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Let $\{a_{n}\}$be a sequence defined by $$a_{0}=0,a_{1}=a_{2}=1,a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$$ Prove that there exist infinitely many positive integers $n$ such that $a_{n}$ is divisible by $n$

I try:let $a,b.c$ be the set of all roots of $X^3-X^2-X-1\in \Bbb{Q}[x]$, we can take $(s,t,u)\in \Bbb{C}^3$, such that $$a_{n}=sa^n+tb^n+uc^n$$ for all $n\in \Bbb{Z}$

J. W. Tanner
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math110
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    What are the first dozen $a_n$? – Empy2 Feb 03 '21 at 14:11
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    Now I can't find the first $n$ such $n|a_{n}$ – math110 Feb 03 '21 at 14:12
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    The first few ones for which $n | a_n$ are $a_1 = 1, a_4 = 4, a_9 = 81$ – peter.petrov Feb 03 '21 at 14:14
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    Why is almost no one upvoting this question? Seems like a very interesting problem. – peter.petrov Feb 04 '21 at 16:56
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    @inequality Where does this problem come from? Is there any chance the sequence is supposed to start with $a_0 = a_1 = 0$ and $a_2 = 1$ instead? If so, it then would be the standard Tribonacci numbers. Also, this question would then be basically a duplicate of Prove or disprove that $2^n$ divides $T_{2^n}$ for $n > 2$.. As your question is currently stated, its indices for equivalent values are just one less than those of the standard sequence, e.g., your $a_3= 2$ and $T_4 = 2$ for the standard one. – John Omielan Feb 04 '21 at 17:33
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    it's from china teacher give me,he say it's old Mathematical Olympiad problem – math110 Feb 05 '21 at 00:35
  • @inequality Does the post given by John answer your question? – Sarvesh Ravichandran Iyer Feb 11 '21 at 10:24
  • @TeresaLisbon I don't think so. While ArtW's proof doesn't mention the initial conditions, they are embedded (hidden) in the expressions used in the recurrence. For instance here, $a_{16} \equiv 8 \pmod{64}$ and not $\equiv 32 \pmod{64}$ as ArtW's formulas predict. In fact, I'm pretty confident that in this problem, it is $2^n-4$ that always divides $a_{2^n-4}$. It doesn't look like it can be reconciled with ArtW's method, and I haven't found a way to prove the claim yet. – user3733558 Feb 11 '21 at 17:33
  • @user3733558 I see. Thanks for pointing it out. We will need to be more creative then. – Sarvesh Ravichandran Iyer Feb 11 '21 at 17:34
  • Well, that's awkward, $2^n-4$ doesn't work. $a_4$, $a_{12}$, $a_{28}$ looked promising, but it fails at $a_{60}$. – user3733558 Feb 11 '21 at 17:45
  • I get that for n < 200,000, the values that work are 1, 4, 9, 12, 28, 39, 52, 476, 1788, 2975, 3471, 4060, 5300, 6031, 7004, 8652, 16796, 16940, 22660, 24075, 24804, 27125, 27156, 28431, 32623, 32812, 34007, 37908, 69020, 77220. My computer doesn't have enough RAM to go much further. Plotting their logarithm does make it plausible that there are infinitely many. They don't seem to be quite randomly distributed either, though it's hard to say with so little data. – Joshua P. Swanson Feb 12 '21 at 11:02
  • @JoshuaP.Swanson I have that list up to $10^6$: ... 77220, 223079, 240236, 249164, 337676, 386725, 393820, 602140, 612300, 641199, 686566, 691652, 719316, 775359, 787244, and 1038180. – user3733558 Feb 12 '21 at 11:25
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    I think $13\cdot 3^m|a_{13\cdot 3^m}$,where $m\equiv 1\pmod 6$ – math110 Feb 12 '21 at 11:37
  • I can confirm math110's claim for $m=1,7,13$, waiting for the verdict for $m=19$. @math110: care to share the proof? :) – user3733558 Feb 12 '21 at 11:54
  • Meanwhile, $m=19$ confirmed. Though, I don't think I have enough CPU power on this laptop to go beyond that. – user3733558 Feb 12 '21 at 11:59
  • If we denote $h(x)$ the Trisano period modulo $x$, then math110's observation is probably related to the fact that $h(3)=13$ and the pattern of $h(13)$ (which is $168$) features the "double-zeroes" separations at $56$ and $112$. I'm perplexed as to how to derive a complete argument though... – user3733558 Feb 12 '21 at 12:39
  • @user3733558,How to find $h(13)=168?$ can you give this proof?because Now I found this problem is the key to show this – math110 Feb 12 '21 at 13:02
  • @inequality: I just looked by using a spreadsheet, so by direct calculation. I'm reading a paper that describes a process using the ring of integers of a splitting field of the Tribonacci polynomial over the field of $p$-adic numbers $\mathbb{Q}_p$. This is very much beyond me, but if you can get the document "Klaška, J. Tribonacci partition formulas modulo m . Acta. Math. Sin.-English Ser. 26, 465–476 (2010)" https://doi.org/10.1007/s10114-010-8433-8 then I suggest you pay attention to Example 4.7 which discusses modulo $3^t$ and derives $h(3)=13$. – user3733558 Feb 12 '21 at 13:58
  • I was wondering what this $\pmod 6$ was all about in math110's formula. It's fairly simple, $3^m$ has period $6$ when taken modulo $h(13)=168$. – user3733558 Feb 12 '21 at 21:27

1 Answers1

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$\DeclareMathOperator*{\lcm}{lcm}$We will need a few theorems from "Some properties of a generalized Fibonacci sequence modulo $m$", M.E. Waddill (1978). Let's denote $h(x)$ the Trisano period of the $a_n$ sequence taken modulo $$. To begin with, a few empirical observations: $$\begin{aligned} h(3) &= 13\\ h(13) &= 168 \end{aligned}$$ Also $$\begin{align*} a_{13\cdot 3} &\equiv 0 \pmod{h(3)}\\ a_{13\cdot 3} &\equiv 0 \pmod{h(13)} \end{align*}$$ This allows us to derive $$\begin{align*} a_{13\cdot 3} &\equiv 0 \pmod{\lcm(h(3),h(13))}\\ &\equiv 0 \pmod{h(13\cdot 3)} &\text{(see Waddill Theorem 4)} \end{align*}$$ And $39|a_{39}$ follows.

The general form for $13\cdot 3^{1+6k}$ with $k$ non-negative integer (as conjectured in the question's comments) poses no major difficulties. We know that

$$a_{13\cdot 3^{1+6k}} \equiv 0 \pmod{h(3^{1+6k})} \tag{1}$$ is true because $$\begin{aligned} h(3^{1+6k}) &= 3^{6k}h(3) &\text{(see Waddill Theorem 8)}\\ &= 13\cdot 3^{6k} &\text{which divides $13\cdot 3^{1+6k}$.} \end{aligned}$$ Taking into account that $3^m$ has period $6$ when taken modulo $168$, we also have $$a_{13\cdot 3^{1+6k}} \bmod{h(13)} = a_{13\cdot 3} \bmod{h(13)} = 0\tag{2}$$ So we can claim that $$\begin{align*} a_{13\cdot 3^{1+6k}} &\equiv 0 \pmod{\lcm(h(3^{1+6k}),h(13))}\\ &\equiv 0 \pmod{h(13\cdot 3^{1+6k})} \end{align*}$$ and conclude that $$13\cdot 3^{1+6k} | a_{13\cdot 3^{1+6k}}$$ for all non-negative integer $k$.

On the topic of finding $h(p)$ for $p$ prime: Trisano periods exist for all $p$ (that's Waddill's Theorem 1), however they are no more predictable than Pisano periods are for Fibonacci sequences. One can only compute as many terms as needed to find the first reoccurrence of the first terms. For small primes, using a spreadsheet is fast and simple:

Trisano formula

Column A for the indices, column B for the sequence itself, cell D1 for the modulus. Then you copy A4:B4 down a few hundred times, and you will find a new occurrence of $0,1,1$ at index $168$.

user3733558
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