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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?

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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 $\mathbb Z \hookrightarrow \mathbb Z_p \hookrightarrow \hat{\mathbb Z}$.

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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 $2$.

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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.

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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.

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