Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On the topic of profinite integers $\hat{\bf Z}$ and Fibonacci numbers $F_n$, Lenstra says (here & here)

For each profinite integer $s$, one can in a natural way define the $s$th Fibonacci number $F_s$, which is itself a profinite integer. Namely, given $s$, one can choose a sequence of positive integers $n_1, n_2, n_3,\dots$ that have more and more initial digits in common with $s$, so that it may be said that $n_i$ converges to $s$ for $i\to\infty$. Then also the numbers $F_{n_1}, F_{n_2}, F_{n_3},\dots$ get more and more initial digits in common, and we define $F_s$ to be their "limit" as $i\to\infty$. This does not depend on the choice of the sequence of numbers $n_i$.

Now, $\hat{\bf Z}\cong\prod_p{\bf Z}_p$ is a direct product of rings of $p$-adic integers, which has FToA and CRT sort of hardcoded into it (via decompositions of the underlying inverse systems). If a sequence of profinite integers $(x_p)$ converges, then in particular each coordinate converges (note I am using the direct product notation for profinite integers, rather than the factorial number system Lenstra uses). This would indicate that the Fibonacci numbers are $p$-adically interpolable.

However, in $p$-adic Interpolation of the Fibonacci Sequence via Hypergeometric Functions,

We say that a sequence $\{a_n\}_{n=1}^\infty$ of rational numbers is $p$-adically interpolatable if there exists a continuous function $f:{\bf Z}_p\to{\bf Q}_p$ such that $f(n)=a_n$ for all nonnegative integers $n$. Since the set of nonnegative integers is dense in ${\bf Z}_p$, for a given sequence $\{a_n\}$ there can be at most one such function, which will only exist under certain strong conditions on $\{a_n\}$. Specifically, an integer sequence is $p$-adically interpolatable if and only if it is purely periodic modulo $p^M$ for all positive integers $M$, with each period a power of $p$. While $\{F_n\}$ is purely periodic modulo $p$ for every prime $p$, its period modulo $p$ is never a power of $p$, which means that the Fibonacci sequence itself can never be $p$-adically interpolated.

This says pretty much the opposite. Evidently there are holes in my understanding. Any ideas?

share|cite|improve this question
up vote 4 down vote accepted

How would you interpolate the Fibonacci sequence from $\mathbb Z$ to $\mathbb Z_p$, using the interpolation to the profinite integers $\hat{\mathbb Z}$? Given a $p$-adic integer $n \in \mathbb Z_p$, the obvious way to define $F_n \in \mathbb Z_p$ would be considering $n$ as an element of $\hat{\mathbb Z}$ via the inclusion $\mathbb Z_p \hookrightarrow \prod_p \mathbb Z_p = \hat{\mathbb Z}$, taking $F_n \in \hat{\mathbb Z}$ and projecting it back down to $\mathbb Z_p$. However, the resulting map $F: \mathbb Z_p \to \mathbb Z_p$ will not be an interpolation of the integer Fibonacci function $\mathbb Z \to \mathbb Z$ since the inclusion $\mathbb Z \hookrightarrow \hat{\mathbb Z}$ is not the same as the composite inclusion $\mathbb Z \hookrightarrow \mathbb Z_p \hookrightarrow \hat{\mathbb Z}$.

share|cite|improve this answer

Here is a more elegant construction of the Fibonacci sequence $F : \widehat{\mathbb{Z}} \to \widehat{\mathbb{Z}}$. The idea is that the $n$-th Fibonacci number is the entry $(1,1)$ of the matrix power $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n$. This is true for $n \in \mathbb{N}$, but we can make it true for $n \in \widehat{\mathbb{Z}}$, too:

The matrix $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ yields a homomorphism of groups $\mathbb{Z} \to \mathrm{GL}_2(\widehat{\mathbb{Z}})$. Since the latter is a profinite group ($\widehat{\mathbb{Z}}$ being a profinite ring), it follows by the universal property of the profinite completion that there is a unique extension to a continuous homomorphism $\widehat{\mathbb{Z}} \to \mathrm{GL}_2(\widehat{\mathbb{Z}})$. The entry $(1,1)$ of this gives the desired map $F$.

For $\mathbb{Z}_p$ this doesn't work since $\mathrm{GL}_2(\mathbb{Z}_p)$ is not a pro-$p$-group.

share|cite|improve this answer

The Fibonacci sequence is not $p$-adically interpolable, because usually for a prime $p$ and an exponent $n$, there is no exponent $m$ such that $a \equiv b \pmod {p^m} \implies F(a) \equiv F(b) \pmod {p^n}$

However it is interpolable on $\hat{\Bbb Z}$ because for every integer $N$ there is an $M$ such that $a \equiv b \pmod M \implies F(a) \equiv F(b) \pmod N$.

For example, pick $N=2$, the fibonacci sequence mod $2$ is $0,1,1,0,1,1,0,\ldots$ and so, it only depends on $n \pmod 3$, so the corresponding $M$ is $3$, which incidentally is not a power of $2$.

share|cite|improve this answer

I think I understand my issue now. Here was my original thought process: Let $\alpha\in{\bf Z}_p$, let $n_i\in\bf Z$ converge to $\alpha$ in $\hat{\bf Z}$ (via ${\bf Z}_p\hookrightarrow\hat{\bf Z}$) but otherwise arbitrary. Then $F_{n_i}$ converges (to say $f$) in $\hat{\bf Z}$ and hence in ${\bf Z}_p$, and in particular $n_i$ converges to $\alpha$ in ${\bf Z}_p$, and so $F_\alpha=\pi_p(f)$ is well-defined and independent of $n_i$. This only considers a subset of the $n_i$ that converge $p$-adically though.

Let $\eta,\kappa\in\hat{\bf Z}$ be distinct profinite integers with equal $p$-adic coordinates, $\pi_p(\eta)=\alpha=\pi_p(\kappa)$. Then let $n_i,m_i\in\bf Z$ respectively converge to $\eta$ and $\kappa$ in $\hat{\bf Z}$; in particular these integers will both converge in ${\bf Z}_p$ to $\alpha$. Both $F_\eta$ and $F_\kappa$ will exist, but there is no guarantee that $\pi_p(F_\eta)=\pi_p(F_\kappa)$, which is required for such a $p$-adic interpolation to be well-defined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.