$ (which equals $A^{nm}+B^{nm}$) can be factored when $m$ is odd. Proving this, we now know that for $L_a$ to be prime, a must be a power of $2$, prime, or $0$.

Lucenaposition
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Aarav Trivedi
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    There is some background info at https://en.wikipedia.org/wiki/Lucas_number#Lucas_primes and https://primes.utm.edu/glossary/page.php?sort=LucasPrime, but surprisingly, nothing to directly address your question. – Chris Culter Aug 18 '25 at 21:04
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    There are a lot of small primes...I suggest doing a much deeper search to see if there is a persistent pattern. It might also be worth trying to generalize the standard fibonacci result that $\gcd(F_n,F_m)=F_{\gcd(m,n)}$ to different seeds. I assume that's been done elsewhere, but worth a look. – lulu Aug 18 '25 at 21:05
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    In general, $F_n\mid F_{nm}.$ This is also true for Lucas numbers when $m$ is odd. This greatly reduces the possible primes. It is not clear this is true for other starts, so it seems possible you can do better. – Thomas Andrews Aug 18 '25 at 22:23
  • https://en.wikipedia.org/wiki/Natural_density – Mason Aug 19 '25 at 03:04
  • https://cs.uwaterloo.ca/journals/JIS/VOL8/Noe/noe5.pdf And https://oeis.org/A000032 – Mason Aug 19 '25 at 03:07
  • Going in the other direction, there are seeds, relatively prime, such that there are no primes at all in the sequence. More information https://oeis.org/A083105 and https://www.primepuzzles.net/problems/prob_031.htm and elsewhere. – Gerry Myerson Aug 19 '25 at 04:19
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    If you go to bigger N, (2, 1) no longer has the most primes. It varies with n. I quickly computed a,b ≤ 10 up to N = 1 million and the biggest was (4, 1), and then N = 10 million, and the maximum is at (9, 2), but all the prime counts approach each other and are well within 1% from the maximum. – Dronir Aug 19 '25 at 08:48
  • Hmm, very interesting. @GerryMyerson I had actually figured this out before, but I'm now wondering if I could subtract these sequences from other Fibonacci sequences to get more prime dense answers. – Aarav Trivedi Aug 19 '25 at 15:20
  • Also @Dronir, thanks for that. My terrible computer took a really look time to even get to 300, so that analytics really helps. – Aarav Trivedi Aug 19 '25 at 15:20
  • Why in the factorization $A^{nm}+B^{nm}=(A^n+B^n)\times S$, $S$ is integer? – Alex Ravsky Aug 22 '25 at 09:58

1 Answers1

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What you are describing is the Lucas number sequence. We commonly take $L_0=2,L_1=1$.

Unlike the Fibonacci sequence, this does not produce a zero term anywhere whether we propagate backwards or forwards. With $L_0=2,L_1=1$ as above we have $L_{-n}=(-1)^nL_n$, and the terms for positive $n$ are positive and monotonically increasing.

This causes not all primes to be factors of Lucas numbers, which is again unlike the Fibonacci ones. For instance, no Lucas numbers are divisible by $5$ or by $13$. Thereby small Lucas numbers tend to have an increased probability of being prime.

For a geometric appearance of Lucas numbers, see here.

Oscar Lanzi
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  • I've found that out, but what I was really trying to get to (which I didn't state too clearly in my question statement, my fault) is if the Lucas Sequence's seeds (2,1) are the seeds that prompt the greatest prime density for the first n elements, with n going to infinity. I've asked this since I have observed many seeds a and b, with a <100 and b<100, and found (2,1) to give the greatest prime density for n = 100. – Aarav Trivedi Aug 18 '25 at 21:42
  • Nine users upvoted this answer. That makes ten who might like to know why it draws a downvote. – Oscar Lanzi Aug 20 '25 at 00:11
  • Can you speak to the density question? – Mason Aug 20 '25 at 00:29
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    It's rarely worth wondering/worrying about an isolated downvote, Oscar. I note that the question also has a single downvote – maybe one user thought it was a bad question, and the same user thought you shouldn't post an answer to a bad question. – Gerry Myerson Aug 20 '25 at 02:31