On the greatest common divisor of n and the nth Fibonacci number, II

Abstract

Let $\mathcal {A}$ be the set of all integers of the form $\gcd (n, F_n)$ , where n is a positive integer and $F_n$ denotes the nth Fibonacci number. Leonetti and Sanna proved that $\mathcal {A}$ has natural density equal to zero, and asked for a more precise upper bound. We prove that

$$ \begin{align*} \#\big(\mathcal{A} \cap [1, x]\big) \ll \frac{x \log \log \log x}{\log \log x} \end{align*} $$

for all sufficiently large x. In fact, we prove that a similar bound also holds when the sequence of Fibonacci numbers is replaced by a general nondegenerate Lucas sequence.

1 Introduction

Let $(u_n)$ be a nondegenerate linear recurrence with integral values. Arithmetic relations between n and $u_n$ have been studied by several authors. For example, the set of positive integers n such that n divides $u_n$ has been studied by Alba González, Luca, Pomerance, and Shparlinski [2], assuming that the characteristic polynomial of $(u_n)$ is separable, and by André-Jeannin [3], Luca and Tron [13], Sanna [22], and Somer [28], when $(u_n)$ is a Lucas sequence. Furthermore, Sanna [24] showed that the set of natural numbers n such that $\gcd (n, u_n) = 1$ has a natural density (see [15] for a generalization). Mastrostefano and Sanna [14, 23] studied the moments of $\log \!\big (\!\gcd (n, u_n)\big )$ and $\gcd (n, u_n)$ when $(u_n)$ is a Lucas sequence, and Jha and Nath [9] performed a similar study over shifted primes. Part of the interest in studying $\gcd (n, u_n)$ resides in the fact that this task can be considered a simpler, albeit nontrivial, case of the general problem of studying the greatest common divisor (GCD) of terms of two linear recurrences, a problem that led to the famous Bugeaud–Corvaja–Zannier bound [5] and the difficult Ailon–Rudnick conjecture [1]. (See the survey of Tron [30] for an extensive treatise on GCDs of terms of linear recurrences, especially Section 3 for these considerations on $\gcd (n, u_n)$ .) Furthermore, more abstractly, when $(u_n)$ is a Lucas sequence of discriminant $\Delta _u$ , we have that $(\gcd (n, u_n))$ is the GCD sequence naturally associated, by Silverman’s correspondence [27], with the algebraic group $\mathbb {G}_a \times \mathbb {G}_m(\sqrt {\Delta _u})$ , which is the product of the additive group with a twist of the multiplicative group.

Let $(F_n)$ be the linear recurrence of Fibonacci numbers, which is defined by $F_1 = F_2 = 1$ and $F_{n + 2} = F_{n + 1} + F_n$ for every positive integer n. Sanna and Tron [26] proved that, for each positive integer k, the set of positive integers n such that $\gcd (n, F_n) = k$ has a natural density, which is given by an infinite series. Kim [11] and Jha [8] obtained formally analogous results in cases of elliptic divisibility sequences and orbits of polynomial maps, respectively. Let $\mathcal {A}$ be the set of numbers of the form $\gcd (n, F_n)$ , for some positive integer n. Leonetti and Sanna [12] provided an effective method to enumerate the elements of $\mathcal {A}$ in increasing order. In particular, the first elements of $\mathcal {A}$ are

$$ \begin{align*} 1, \quad 2, \quad 5, \quad 7, \quad 10, \quad 12, \quad 13, \quad 17, \quad 24, \quad 25, \quad 26, \quad 29, \quad 34, \quad 35, \quad \dots \end{align*} $$

(see [17, A285058] for more terms). Then they proved that

(1.1) $$ \begin{align} \#\mathcal{A}(x) \gg \frac{x}{\log x} \end{align} $$

for all $x \geq 2$ . Their approach relied on a result of Cubre and Rouse [6], which in turn follows from Galois theory and the Chebotarev density theorem. Later, Jha and Sanna [10, Proposition 1.4] obtained an elementary proof as an application of related arithmetic problem over shifted primes. Leonetti and Sanna [12] also gave the upper bound $\#\mathcal {A}(x) = o(x)$ as $x \to +\infty $ , and asked for a more precise estimate. We prove the following upper bound on $\#\mathcal {A}(x)$ .

Theorem 1.1 We have

$$ \begin{align*} \#\mathcal{A}(x) \ll \frac{x \log \log \log x}{\log \log x} \end{align*} $$

for all sufficiently large x.

In fact, we prove that an upper bound similar to that of Theorem 1.1 also holds when the sequence of Fibonacci numbers is replaced by a general nondegenerate Lucas sequence (see Theorem 3.1).

In light of the gap between the upper bound of Theorem 1.1 and the lower bound (1.1), it is natural to wonder which is the true order of $\#\mathcal {A}(x)$ . By performing some numerical experiments (see Section 4), we found that $\#\mathcal {A}(x)$ appears to be asymptotic to $x / (\log x)^c$ , as $x \to +\infty $ , for some constant $c \approx 0.63$ (see Figure 1).

Figure 1 A plot of $\#\mathcal {A}(x) / (x / (\log x)^c)$ for x up to $10^6$ .

1.1 Notation

For every set of positive integers $\mathcal {S}$ and for every $x> 0$ , we define $\mathcal {S}(x) := \mathcal {S} \cap [1, x]$ . Throughout, we reserve the letter p for prime numbers. We employ the Landau–Bachmann “Big Oh” and “little oh” notation O and o, as well as the associated Vinogradov symbols $\ll $ and $\gg $ . Moreover, we write $f \asymp g$ to denote that $f \ll g$ and $f \gg g$ . Any dependence of the implied constants is explicitly stated or indicated with subscripts. In particular, $O_u$ is a shortcut for $O_{a_1, a_2}$ , and similarly for $\ll _u$ , $\gg _u$ , and $\asymp _u$ . We let $\operatorname {\mathrm {Li}}(x) := \int _2^x (\log t)^{-1}\,\mathrm {d}t$ denote the integral logarithm.

2 Preliminaries on Lucas sequences

In what follows, let $(u_n)$ be a Lucas sequences, that is, a sequence of integers satisfying $u_0 = 0$ , $u_1 = 1$ , and $u_n = a_1 u_{n - 1} + a_2 u_{n - 2}$ for every integer $n \geq 2$ , where $a_1, a_2$ are fixed nonzero relatively prime integers. Furthermore, assume that $(u_n)$ is nondegenerate, which means that the ratio of the roots of the characteristic polynomial $X^2 - a_1 X - a_2$ is not a root of unity. In particular, the discriminant $\Delta _u := a_1^2 + 4a_2$ is nonzero.

For each positive integer m that is relatively prime with $a_2$ , let $z_u(m)$ be the rank of appearance of m in the Lucas sequence, that is, the smallest positive integer n such that m divides $u_n$ . It is well known that $z_u(m)$ exists (see, e.g., [20]). Furthermore, put $\ell _u(m) := \operatorname {\mathrm {lcm}}\!\big (m, z_u(m)\big )$ and, for each positive integer n, let $g_u(n) := \gcd (n, u_n)$ and $\mathcal {A}_u := \big \{g_u(n) : n \in \mathbb {N}\big \}$ .

The next lemma collects some elementary properties of $z_u$ , $\ell _u$ , $g_u$ , and $\mathcal {A}_u$ .

Lemma 2.1 For all positive integers $m,n$ and all prime numbers p, with $p \nmid a_2$ , we have:

1. (i) $m \mid u_n$ if and only if $\gcd (m, a_2) = 1$ and $z_u(m) \mid n$ .

2. (ii) $z_u(m) \mid z_u(n)$ whenever $\gcd (mn, a_2) = 1$ and $m \mid n$ .

3. (iii) $z_u(p)\mid p - (-1)^{p-1}\eta _u(p)$ , where

   $$ \begin{align*} \eta_u(p) := \begin{cases} +1, & \text{ if } p \nmid \Delta_u \text{ and } \Delta_u \equiv x^2 \ \ \pmod p \text{ for some } x \in \mathbb{Z}, \\ -1, & \text{ if } p \nmid \Delta_u \text{ and } \Delta_u \not\equiv x^2 \ \ \pmod p \text{ for all } x \in \mathbb{Z},\\ 0, & \text{ if } p \mid \Delta_u. \end{cases} \end{align*} $$

4. (iv) $z_u(p^n)=p^{e_u(p, n)}\,z_u(p)$ , where $e_u(p, n)$ is some nonnegative integer less than n.

5. (v) $\ell _u(p^n) = p^n z_u(p)$ if $p \nmid \Delta _u$ , and $\ell _u(p^n) = p^n$ if $p \mid \Delta _u$ .

6. (vi) $g_u(m) \mid g_u(n)$ whenever $m \mid n$ .

7. (vii) $n \mid g_u(m)$ if and only if $\gcd (n, a_2) = 1$ and $\ell _u(n) \mid m$ .

8. (viii) $n \in \mathcal {A}_u$ if and only if $\gcd (n, a_2) = 1$ and $n = g_u\big (\ell _u(n)\big )$ .

9. (ix) $m \mid n$ whenever $\gcd (mn, a_2) = 1$ , $\ell _u(m) \mid \ell _u(n)$ , and $n \in \mathcal {A}_u$ .

Proof (i)–(iv) are well-known properties of the rank of appearance of a Lucas sequence (see, e.g., [20], [21, Chapter 1], or [22, Section 2]), whereas (v)–(vii) follow easily from (i)–(iv) and from the definitions of $\ell _u$ and $g_u$ . Let us prove (viii). If $n \in \mathcal {A}_u$ , then there exists a positive integer m such that $n = g_u(m)$ . In particular, $n \mid g_u(m)$ and, by (vii), we get that $\gcd (n, a_2) = 1$ and $\ell _u(n) \mid m$ . Therefore, by (vi), we have that $g_u\big (\ell _u(n)\big ) \mid n$ . Since n divides both $\ell _u(n)$ and $u_{\ell _u(n)}$ (by (i)), it follows that $n \mid g_u\big (\ell _u(n)\big )$ . Hence, $n = g_u\big (\ell _u(n)\big )$ . The other implication is straightforward. Finally, (ix) follows quickly from (vii) and (viii).

For each positive integer d, let $\mathcal {P}_{u,d}$ be the set of prime numbers p not dividing $a_2$ and such that d divides $z_u(p)$ . Cubre and Rouse [6] proved an asymptotic formula for $\#\mathcal {P}_d(x)$ in the special case in which $(u_n)$ is the sequence of Fibonacci numbers. Sanna [25] extended this result to Lucas sequences (under some mild restrictions) and also provided an error term. In particular, as a consequence of [25, Theorem 1.1], we have the following asymptotic formula.

Lemma 2.2 There exists a constant $B_u> 0$ such that for all $x \geq \exp (B_u d^{40})$ and for all odd positive integers d, with $3 \nmid d$ if the square-free part of $\Delta _u$ is equal to $-3$ , we have that

(2.1) $$ \begin{align} \#\mathcal{P}_d(x) = \delta_u(d) \operatorname{\mathrm{Li}}(x) + O_u\!\left(\frac{x}{(\log x)^{12/11}}\right) , \end{align} $$

where $\delta _u(d)$ is a quantity satisfying $\delta _u(d) \asymp _u 1 / d$ .

Proof If $\Delta _u$ is not a square, then, from [25, Theorem 1.1], we have that there exists a constant $B_u> 0$ such that

(2.2) $$ \begin{align} \#\mathcal{P}_d(x) = \delta_u(d) \operatorname{\mathrm{Li}}(x) + O_u\!\left(\frac{d}{\varphi(d)} \cdot \frac{x \, (\log \log x)^{\omega(d)}}{(\log x)^{9/8}}\right) , \end{align} $$

for all odd positive integers d, with $3 \nmid d$ if the square-free part of $\Delta _u$ is equal to $-3$ , and for all $x \geq \exp (B_u d^{40})$ , where $\delta _u(d) \asymp _u 1 / d$ , whereas $\varphi (d)$ and $\omega (d)$ are the Euler totient function and the number of prime factors of d, respectively.

If $\Delta _u$ is a square, then, by the Binet formula, we have that

$$ \begin{align*} u_n = \frac{\alpha^n - \beta^n}{\alpha - \beta} \end{align*} $$

for every positive integer n, where $\alpha := (a_1 - \sqrt {\Delta _u}) / 2$ and $\beta := (a_1 + \sqrt {\Delta _u}) / 2$ are integers. Consequently, for every prime number p not dividing $a_2 \Delta _u$ , we have that $z_u(p)$ is equal to the multiplicative order of $\alpha / \beta $ modulo p. Then (2.2) follows from a result of Moree [16, Lemma 1]. (Moree did not make explicit the factor $d/\varphi (d)$ , but this can be easily done [cf. [25, Lemma 6.3]].)

Note that we can assume that $B_u$ (and consequently x) is sufficiently large. In particular, we have that $d \leq (\log x)^{1/40}$ . Put $\varepsilon := 1/330$ . By the classic lower bound for $\varphi (d)$ (see, e.g., [29, Chapter I.5, Theorem 5.6]), we have that

$$ \begin{align*} \frac{d}{\varphi(d)} \ll \log \log d \ll \log \log \log x \leq (\log x)^{\varepsilon}. \end{align*} $$

Recall that $\omega (d) \leq \big (1 + o(1)\big ) \log d / \log \log d$ as $d \to +\infty $ (see, e.g., [29, Chapter I.5, Theorem 5.5]). Therefore, there exists an absolute constant $C> 0$ such that if $d> C$ , then

$$ \begin{align*} \omega(d) \leq (1 + \varepsilon) \frac{\log d}{\log \log d} \leq \left(\frac1{40} + 2\varepsilon\right) \frac{\log \log x}{\log \log \log x} , \end{align*} $$

and consequently $(\log \log x)^{\omega (d)} \leq (\log x)^{\frac 1{40} + 2\varepsilon }$ . Furthermore, if $d \leq C$ , then

$$ \begin{align*} (\log \log x)^{\omega(d)} \leq (\log x)^\varepsilon. \end{align*} $$

Thus, (2.1) follows.

Remark 2.3 In Lemma 2.2, the exponent $12/11$ can be replaced by $11/10 + \varepsilon $ , for every $\varepsilon> 0$ , assuming that x is sufficiently large depending on $\varepsilon $ .

We also need an upper bound for $\#\mathcal {P}_{u,d}(x)$ .

Lemma 2.4 We have

$$ \begin{align*} \#\mathcal{P}_{u,d}(x) \ll_u \frac{x}{\varphi(d) \log (x / d)} \end{align*} $$

for all positive integers d and for all $x> d$ .

Proof By Lemma 2.1 (iii), we have that

$$ \begin{align*} \#\mathcal{P}_{u,d}(x) \leq O_u(1) + \#\big\{p \leq x : p \equiv \pm 1 \ \ \pmod d \big\} \ll_u \frac{x}{\varphi(d) \log (x / d)} , \end{align*} $$

where we applied the Brun–Titchmarsh inequality [29, Chapter I.4, Theorem 4.16].

Now, we give an upper bound for the sum of reciprocals of primes in $\mathcal {P}_{u,d}$ .

Lemma 2.5 We have

$$ \begin{align*} \sum_{p \,\in\, \mathcal{P}_{u,d}(x)} \frac1{p} = \delta_u(d)\log \log x + O_u\!\left(\frac{\log(2d)}{\varphi(d)}\right) \end{align*} $$

for all odd positive integers d, with $3 \nmid d$ if the square-free part of $\Delta _u$ is equal to $-3$ , and for all $x \geq 3$ .

Proof First, suppose that $x < \exp (B_u d^{40})$ , where $B_u$ is the constant of Lemma 2.2. Hence, we have that

$$ \begin{align*} \delta_u(d)\log \log x \ll_u \frac{\log \log x}{d} \ll_u \frac{\log(2d)}{d}. \end{align*} $$

Moreover, by [19, Theorem 1 and Remark 1], we have that

$$ \begin{align*} \sum_{\substack{p \,\leq\, x \\[1pt] p \,\equiv\, \pm 1 \ \ \pmod d}} \frac1{p} = \frac{2\log \log x}{\varphi(d)} + O\!\left(\frac{\log(2d)}{\varphi(d)}\right). \end{align*} $$

This together with Lemma 2.1 (iii) yields that

$$ \begin{align*} \sum_{p \,\in\, \mathcal{P}_{u,d}(x)} \frac1{p} \leq \frac{O_u(1)}{d} + \sum_{\substack{p \,\leq\, x \\[1pt] p \,\equiv\, \pm 1 \ \ \pmod d}} \frac1{p} \ll_u \frac1{d} + \frac{\log (2d)}{\varphi(d)}. \end{align*} $$

Hence, the claim follows. Now, suppose that $x \geq \exp (B_u d^{40})$ . By partial summation, we have that

$$ \begin{align*} \sum_{p \,\in\, \mathcal{P}_{u,d}(x)} \frac1{p} = \frac{\#\mathcal{P}_{u,d}(x)}{x} + \int_1^x \frac{\#\mathcal{P}_{u,d}(t)}{t^2} \,\mathrm{d} t. \end{align*} $$

From Lemma 2.1 (iii), we get easily that $\#\mathcal {P}_{u,d}(x) / x \ll _u 1/d$ . Thus, it remains to bound the integral. By Lemma 2.1 (iii) again, we have that $\#\mathcal {P}_{u,d}(t)>0$ only if $t\ge d-1$ , and that $\#\mathcal {P}_{u,d}(t) \ll _u 1$ for $t \in [1, 2d]$ . Hence, we have

$$ \begin{align*} \int_1^{2d} \frac{\#\mathcal{P}_{u,d}(t)}{t^2} \,\mathrm{d} t \ll_u \int_{d-1}^{2d} \frac{\mathrm{d} t}{t^2} \ll \frac1{d}. \end{align*} $$

By Lemma 2.4, we have that

$$ \begin{align*} \int_{2d}^{\exp(Bd^{40})} \frac{\#\mathcal{P}_{u,d}(t)}{t^2} \,\mathrm{d} t &\ll_u \int_{2d}^{\exp(B_u d^{40})} \frac{\mathrm{d} t}{\varphi(d) \,t \log(t / d)} \\ &= \left[ \frac{\log\log(t/d)}{\varphi(d)}\right]_{t \,=\, 2d}^{\exp(B_u d^{40})} \ll \frac{\log (2d)}{\varphi(d)}. \end{align*} $$

By Lemma 2.2, we have that

$$ \begin{align*} \int_{\exp(B_u d^{40})}^x \frac{\#\mathcal{P}_{u,d}(t)}{t^2} \,\mathrm{d} t &= \int_{\exp(B_u d^{40})}^x \frac{\delta_u(d) \operatorname{\mathrm{Li}}(t)}{t^2} \,\mathrm{d} t + O_u\!\left(\int_{\exp(B_u d^{40})}^x \frac{\mathrm{d}t}{t (\log t)^{12/11}} \right) \\ &= \delta_u(d) \left[\log \log t - \frac{\operatorname{\mathrm{Li}}(t)}{t} \right]_{t \,=\, \exp(B_u d^{40})}^x + O_u\!\left( \frac1{d^{40/11}} \right) \\ &= \delta_u(d) \left(\log \log x + O\big(\log (2d)\big) \right) + O_u\!\left( \frac1{d^{40/11}} \right) \\ &= \delta_u(d) \log \log x + O_u\!\left(\frac{\log (2d)}{d} \right). \end{align*} $$

Putting these together, the claim follows.

The following sieve result is a special case of [7, Theorem 7.2] (cf. [18, Lemma 2.2]).

Lemma 2.6 We have

$$ \begin{align*} \#\{n\le x : p\mid n \Rightarrow p \notin \mathcal{P}\} \ll x \prod_{p \,\in\, \mathcal{P}(x)} \left(1 - \frac{1}{p}\right) , \end{align*} $$

for all $x \ge 2$ and for all sets of prime numbers $\mathcal {P}$ .

We also need the so-called Primitive Divisor Theorem for Lucas sequence [4].

Theorem 2.7 For every integer $r \geq 31$ , there exists a prime number p such that $z_u(p) = r$ .

3 Proof of Theorem 1.1

We shall prove the following more general result, of which Theorem 1.1 is a particular case.

Theorem 3.1 We have

$$ \begin{align*} \#\mathcal{A}_u(x) \ll_u \frac{x \log \log \log x}{\log \log x} \end{align*} $$

for all sufficiently large x.

Proof Let $r := 30\prod _{p \mid 6\Delta _u} z_u(p) + 1$ . Since $r \geq 31$ , it follows from Theorem 2.7 that there exists a prime number q such that $z_u(q) = r$ . Furthermore, by Lemma 2.1 (i), we have that $\gcd (6\Delta _u, u_r) = 1$ and so $q \nmid 6\Delta _u$ . Note also that $\gcd (6, r) = 1$ .

From Lemma 2.2, we know that $\delta _u(dr) \geq C_u / d$ for every positive integer d with $\gcd (6, d) = 1$ , where $C_u> 0$ is a constant depending only on $a_1, a_2$ . (Note that r is completely determined by $a_1, a_2$ .) Suppose that $x> 0$ is sufficiently large, and put

$$ \begin{align*} k := \left\lfloor \frac1{\log q} \log\!\left(\min(1, C_u) \, \frac{\log \log x}{\log \log \log x}\right)\right\rfloor \end{align*} $$

and $d := q^k$ . Hence, we get that

$$ \begin{align*} d \leq \min(1, C_u) \, \frac{\log \log x}{\log \log \log x} \end{align*} $$

and so

$$ \begin{align*} \frac{\log d}{\delta_u(dr)} \leq \frac{d}{C_u} \log d \leq \frac{\log \log x}{\log \log \log x} \log d \leq \log \log x. \end{align*} $$

Therefore, we have that

(3.1) $$ \begin{align} (\log x)^{\delta_u(dr)} \geq d \gg_u \frac{\log \log x}{\log \log \log x}. \end{align} $$

We split $\mathcal {A}_u$ into two subsets: $\mathcal {A}_u^\prime $ is the subset of $\mathcal {A}_u$ consisting of integers without prime factors in $\mathcal {P}_{u,dr}$ , and $\mathcal {A}_u^{\prime \prime } := \mathcal {A}_u \setminus \mathcal {A}_u^\prime $ .

First, we give an upper bound on $\#\mathcal {A}_u^\prime (x)$ . Note that $\gcd (6, dr) = 1$ . By Lemmas 2.5 and 2.6, we get that

(3.2) $$ \begin{align} \#\mathcal{A}_u^\prime(x) \ll x \prod_{p \,\in\, \mathcal{P}_{u,dr}(x)} \left(1-\frac{1}{p}\right) \leq x \exp\!\left(-\!\!\sum_{p \,\in\, \mathcal{P}_{u,dr}(x)} \frac1{p}\right) \ll_u \frac{x}{(\log x)^{\delta_u(dr)}} , \end{align} $$

where we also used the inequality $1 - x \leq \exp (-x)$ , which holds for $x \geq 0$ .

Now, we give an upper bound on $\#\mathcal {A}_u^{\prime \prime }(x)$ . If $n \in \mathcal {A}_u^{\prime \prime }$ , then n has a prime factor $p \in \mathcal {P}_{u,dr}$ . Hence, we have that $p \mid n$ and $dr \mid z_u(p)$ . Thus, by Lemma 2.1 (ii), we get that $z_u(p) \mid z_u(n)$ and so $dr \mid \ell _u(n)$ . Recalling that $d = q^k$ , $z_u(q) = r$ , and $q \nmid \Delta _u$ , by Lemma 2.1 (v), we have that $\ell _u(d) = dr$ . Hence, we get that $\ell _u(d) \mid \ell _u(n)$ and, by Lemma 2.1 (ix), it follows that $d \mid n$ . Thus, all the elements of $\mathcal {A}_u^{\prime \prime }$ are multiples of d. Consequently, we have that

(3.3) $$ \begin{align} \#\mathcal{A}_u^{\prime\prime}(x) \leq \frac{x}{d}. \end{align} $$

Therefore, putting together (3.2) and (3.3), and using (3.1), we obtain that

$$ \begin{align*} \#\mathcal{A}_u(x) = \#\mathcal{A}_u^\prime(x) + \#\mathcal{A}_u^{\prime\prime}(x) \ll_u \frac{x}{(\log x)^{\delta_u(dr)}} + \frac{x}{d} \ll_u \frac{x \log \log \log x}{\log \log x} , \end{align*} $$

as desired. The proof is complete.

4 Numerical computations

We computed the elements of $\mathcal {A} \cap [1, 10^6]$ by using a program written in C that employs Lemma 2.1 (viii). Note that computing $g\big (\ell (n)\big )$ directly as $\gcd \!\big (\ell (n), F_{\ell (n)}\big )$ would be prohibitive, in light of the exponential grown of Fibonacci numbers. Instead, we used the fact that

$$ \begin{align*} g\big(\ell(n)\big) = \gcd\!\big(\ell(n), F_{\ell(n)} \bmod \ell(n)\big), \end{align*} $$

and we computed Fibonacci numbers modulo an integer by efficient matrix exponentiation.