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considering a starting sequence of integer $(a_1,a_2,...,a_n) \in \mathbb{Z^n}$ let it apply the ducci operator $D$ that act like this \begin{equation} \mathbb{Z^n}\rightarrow \mathbb{Z^n} \\ (a_1,a_2,...,a_n)\mapsto (|a_1-a_2|,|a_2-a_3|,...,|a_n-a_1|) \end{equation} Let $D^k$ the Ducci operator acting $k$ times. It is know that if $n=2^j$ then $D^k$ converge to the zero sequence $(0,0,...,0)$, in a finite number of steps. Otherwise it will converge to zero ot to a periodic sequence. I was wondering if there is any results in the p-adic integers. Let rephase: $(a_1,a_2,...,a_n) \in \mathbb{Z_p^n}$ where $\mathbb{Z_p}$ is the p-adic integers ring. Let $D_p$ the p-adic ducci operator acting \begin{equation} \mathbb{Z_p^n}\rightarrow \mathbb{Z_p^n} \\ (a_1,a_2,...,a_n)\mapsto (|a_1-a_2|_p,|a_2-a_3|_p,...,|a_n-a_1|_p) \end{equation} Where $|x|_p$ is the p-adic valuation. Is is obvius that if $|a_i-a_{i+1}|_p$ is in the set $0,1,p^2,..p^{p-1}$ but I cannot go further, in deciding if applying the p-adic Ducci operator will create a sequence in $\mathbb{Z^p}$ that converges or becomes periodic. I google a while but didn't find any meaningfull reply. Does someone faced this problem?

Thanks in advance

Paolo

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    $\begingroup$ You are slightly mistaken in the last paragraph: for every $p$-adic integer $a$ we have $|a|=p^{-n}$ with $n\geq 0$. You also seem to be confusing the ring of $p$-adic integers with the ring of integers modulo $p$. Anyway, I guess that you should be able to prove what you want using the ultrametric inequality: for every pair $x,y$ of $p$-adic integers we have $|x+y|_p\leq\max(|x|_p,|y|_p)$, with equality unless $|x|_p=|y|_p$. $\endgroup$
    – A.P.
    Commented Jan 19, 2016 at 22:48
  • $\begingroup$ But the map doesn’t send $\Bbb Z_p^n$ to itself. $\endgroup$
    – Lubin
    Commented Jan 21, 2016 at 4:05
  • $\begingroup$ @A.P. for me $Z_p$ is the p-adic integers ring where $|x|_p \leq 1$, $|x|_p = \frac{1}{p^{ord_px}}$ where $ord_p(x) $ is the max power of $p$ that divide $x$ $\endgroup$
    – user730712
    Commented Jan 21, 2016 at 10:21
  • $\begingroup$ @Lubin yes you're right but I can easily extend the problem to $Q_p$ the filed of p-adics. In this case the Ducci p-adic operator acts from $Q_p$ to $Q_p$ (seems to me) $\endgroup$
    – user730712
    Commented Jan 21, 2016 at 10:24
  • $\begingroup$ Well, you say that it is obvious that $|a_i - a_{i+1}|_p \in \{0,1,p,\dotsc,p^{p-1}\}$, but not only this isn't obvious... in general it is outright false! As a simple example, considering $a_i = p^n$ and $a_{i+1} = 0$ you see that $|a_i - a_{i+1}|_p$ can take any value in $\left\{ \frac{1}{p^n} : n \in \Bbb{Z}_{\geq 0}\right\} \not \subset \Bbb{Z}_p$... $\endgroup$
    – A.P.
    Commented Jan 21, 2016 at 11:17

1 Answer 1

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Things go very smoothly once we formalize the setting. First of all, fix a positive (rational) integer $n$, define the set $P = \{0\} \cup \{p^k : k \in \Bbb{Z}\}$, and consider the map $$ \begin{align} &D_p \colon &\Bbb{Q}_p^n &\to P^n \subset \Bbb{Q}_p^n\\ &&(a_1,\dotsc,a_n) &\mapsto (|a_1 - a_2|_p, |a_2 - a_3|_p,\dotsc, |a_n - a_1|_p). \end{align} $$ Then, given a tuple $\alpha = \alpha^{(0)} \in \Bbb{Q}_p^n$, consider the $p$-adic Ducci sequence $\{\alpha^{(k)}\}_{k = 0}^{\infty}$ with $\alpha^{(k)} = D_p^k(\alpha)$ for every $k \geq 1$. Now observe that, up to dropping the first term of the sequence, we may assume that $\alpha \in P^n$. Furthermore, by the ultrametric inequality we know that for every $x,y \in P$ we have $$ \begin{cases} |x - y|_p = \max(|x|_p,|y|_p) & \text{if } x \neq y \\ 0 & \text{if } x = y \end{cases} $$ thus $p^{-\nu} \leq a_i^{(k)} \leq p^{\nu}$ for every $i \in \{1,\dotsc,n\}$ and $k \geq 0$, where $\nu = \max|v_p(a_i^{(0)})|$. In particular, this means that there are only finitely many possible values for $\alpha^{(k)}$, so by the pidgeonhole principle there must be some $r,s > 0$ such that $\alpha^{(r)} = \alpha^{(r+s)}$. But then $$ \alpha^{(r + i + hs)} = \alpha^{(r+i)} \quad \text{for every } 0 \leq i < s \text{ and } h \geq 0 $$ i.e. $\{\alpha^{(k)}\}_{k = 0}^{\infty}$ is ultimately periodic.

We can actually go much further than that: with little work we can adapt the proof of lemma 3 in [2] to $D_p$, so the study of the periods of $p$-adic Ducci sequences reduces to that of the periods of Ducci sequences in $\Bbb{F}_2^n$. In particular, the proofs in section 3 of [1] work in this case without modification. For example, this means that if $n$ is a power of $2$, then for every $p$-adic Ducci sequence $\{\alpha^{(k)}\}$ there is a $K \geq 0$ such that$\alpha^{(k)} = 0$ for all $k \geq K$.


[1] Ehrlich, Amos. Periods in Ducci’s n-number game of differences. The Fibonacci Quart 28 (1990): 302-305.

[2] Furno, Anne Ludington. Cycles of differences of integers. Journal of Number Theory 13.2 (1981): 255-261.

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  • $\begingroup$ A.P. You've got it!!! $\endgroup$
    – user730712
    Commented Jan 23, 2016 at 10:23

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