Polynomials Generated by the Fibonacci Sequence

David Garth^{1}

Division of Mathematics & Computer Science Truman State University Kirksville, MO 63501

USA

dgarth@truman.edu

Donald Mills Department of Mathematics

Rose-Hulman Institute of Technology

Terre Haute, IN 47803-3999

USA

mills@rose-hulman.edu

Patrick Mitchell Department of Mathematics Midwestern State University

Wichita Falls, TX 76308

USA

patrick.mitchell@mwsu.edu

Abstract

The Fibonacci sequence’s initial terms are F_{0} = 0 and F_{1} = 1, with F_{n} = F_{n−1} + F_{n−2} for n ≥ 2. We define the polynomial sequence p by setting p_{0}(x) = 1 and p_{n}(x) = xp_{n−1}(x) + F_{n+1} for n ≥ 1, with p_{n}(x) = ∑_{k=0}^{n} F_{k+1}x^{n−k}. We call p_{n}(x) the Fibonacci-coefficient polynomial (FCP) of order n. The FCP sequence is distinct from the well-known Fibonacci polynomial sequence.

1Corresponding author.

We answer several questions regarding these polynomials. Specifically, we show that each even-degree FCP has no real zeros, while each odd-degree FCP has a unique, and (for degree at least 3) irrational, real zero. Further, we show that this sequence of unique real zeros converges monotonically to the negative of the golden ratio. Using Rouché’s theorem, we prove that the zeros of the FCP’s approach the golden ratio in modulus. We also prove a general result that gives the Mahler measures of an infinite subsequence of the FCP sequence whose coefficients are reduced modulo an integer m ≥ 2. We then apply this to the case that m = L_{n}, the n^{th} Lucas number, showing that the Mahler measure of the subsequence is ϕ^{n−1}, where ϕ = \frac{1+\sqrt{5}}{2}.

1 Introduction

The Fibonacci sequence’s initial terms are F_{0} = 0 and F_{1} = 1, with F_{k} = F_{k−1} + F_{k−2} for k ≥ 2. We define the polynomial sequence {p_{n}(x)} by setting p_{0}(x) = 1 and p_{n}(x) = xp_{n−1}(x) + F_{n+1} for n ≥ 1. Equivalently, we have that

p_{n}(x) = ∑_{k=0}^{n} F_{k+1}x^{n−k}.

Thus, p_{1}(x) = x + 1, p_{2}(x) = x^{2} + x + 2, p_{3}(x) = x^{3} + x^{2} + 2x + 3, and so forth. We call p_{n}(x) the Fibonacci-coefficient polynomial (FCP) of order n.

The FCP sequence is distinct from the well-known Fibonacci polynomial sequence, given by P_{1}(x) = 1 and P_{2}(x) = x with P_{n+1}(x) = xP_{n}(x) + P_{n−1}(x) for n ≥ 2. For, while P_{1}(x) = p_{0}(x) and P_{2}(x) = p_{1}(x−1), there does not exist a ∈ ℂ such that x^{2} +1 = P_{3}(x) = p_{2}(x−a).

Next, for m ≥ 2, let {p_{n}^{(m)}(x)} be the sequence obtained from the Fibonacci coefficient polynomials by reducing the coefficients modulo m and adjusting the residue class, using (F_{k} mod m) − m if F_{k} mod m > \frac{m}{2}. Thus, for odd m the residue class is centered about 0, and for even m the residue class ranges from −m/2 + 1 to m/2. As an example, p_{5}^{(2)}(x) = x^{5} + x^{4} + x^{2} + x, p_{5}^{(3)}(x) = x^{5} + x^{4} − x^{2} − x − 1, and p_{5}^{(4)}(x) = x^{5} + x^{4} + 2x^{3} − x^{2} + x.

In this paper, we answer several questions regarding these sequences of polynomials. In Section 3 we show that the number of real zeros of p_{n}(x) is either 0 or 1, according to whether n is even or odd, respectively. Further, we prove that the sequence of unique real zeros for the odd-degree FCP’s converges monotonically to the negative of the golden ratio, and that the zeros of the FCP’s approach the golden ratio in modulus. In Section 4 we consider the polynomial sequence {p_{n}^{(m)}(x)}, showing that there is an infinite subsequence of these polynomials whose Mahler measure equals the Mahler measure of p_{t−2}^{(m)}(x), where t is the period of the Fibonacci sequence reduced modulo m. We then consider the specific case when m = L_{n}, the n^{th} Lucas number. In this case the coefficients of the polynomials in the aforementioned infinite subsequence are Fibonacci numbers, and we show that the Mahler measure of the polynomials is φ^{n−1}, where φ = \frac{1+\sqrt{5}}{2}. We start, however, by proving some necessary Lemmas.

2 Preliminary Lemmas

In this section we prove some technical lemmas that will be needed later. We will make use of the close relationship between the Fibonacci numbers, the Lucas numbers, and the golden ratio φ = \frac{1+\sqrt{5}}{2}. Recall that the Lucas numbers are defined by L_{0} = 2, L_{1} = 1, and L_{n} = L_{n−1} + L_{n−2} for n ≥ 2. We first remind the reader of Binet’s formula ([1, Chapter 5]).

Lemma 2.1 (Binet’s Formula). Let φ = (1 + \sqrt{5})/2 and τ = (1 − \sqrt{5})/2. For all natural numbers n, we have

F_{n} = \frac{φ^{n} − τ^{n}}{φ − τ} = \frac{1}{\sqrt{5}}φ^{n} + \frac{1}{\sqrt{5}} · (−\frac{1}{φ})^{n} = ⌊\frac{φ^{n}}{\sqrt{5}}⌋

and

L_{n} = φ^{n} + τ^{n} = φ^{n} + (−\frac{1}{φ})^{n} .

Recall that the sequence of Fibonacci numbers, reduced modulo an integer m ≥ 2, is purely periodic. Renault [4] gave a nice discussion of the properties of the reduced Fibonacci sequence, including a list of the periods for each m. We will be interested in the period when reducing modulo a Lucas number. The following lemmas show that in this case the period has a particularly nice form.

Lemma 2.2. If n is odd, then the period of the Fibonacci sequence modulo L_{n}, with residue classes adjusted to range between −\frac{L_{n}−1}{2} to \frac{L_{n}−1}{2} for odd L_{n} and −\frac{L_{n}}{2} + 1 to \frac{L_{n}}{2} for even L_{n}, is

F_{0}, F_{1}, F_{2}, F_{3}, …, F_{n}, (−F_{n−1}), F_{n−2}, (−F_{n−3}), F_{n−4}, …, (−F_{2}), F_{1}.  (1)

Proof. Recall the well-known identity L_{n} = F_{n−1} + F_{n+1} ([1], page 80). Notice that for 0 ≤ i ≤ n

L_{n} = F_{n+1} + F_{n−1}
= F_{n} + F_{n−1} + F_{n−1}
≥ F_{n} + F_{n−1} + F_{n−2}
= 2F_{n}
≥ 2F_{i} for 1 ≤ i ≤ n

Thus, the first n + 1 elements of the adjusted period mod L_{n} are as in the list (1), since F_{i} = 0 · L_{n} + F_{i} and 0 ≤ F_{i} ≤ \frac{L_{n}}{2} for 0 ≤ i ≤ n. The remaining entries in the list follow from the identity

F_{n+k} = F_{k}L_{n} + (−1)^{k}F_{n−k} for 1 ≤ k ≤ n.  (2)

To verify this identity, notice that by Binet’s Theorem we have

F_{n+k} − F_{k}l_{n} = \frac{φ^{n+k} − τ^{n+k}}{φ − τ} − \frac{(φ^{k} − τ^{k})(φ^{n} + τ^{n})}{φ − τ}
= \frac{φ^{n+k} − τ^{n+k} − φ^{n+k} − φ^{k}τ^{n} + φ^{n}τ^{k} + τ^{n+k}}{φ − τ}
= \frac{φ^{k}τ^{k}(φ^{n−k} − τ^{n−k})}{φ − τ}
= (−1)^{k}F_{n−k}

The last equality follows from Binet’s formula and the fact that φτ = −1. This establishes the remaining entries in the list (1). To see that there are no further entries in the period, it is easy to see from Binet’s formulas that F_{2n} = F_{n}L_{n}, and so the next entry in the Fibonacci sequence reduced modulo m is 0. Also, since n is odd, it can be shown in using Binet’s formulas that F_{2n+1} = F_{n+1}L_{n} + 1. Thus, the next entry in the list is 1. Since the next pair of entries in the sequence after those in the list (1) are 0 and 1, and the list (1) repeats.

Lemma 2.3. If n is even, then the period of the Fibonacci sequence modulo L_{n}, with residue classes adjusted as in Lemma 2.2 is

F_{0} F_{1} F_{2} F_{3} ⋯ F_{n} (−F_{n−1}) F_{n−2} (−F_{n−3}) ⋯ F_{2} (−F_{1}) F_{0}

(−F_{1}) (−F_{2}) (−F_{3}) ⋯ (−F_{n}) F_{n−1} (−F_{n−2})F_{n−3} ⋯ (−F_{2}) (F_{1}).

Proof. The first 2n terms of the period are established in the same manner as in the proof of Lemma 2.3. The rest of the period follows from the formulas

F_{2n+k} = F_{n+k}L_{n} − F_{k} for 1 ≤ k ≤ n

and

F_{3n+k} = (F_{2n+k} − F_{k})L_{n} + (−1)^{k+1}F_{n−k},

both of which can be proved using Binet’s formulas in a manner similar to equation (2). After this it is easy to show that the pair 0, 1 are the next terms of the sequence, and this proves the Lemma.

We close this section with the following technical lemma which will prove crucial in the proof of Theorem 4.2.

Lemma 2.4. For n ≥ 13, the sequence φ − \frac{1}{n^{2}} − (\frac{2}{\sqrt{5}})^{1/n} φ is monotone decreasing and approaches 0 as n → ∞.

Proof. It is clear that the sequence approaches 0 as n → ∞. We now show it is monotone decreasing. Define d : (0, ∞) → ℝ by

d(x) = φ − \frac{1}{x^{2}} − (\frac{2}{\sqrt{5}})^{1/x} φ.

Notice that

d′(x) = \frac{2}{x^{3}} − (\frac{2}{\sqrt{5}})^{1/x} log (\frac{2}{\sqrt{5}}) (−\frac{1}{x^{2}} ) φ

= \frac{1}{x^{2}} (\frac{2}{x} + (\frac{2}{\sqrt{5}})^{1/x} log (\frac{2}{\sqrt{5}}) φ)

Now, when x = 13, d′(x) < 0. Also, notice that

\frac{2}{x} + (\frac{2}{\sqrt{5}})^{1/x} log (\frac{2}{\sqrt{5}}) φ → log (\frac{2}{\sqrt{5}}) φ < 0

as x → ∞. Now, since \frac{2}{x} decreases to 0, and (\frac{2}{\sqrt{5}})^{1/x} increases to 1, it follows that d′(x) < d′(13) < 0 for all x ≥ 13. Thus, d(x) is decreasing for x ≥ 13.

3 Zeros of FCPs

In this section we investigate the zeros of the FCP sequence. For the remainder of this paper let g_{n}(x) = p_{n}(x)(x^{2} − x − 1) for n ∈ ℤ^{+}. We will make use of the following identities, which are straightforward to verify. For each identity, n ∈ ℤ^{+}.

g_{n}(x) := p_{n}(x)(x^{2} − x − 1) = x^{n+2} − F_{n+2}x − F_{n+1}  (3)
x^{n+1} − F_{n+1} = (x − 1)p_{n}(x) − xp_{n−2}(x)  (4)
x^{n} + F_{n} = p_{n}(x) − (x + 1)p_{n−2}(x)  (5)

We start by characterizing the real zeros of the polynomials in the FCP sequence.

Theorem 3.1. For even n, p_{n}(x) has no real zeros. For odd n, p_{n}(x) has exactly one real zero, which lies in the interval (−φ, −1]. Moreover, the sequence of real zeros of the p_{n}(x), with n odd, decreases monotonically to zero.

Proof. First assume n is even. By Descartes’ rule of signs g_{n}(x) has exactly one positive real zero and one negative real zero. Since φ > 0 and τ < 0 are zeros of x^{2} − x − 1, a factor of g_{n}(x), it follows that p_{n}(x) has no real zeros.

Now assume n is odd. By similar reasoning it follows from Descartes’ rule of signs applied to g_{n}(x) that p_{n}(x) has exactly one real zero, which is negative. Notice that g_{1}(−1) = 0, and g_{n}(−1) = (−1)^{n+2} + F_{n+2} − F_{n+1} = F_{n} − 1 > 0 for n > 1. Also, since n is odd, for n > 1 we have

g_{n}(−φ) = −φ(φ^{n+1} − F_{n+2}) − F_{n+1}.  (6)

By Binet’s formulas, for odd n > 1 we have that

φ^{n+1} − F_{n+2} = φ^{n+1} − \frac{φ^{n+2}}{\sqrt{5}} − \frac{1}{\sqrt{5}φ^{n+2}}
= φ^{n+2} (\frac{1}{φ} − \frac{1}{\sqrt{5}} − \frac{1}{\sqrt{5}φ^{2n+4}} ) > φ^{n+2} (\frac{1}{φ} − \frac{1}{\sqrt{5}} − \frac{1}{\sqrt{5}φ^{10}} ) > 0.

It follows that the right hand side of (6) is negative, and thus g_{n}(x) has a zero in the interval (−φ, −1] for odd n ≥ 1. Since x^{2} − x − 1 has no zero in this interval, it follows that this zero must be the negative zero of p_{n}(x).

Now let α_{n} be the real zero of p_{n}(x), for odd n. Taking the derivative of g_{n}(x) gives g′_{n}(x) = (n + 2)x^{n+1} − F_{n+2}. As n is odd, the critical points of g_{n}(x) are

x_{n,±} = ± (\frac{F_{n+2}}{n + 2})^{1/(n+1)} .

Since α_{n}, φ and τ are the zeros of g_{n}, with α_{n} ∈ (−φ, −1) for n > 1, it follows that g_{n}(x) achieves a local maximum at x_{n,−}, with −φ < α_{n} ≤ x_{n,−}. As F_{n+2}^{1/(n+1)} → φ as n → ∞, it follows that α_{n} → −φ as n → ∞ through odd values of n.

All that remains is to show that the convergence is monotonic. That is, we need to show that α_{n} ≤ α_{n−2} for all odd n. We assume n ≥ 3, as an easy check shows that α_{3} < α_{1} = −1. We saw earlier that p_{n}(−1) > 0 and p_{n}(−φ) < 0 for all odd n > 1. The result will follow if we can show that p_{n}(α_{n−2}) > 0 for all odd n, since each p_{n}(x) has a unique real zero.

For convenience of notation, let a = α_{n−2} in the remainder of the proof. By identity (5), we have that p_{n}(a) = a^{n} + F_{n}. So we need to show that a^{n} > −F_{n}, i.e. that a > −\sqrt[n]{F_{n}}. This will follow if we can show that p_{n−2}(−\sqrt[n]{F_{n}}) < 0, since p_{n−2}(−1) > 0. It will follow then that a ∈ (−\sqrt[n]{F_{n}}, −1).

Using identity (3), we have that

p_{n−2}(−\sqrt[n]{F_{n}}) = \frac{(−\sqrt[n]{F_{n}})^{n} + F_{n}(\sqrt[n]{F_{n}}) − F_{n−1}}{(−\sqrt[n]{F_{n}})^{2} + \sqrt[n]{F_{n}} − 1}

Notice that the denominator of this last expression is positive, since −\sqrt[n]{F_{n}} < −1, and x^{2} − x − 1 > 0 whenever x ≤ −1. So p_{n−2}(−\sqrt[n]{F_{n}}) has the same sign as the numerator of the previous expression.

Now notice that

(−\sqrt[n]{F_{n}})^{n} + F_{n}(\sqrt[n]{F_{n}}) − F_{n−1} = −F_{n} + F_{n}(\sqrt[n]{F_{n}}) − F_{n−1}
= −F_{n+1} + F_{n}(\sqrt[n]{F_{n}})

We want to show that −F_{n+1} + F_{n}(\sqrt[n]{F_{n}}) < 0, or equivalently, that

\sqrt[n]{F}_{n} < \frac{F_{n+1}}{F_{n}} .

One verifies this numerically for 3 ≤ n ≤ 11. For n ≥ 13, we approach the matter as follows.

It is well-known from the theory of continued fractions that for odd n, \frac{F_{n+1}}{F_{n}} < φ (see Theorem 7.10 of [3]). We also have that

|φ − \frac{F_{n+1}}{F_{n}} | < \frac{1}{F_{n}F_{n+1}} ,  (7)

(see Theorem 7.11 of [3]). It follows from (7) that

\frac{F_{n+1}}{F_{n}} > φ − \frac{1}{F_{n}F_{n+1}} > φ − \frac{1}{n^{2}} .

On the other hand, by Binet’s formula, since n is odd, we have that

F_{n} = \frac{1}{\sqrt{5}}φ^{n} + \frac{1}{\sqrt{5}} · \frac{1}{φ^{n}} .

Thus,

\sqrt[n]{F}_{n} = \sqrt[n]{\frac{1}{\sqrt{5}}φ^{n} + \frac{1}{\sqrt{5}} · \frac{1}{φ^{n}}}

≤ \sqrt[n]{\frac{1}{\sqrt{5}}φ^{n} + \frac{1}{\sqrt{5}}φ^{n}}

= \sqrt[n]{\frac{2}{\sqrt{5}}} · φ

By Lemma 2.4, we then have that for n ≥ 13,

\sqrt[n]{F}_{n} ≤ \sqrt[n]{\frac{2}{\sqrt{5}}} · φ < φ − \frac{1}{n^{2}} < \frac{F_{n+1}}{F_{n}} .

This completes the proof of the theorem.

We remark that it follows immediately from Theorem 3.1 and Gauss’ Polynomial Theorem that p_{n}(x) has no rational zeros for n ≥ 2. Based on numerical tests we are led to conjecture that p_{n}(x) is irreducible for all n.

Theorem 3.2. As n increases without bound, the roots of p_{n}(x) approach φ in modulus.

Before proving this, we recall some facts. First, we remind the reader of Rouché’s Theorem. This version of the theorem is found page 2 of [2].

Theorem 3.3 (Rouché’s Theorem). If P (z) and Q(z) are analytic interior to a simple closed Jordan curve C, and if they are continuous on C and

|P (z)| < |Q(z)|, z ∈ C,  (8)

then the function H(z) = P (z) + Q(z) has the same number of zeros interior to C as does Q(z).

We also need the following Lemma, which follows easily from what we have shown of the function x^{n+2} − F_{n+2}x − F_{n+1}.

Lemma 3.4. For all n ∈ ℤ^{+}, we have φ^{n} = φF_{n} + F_{n−1}. Moreover, if c > φ then c^{n+2} > F_{n+2}c + F_{n+1}, and if 0 < c < φ then c^{n+2} < F_{n+2}c + F_{n+1}.

Proof of Theorem 3.2. Consider the circle |z| = c in the complex plane, where c > φ. Then for |z| = c we have that

|F_{n+2}z + F_{n+1}| ≤ F_{n+2}|z| + F_{n+1}
= F_{n+2}c + F_{n+1}
< c^{n+2}
= |z^{n+2}|.

Notice that the inequality in the third line above follows from Lemma 3.4. It then follows from Rouche’s theorem, with P_{n}(z) = −F_{n+2}z − F_{n+1} and Q_{n}(z) = z^{n+2}, that p_{n}(z)(z^{2} − z − 1) = z^{n+2} − F_{n+2}z − F_{n+1} has n + 2 zeros inside |z| = c. Since c > φ was artibrary, it follows that p_{n}(z)(z^{2} − z − 1) has n + 2 zeros inside or on |z| = φ.

Now consider the circle |z| = c, where \frac{1}{φ} < c < φ. Then for |z| = c we have that

|F_{n+2}z + F_{n+1}|^{2} = (F_{n+2}z + F_{n+1})(F_{n+2}\overline{z} + F_{n+1})
= F_{n+2}^{2}|z|^{2} + F_{n+2}F_{n+1}(z + \overline{z}) + F_{n+1}^{2}
= F_{n+2}^{2}c^{2} + 2F_{n+2}F_{n+1}Re(z) + F_{n+1}^{2}
≥ F_{n+2}^{2}c^{2} − 2F_{n+2}F_{n+1}c + F_{n+1}^{2}
= (F_{n+2}c − F_{n+1})^{2}.

Therefore

|F_{n+2}z + F_{n+1}| ≥ |F_{n+2}c − F_{n+1}| for |z| = c, \frac{1}{φ} < c < φ.  (9)

We now apply Binet’s formula to get a lower bound for |F_{n+2}c−F_{n+1}|. Recall that Binet’s formula is

F_{n} = \frac{φ^{n}}{\sqrt{5}} − \frac{1}{\sqrt{5}(−φ)^{n}} .

Thus, we have that

F_{n+2}c − F_{n+1} = (\frac{φ^{n+2}}{\sqrt{5}} − \frac{(−1)^{n+2}}{\sqrt{5}φ^{n+2}} ) c − (\frac{φ^{n+1}}{\sqrt{5}} − \frac{(−1)^{n+1}}{\sqrt{5}φ^{n+1}} )
= φ^{n+2} (\frac{c}{\sqrt{5}} − \frac{1}{φ\sqrt{5}}) + \frac{(−1)^{n+1}}{\sqrt{5}φ^{n+1}} (\frac{c}{φ} + 1) .

Since \frac{1}{φ} < c < φ, it then follows that

F_{n+2}c − F_{n+1} > c^{n+2} for n large enough.  (10)

Therefore, by (9) and (10), for |z| = c, with \frac{1}{φ} < c < φ,

|F_{n+2}z + F_{n+1}| ≥ |F_{n+2}c − F_{n+1}| > c^{n+2} for large n.

By Rouche’s theorem then applied to P_{n}(z) = z^{n+2} and Q_{n}(z) = −F_{n+2}z − F_{n+1}, it follows that if \frac{1}{φ} < c < φ then for n large enough p_{n}(z)(z^{2} − z − 1) has one zero in |z| = c. Since −\frac{1}{φ} is one of these zeros, it follows that p_{n}(z) has no zeros inside |z| = c. Since c was arbitrary, it follows that as n → ∞, the zeros of p_{n}(z) approach the circle |z| = φ.

4 Mahler Measures of Reduced-Coefficient FCPs

Theorem 3.2 shows that the Mahler measure of the FCP sequence is asymptotic to φ^{n}. In this section we consider is the Mahler measure of the reduced coefficient FCP’s p_{k}^{(m)}(x) mentioned in the introduction. Recall that the Mahler measure of an nth-degree polynomial g(x) ∈ ℝ[x] with leading coefficient a_{n} and complex roots α_{1}, ..., α_{n}, is given by

M(g(x)) = |a_{n}| ∏_{j=1}^{n} max{1, |α_{j}|}.

Theorem 4.1. Let m ≥ 2, and let t be the number of terms in one period of the Fibonacci sequence reduced modulo m. If k ≡ −2 or −1 mod t, then

M (p_{k}^{(m)}(x)) = M (p_{t−2}^{(m)}(x)) .

Proof. Let a_{1}, …, a_{t} be the period of the Fibonacci sequence modulo m, adjusted as described in the introduction. Since a_{1} = 0 and a_{2} = 1, notice that

p_{t−2}^{(m)}(x) = x^{t−2} + a_{3}x^{t−3} + ⋯ + a_{t−1}x + a_{t}.

Now let C_{j}(x) = x^{j} + x^{j−1} + ⋯ + x + 1, where j ≥ 1. If k ≡ −2 mod t, then

p_{k}^{(m)}(x) = p_{t−2}^{(m)}(x)C_{l}(x^{t}),

where l = (k − t + 2) mod t. Now suppose k ≡ −1 mod t. Then a_{k+1} = a_{1} = 0, and so

p_{k}^{(m)}(x) = xp_{t−2}^{(m)}(x)C_{l}(x^{t}).

Notice that for every j, all the the roots of C_{j}(x^{t}) have modulus 1. Since the Mahler measure is multiplicative the Theorem follows.

We now turn our attention to the particular case in which m is a Lucas number. As Lemmas 2.2 and 2.3 show, the period of the Fibonacci sequence reduced modulo L_{n} is 2n if n is odd and 4n if n is even. Moreover, for odd n,

p_{2n−2}^{(L_{n})}(x) = x^{2n−2} + F_{2}x^{2n−3} + ⋯ + F_{n}x^{n−1}
−F_{n−1}x^{n−2} + F_{n−2}x^{n} − 3 − ⋯ + F_{3}x^{2} − F_{2}x + F_{1}
= x^{n−1}p_{n−1}(x) + x^{n−2}p_{n−2} (−\frac{1}{x}) ,

and by Theorem 4.1, if k ≡ −2 or −1 mod 2n, then M (p_{k}^{(L_{n})}(x)) = M (p_{2n−2}^{(L_{n})}(x)). If n is even, then notice that Lemma 2.3 gives that the period mod L_{n} is divided into two halves, the second of which is the negation of the first. By an argument similar to the proof of Theorem 4.1, we can show that if n is even and k ≡ −2 or −1 mod 2n, then M (p_{k}^{(L_{n})}(x)) = M (p_{2n−2}^{(L_{n})}(x)).

For the next Theorem, let Φ_{k}(x) be the k^{th} cyclotomic polynomial. Recall the following well-known fact about cyclotomic polynomials:

∏_{d|n} Φ_{d}(x) = x^{n} − 1.  (11)

Theorem 4.2. If k ≡ −2 or −1 mod 2n then M (p_{k}^{(L_{n})}(x)) = φ^{n−1}.

Proof. By the preceding discussion, it follows that all we need to show is that

M (p_{2n−2}^{(L_{n})}(x)) = φ^{n−1}.  (12)

We first consider the case when n is odd. It is straightforward to show that

p_{2n−2}^{(L_{n})}(x)(x^{2} − x − 1) = x^{2n} − L_{n}x^{n} − 1, for n odd.  (13)

Now, using identity (11), notice that

∏_{\begin{matrix}d|n \\ d≠1 \end{matrix}} Φ_{d}(−φx)Φ_{d}(x/φ) = \frac{−φ^{n}x^{n} − 1}{−φx − 1} · \frac{\frac{x^{n}}{φ^{n}} − 1}{\frac{x}{φ} − 1}

= \frac{x^{2n} − (φ^{n} − \frac{1}{φ^{n}} ) x − 1}{x^{2} − (φ − \frac{1}{φ}) x − 1}

= \frac{x^{2n} − L_{n}x − 1}{x^{2} − x − 1}.  (14)

Notice that the last equality above follows from Binet’s formula, since n is odd. It now follows from (13) and (14)

p_{2n−2}^{(L_{n})}(x)(x^{2} − x − 1) = (x^{2} − x − 1) ∏_{\begin{matrix}d|n \\ d≠1 \end{matrix}} Φ_{d}(−φx)Φ_{d}(x/φ),

and therefore

p_{2n−2}^{(L_{n})}(x) = ∏_{\begin{matrix}d|n \\ d≠1 \end{matrix}} Φ_{d}(−φx)Φ_{d}(x/φ).

Now, it is clear that the roots of Φ_{d}(φx) all have modulus φ, and those of Φ_{d}(xφ) all have modulus \frac{1}{φ}. Thus, it follows that M (p_{2n−2}(x)) = φ^{n−1}.

Now suppose n ≥ 2 is even. Let

r_{n}(x) = x^{2n−4} + F_{4}x^{2n−6} + F_{6}x^{2n−8} + ⋯ + F_{n−2}x^{n} + F_{n}x^{n−2} + F_{n−2}x^{n−4} + ⋯ + F_{4}x^{2} + 1.

A straightforward calculation shows that

p_{2n−2}^{(L_{n})}(x) = (x^{2} + x − 1)r_{n}(x).  (15)

It is also straightforward to show that

r_{n}(x)(x^{4} − 3x^{2} + 1) = x^{2n} − L_{n}x^{n} + 1.  (16)

In a manner similar to that which was performed in the case when n is odd, we can show that

∏_{\begin{matrix}d|n \\ d≠1,2 \end{matrix}} Φ_{d}(−φx)Φ_{d}(x/φ) = \frac{x^{2n} − L_{n}x^{n} + 1}{x^{4} − 3x^{2} + 1}.

(This identity made use of the fact that n is even). Combining this last result with (15) and (16) we see that

p_{2n−2}^{(L_{n})}(x) = (x^{2} + x − 1)r_{n}(x)
= (x^{2} + x − 1) (\frac{x^{2n} − L_{n}x^{n} + 1}{x^{4} − 3x^{2} + 1})
= (x^{2} + x − 1) ∏_{\begin{matrix}d|n \\ d≠1,2 \end{matrix}} Φ_{d}(−φx)Φ_{d}(x/φ).

From this it is clear that M (p_{2n−2}(x)) = φ^{n−1}.

It is worth mentioning that the proof of Theorem 4.2 shows that for k ≡ −1 or −2 mod L_{n} the zeros of p_{k}^{(L_{n})}(x) which lie outside the unit circle are φe^{2jπi/n} for 1 ≤ j ≤ n if n is odd, and for 1 ≤ j ≤ n and j ≠ n/2 if n is even.

References

[1] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley-Interscience, 2001.

[2] M. Marden, Geometry of Polynomials, Second Ed., Math. Surveys 3, Amer. Math. Soc., Providence, Rhode Island, 1966.

[3] I. Niven, H. Zuckerman, H. Montgomery, An Introduction to the Theory of Numbers, Fifth Ed., John Wiley and Sons, 1991.

[4] M. Renault, Properties of the Fibonacci Sequence Under Various Moduli, Master’s Thesis, Wake Forest University, 1996.

2000 Mathematics Subject Classification: Primary 11C08; Secondary 13B25. Keywords: Fibonacci, sequence, polynomial, zero, root, Rouchés theorem, Mahler measure. (Concerned with sequences \underline{A000045} and \underline{A019523}.)

Received March 8 2007; revised version received June 11 2007. Published in Journal of Integer Sequences, June 19 2007. Return to Journal of Integer Sequences home page.