On Inequalities Related to a Generalized Euler Totient Function and Lucas Sequences

Ashish Kumar Pandey and B. K. Sharma Department of Mathematics University of Allahabad Prayagraj 211002 India

                   apandey.aumaths@gmail.com
                    bksharma@allduniv.ac.in

Abstract

Let φ(n) be the Euler totient function of n, defined as the number of positive integers less than or equal to n that are co-prime with n. In this paper, we consider the function φ_{k}, a generalization of φ, and establish some inequalities related to Lucas sequences of the first kind (U_{n})_{n≥1} with characteristic equation having real roots. As an application to these inequalities, we further establish inequalities related to Fibonacci, Pell, and balancing sequences.

1 Introduction

Mathematicians have long been interested in studying arithmetic functions and the associated problems in number theory. These problems include solving Diophantine equations, exploring generalizations, and investigating divisibility concerned with arithmetic functions (see [3, 4, 5, 9, 10, 15, 16] and references therein). The Euler totient function φ(n) counts the number of positive integers less than or equal to n that are co-prime with n. Additionally, σ_{k}(n) represents the sum of the k^{th} power of the positive divisors of n, where k is a non-negative integer. Also σ_{k}(n) reduces to τ(n) for k = 0, where τ(n) denotes the number of positive divisors of n. These arithmetic functions are primarily utilized in the field of number theory. In recent times, numerous researchers have worked towards examining arithmetic functions connected to binary recurrence sequences like the Fibonacci sequence, Lucas sequence, Pell sequence, and balancing sequence (see [1, 5, 9, 10, 11, 12, 13, 14]).

The study of inequalities related to arithmetic functions and binary recurrence sequences is one of the interesting problems, and many researchers have contributed in this direction. Before mentioning some recent results, we first define Lucas sequence of the first kind.

Let r = α + β and s = −αβ be two non-zero co-prime integers with Δ = r^{2} + 4s > 0. Then α and β are roots of the quadratic equation x^{2} − rx − s = 0. For any non-negative integer n, define

U_{n} := U_{n}(r, s) = \frac{α^{n} − β^{n}}{α − β} .

The sequence (U_{n})_{n≥0} is called a Lucas sequence of the first kind and quadratic equation x^{2} − rx − s = 0 is the corresponding characteristic equation of (U_{n})_{n≥0}. From the definition, it is clear that U_{0} = 0, U_{1} = 1 and that

U_{n+2} = rU_{n+1} + sU_{n}, n = 0, 1, 2, ….

In 1997, Luca [7] showed that the Euler totient function for Lucas sequence of the first kind (U_{n})_{n≥1} satisfies the inequality φ(|U_{n}|) ≥ |U_{φ(n)}| for those sequences whose characteristic equation has real roots, and the inequality is not valid for the sequences with characteristic equation having complex roots. In [8], Luca proved that the n^{th} Fibonacci number satisfies σ_{k}(F_{n}) ≤ F_{σ_{k}(n)} and τ(F_{n}) ≥ F_{τ(n)} which are extended to the case of balancing numbers by Sahukar and Panda [17]. There are many generalizations and analogs of the Euler totient function [18]. For example, φ_{k} is defined as

φ_{k}(n) = ∑_{1≤m<n,(m,n)=1} m^{k},

where k ≥ 0 is an integer, and the Jordan totient function J_{k} is defined as

J_{k}(n) = n^{k} ∏_{p|n} (1 − \frac{1}{p^{k}} ) ,

where k is a positive integer and p is a prime number (see [18, 2]). Clearly, we have φ_{0}(n) = φ(n) = J_{1}(n).

In 2019, Jaidee and Pongsriiam [6] showed that for every natural number n, the n^{th} Fibonacci number satisfies φ_{k}(F_{n}) ≤ F_{φ_{k}(n)} except for (n, k) = (6, 1) for which the inequality is reversed, and J_{k}(F_{n}) ≤ F_{J_{k}(n)} for all n ≥ 1. They also established similar results for Lucas sequence in the same article. Motivated by the work of [6] and [7], we establish inequalities related with φ_{k} and Lucas sequences of the first kind (U_{n})_{n≥1} with characteristic equation having real roots. As an application, we also establish inequalities related to Fibonacci, Pell, and balancing sequences. More precisely, we prove the following result:

Theorem 1. Let n and k be positive integers, φ_{k} be the generalized Euler totient function, and (U_{n})_{n≥1} be Lucas sequence of the first kind with α, β as non-zero real roots of the corresponding characteristic equation. Then the following statements hold.

(i) For every k ≥ 1, φ_{k}(|U_{1}|) = |U_{φ_{k}(1)}|, and if |U_{2}| = 1 or 2, then φ_{k}(|U_{2}|) = |U_{φ_{k}(2)}| otherwise φ_{k}(|U_{2}|) > |U_{φ_{k}(2)}|.

(ii) φ_{1}(|U_{n}|) < |U_{φ_{1}(n)}| whenever n ≥ 16.

(iii) φ_{k}(|U_{n}|) < |U_{φ_{k}(n)}| for every n ≥ 3 and k ≥ 2 except for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2).

Remark 2. In Theorem 1, the inequality may or may not hold for (n, 1) where 3 ≤ n ≤ 15, and (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). Therefore, the inequality can be checked for every Lucas sequence of the first kind just by computing for the values mentioned above.

As a consequence of Theorem 1, we obtain the result of Jaidee and Pongsriiam [6, Theorem 3.2(i)] concerning the Fibonacci sequence. Additionally, we establish a set of inequalities associated with the Pell sequence and the balancing sequence within the theorem, which can be expressed as follows:

Theorem 3. Let n and k be positive integers, φ_{k} be the generalized Euler totient function, and (F_{n})_{n≥1}, (P_{n})_{n≥1}, and (B_{n})_{n≥1} are Fibonacci sequence, Pell sequence and balancing sequence, respectively. Then the following statements hold.

(i) φ_{k}(F_{n}) ≤ F_{φ_{k}(n)} for all n ≥ 1 except for (n, k) = (6, 1) for which the inequality reverses, and equality holds for (n, k) = (1, k), (2, k), (4, 1), where k ≥ 1.

(ii) φ_{k}(P_{n}) ≤ P_{φ_{k}(n)} for all n ≥ 1 except for (n, k) = (3, 1), (4, 1), (6, 1), (3, 2) for which the inequality reverses, and equality holds for (n, k) = (1, k), (2, k), where k ≥ 1.

(iii) φ_{k}(B_{n}) ≤ B_{φ_{k}(n)} for all n ≥ 1 except for (n, k) = (2, k), (3, 1), (4, 1), (6, 1), (12, 1), (3, 2) for which the inequality reverses, and equality holds for (n, k) = (1, k), where k ≥ 1.

2 Some examples and related results

In this section, we first give some examples of Lucas sequences of the first kind and some results which are needed to prove our main theorems.

Example 4 (Lucas sequences of the first kind). Some examples of Lucas sequences of the first kind are as follows:

(i) The Fibonacci sequence (F_{n})_{n≥0} defined as

F_{0} = 0, F_{1} = 1 and F_{n+2} = F_{n+1} + F_{n}, n ≥ 0,

is the Lucas sequence of the first kind with r = 1 and s = 1 (see [19, \underline{A000045}]). The characteristic equation of (F_{n})_{n≥0} is x^{2} −x−1 = 0 with roots α = \frac{1+\sqrt{5}}{2} and β = \frac{1−\sqrt{5}}{2}.

(ii) The Pell sequence (P_{n})_{n≥0} defined as

P_{0} = 0, P_{1} = 1 and P_{n+2} = 2P_{n+1} + P_{n}, n ≥ 0,

is the Lucas sequence of the first kind with r = 2 and s = 1 (see [19, \underline{A000129}]). The characteristic equation of (P_{n})_{n≥0} is x^{2} − 2x − 1 = 0 with roots α = 1 + \sqrt{2} and β = 1 − \sqrt{2}.

(iii) The balancing sequence (B_{n})_{n≥0} defined as

B_{0} = 0, B_{1} = 1 and B_{n+2} = 6B_{n+1} − B_{n}, n ≥ 0,

is the Lucas sequence of the first kind with r = 6 and s = −1 (see [19, \underline{A001109}]). The characteristic equation of (B_{n})_{n≥0} is x^{2} − 6x + 1 = 0 with roots α = 3 + \sqrt{2} and β = 3 − \sqrt{2}.

Throughout this article, we consider the Lucas sequences of the first kind for which α, β are real roots of the corresponding characteristic equation such that |α| > |β| > 0. Further, we obtain

|α| = \frac{|r| + \sqrt{Δ}}{2}, |β| = \frac{||r| − \sqrt{Δ}|}{2}.

We now give results related with Lucas sequences of the first kind which will be used to prove our main results:

Lemma 5. [7, Lemma 1] Let (U_{n})_{n≥1} be Lucas sequence of the first kind, where n is positive integer and α, β be non-zero real roots of the corresponding characteristic equation. Then the following statements hold.

(i) If |α| ≠ \frac{1+\sqrt{5}}{2}, then |α| ≥ 2.

(ii) |U_{n}| < 2|α|^{n}.

(iii) If 0 < m < n and m is even, then |\frac{U_{n}}{U_{m}} | > |α|^{n−m}.

Lemma 6. Let Fibonacci sequence (F_{n})_{n≥0} as defined in Example 4 with α = \frac{1+\sqrt{5}}{2} and

β = \frac{1−\sqrt{5}}{2} as real roots of the corresponding characteristic equation. Then the following statements hold.

(i) (Binet formula) F_{n} = \frac{α^{n}−β^{n}}{α−β} for all n ≥ 0.

(ii) α^{n−2} ≤ F_{n} ≤ α^{n−1} for all n ≥ 0.

Proof. The proof of statement (i) follows directly from the definition of the Lucas sequence of the first kind and statement (ii) can be proved by mathematical induction.

Using the Lemma 5 and Lemma 6, we establish the following result: Lemma 7. Let (U_{n})_{n≥1} be Lucas sequence of the first kind and α, β be non-zero real roots of the corresponding characteristic equation. Then the following statements hold.

(i) If |α| = \frac{1+\sqrt{5}}{2}, then |U_{n}| = F_{n} and |α|^{n−2} ≤ |U_{n}| ≤ |α|^{n−1}. (ii) If |α| ≥ 2, then |α|^{n−2} < |U_{n}| < |α|^{n+1}. Proof.

(i) Let |α| = \frac{1+\sqrt{5}}{2}. Since α is a real number, either α = \frac{1+\sqrt{5}}{2} or α = − (\frac{1+\sqrt{5}}{2}). Now if α = \frac{1+\sqrt{5}}{2}, then β = \frac{1−\sqrt{5}}{2} is also a root of the characteristic equation. Therefore, |U_{n}| = |F_{n}| = F_{n}. Similarly, if α = − (\frac{1+\sqrt{5}}{2}), then β = \frac{−1+\sqrt{5}}{2} = − (\frac{1−\sqrt{5}}{2}) is also a root of the characteristic equation. Therefore, in this case we have

|U_{n}| = |(−1)^{n−1}F_{n}| = |F_{n}| = F_{n}. Hence, for either case using Lemma 6 (ii), we have |α|^{n−2} ≤ |U_{n}| ≤ |α|^{n−1}. This completes the proof of (i).

(ii) Consider |α| ≥ 2. Using Lemma 5 (ii), we have |U_{n}| < 2|α|^{n} ≤ |α||α|^{n} = |α|^{n+1}. Now using Lemma 5 (iii) for m = 2, we have |U_{n}| ≥ |\frac{U_{n}}{U_{2}} | > |α|^{n−2}. On combining the above two inequalities, we conclude that |α|^{n−2} < |U_{n}| < |α|^{n+1}.

Now we establish some results related with the Euler totient function and the generalized Euler totient function φ_{k}. Lemma 8. [6, Lemma 2.5] Let n ≥ 3. Then the following statements hold.

(i) If k = 1, then φ_{k}(n) = \frac{nφ(n)}{2}.

(ii) If k ≥ 2, then \frac{n^{k}φ(n)}{2^{k}} < φ_{k}(n) < \frac{n^{k}φ(n)}{2}.

Lemma 9. [7, Lemma 3 (iv)] Let φ be the Euler totient function and n be any natural number. Then φ(n) ≥ 2\sqrt{n/3} whenever n ≠ 1, 2, 6. Lemma 10. Let n and k be positive integer such that 3 ≤ n ≤ 6 and k ≥ 2. Then for the generalized Euler totient function φ_{k}, the following statements hold.

(i) φ_{k}(3) ≥ 4(k + 1) + 2 whenever k ≥ 5. (ii) φ_{k}(4) ≥ 5(k + 1) + 2 whenever k ≥ 3.

(iii) φ_{k}(5) ≥ 6(k + 1) + 2 whenever k ≥ 2.

(iv) φ_{k}(6) ≥ 7(k + 1) + 2 whenever k ≥ 2.

Proof. By definition of φ_{k}, φ_{k}(3) = 1 + 2^{k}. Therefore, φ_{k}(3) ≥ 4(k + 1) + 2 if and only if 2^{k} ≥ 4k + 5 and by induction, 2^{k} ≥ 4k + 5 is true whenever k ≥ 5. This implies that φ_{k}(3) ≥ 4(k + 1) + 2 whenever k ≥ 5. This completes the proof of (i). The proof of (ii), (iii), and (iv) follows similarly using definition of function φ_{k} and mathematical induction.

3 Proof of main results

3.1 Proof of Theorem 1

(i) Using the definition of φ_{k}, we have φ_{k}(1) = 1 and φ_{k}(2) = 1 for every k ≥ 1. Therefore, using the definition of the Lucas sequence of the first kind for n = 1, we have φ_{k}(|U_{1}|) = φ_{k}(1) = 1 = |U_{1}| = |U_{φ_{k}(1)}| for every k ≥ 1. Similarly, using the definition of the Lucas sequence of the first kind for n = 2, we have |U_{2}| = |α + β| = |r| ≥ 1. Therefore, φ_{k}(|U_{2}|) = φ_{k}(|r|) ≥ φ_{k}(1) = 1 = |U_{1}| = |U_{φ_{k}(2)}| for every k ≥ 1. Hence, if |U_{2}| = |r| = 1 or 2, then φ_{k}(|U_{2}|) = |U_{φ_{k}(2)}| otherwise φ_{k}(|U_{2}|) = φ_{k}(|r|) > 1 = |U_{φ_{k}(2)}| for every k ≥ 1.

(ii) Let n be positive integer such that n ≥ 16. By Lemma 5(i), we divide our problem in two cases:

Case I: |α| = \frac{1+\sqrt{5}}{2}.

If |α| = \frac{1+\sqrt{5}}{2}, then using Lemma 8 and Lemma 7, we obtain

φ_{1}(|U_{n}|) = \frac{|U_{n}|φ(|U_{n}|)}{2} < \frac{|U_{n}||U_{n}|}{2} = \frac{|U_{n}|^{2}}{2} ≤ \frac{|α|^{2(n−1)}}{2} ≤ |α|^{2(n−1)−1},

and |U_{φ_{1}(n)}| ≥ |α|^{φ_{1}(n)−2}. Therefore, in order to prove φ_{1}(|U_{n}|) < |U_{φ_{1}(n)}|, it is sufficient to show that φ_{1}(n) − 2 ≥ 2(n − 1) − 1, i.e.,

φ_{1}(n) ≥ 2n − 1.  (1)

Case II: |α| ≥ 2.

If |α| ≥ 2, again using Lemma 8 and Lemma 7, we obtain

φ_{1}(|U_{n}|) = \frac{|U_{n}|φ(|U_{n}|)}{2} < \frac{|U_{n}||U_{n}|}{2} = \frac{|U_{n}|^{2}}{2} < \frac{|α|^{2(n+1)}}{2} ≤ |α|^{2(n+1)},

and |U_{φ_{1}(n)}| > |α|^{φ_{1}(n)−2}. Hence, in order to prove φ_{1}(|U_{n}|) < |U_{φ_{1}(n)}|, it is sufficient to show that φ_{1}(n) − 2 ≥ 2(n + 1), i.e.,

φ_{1}(n) ≥ 2n + 4.  (2)

Since 2n + 4 > 2n − 1 for every n ≥ 16, therefore, in order to prove (1) and (2), it is

sufficient to prove (2), i.e., φ_{1}(n) ≥ 2n + 4. By Lemma 8, φ_{1}(n) = \frac{nφ(n)}{2} and using

Lemma 9, for n ≥ 16, φ(n) ≥ 2\sqrt{n/3}. Hence, in order to prove (2), it is sufficient to prove that

n\sqrt{n/3} ≥ 2n + 4.  (3)

Consider the function f : [16, ∞) → ℝ defined by

f(x) = x\sqrt{x/3} − 2x − 4.

If x ≥ 16, then f′(x) = \sqrt{3x}/2 − 2 > 2\sqrt{3} − 2 > 0, and therefore, f is an increasing function on [16, ∞). Therefore, f(x) ≥ f(16) = 16\sqrt{16/3} − 2 · 16 − 4 = 0.95 > 0. This implies that

x\sqrt{x/3} − 2x − 4 ≥ 0 for x ≥ 16.

Hence, (3) holds and this proves that φ_{1}(|U_{n}|) < |U_{φ_{1}(n)}| whenever n ≥ 16.

(iii) We proceed in similar manner as in proof of (i). Let n and k be positive integer such that n ≥ 7 and k ≥ 2. By Lemma 5(i), we divide our problem in two cases:

Case I: |α| = \frac{1+\sqrt{5}}{2}.

If |α| = \frac{1+\sqrt{5}}{2}, then using Lemma 8 and Lemma 7, we obtain

φ_{k}(|U_{n}|) < \frac{|U_{n}|^{k}φ(|U_{n}|)}{2}

< \frac{|U_{n}|^{k}|U_{n}|}{2} = \frac{|U_{n}|^{k+1}}{2}

≤ \frac{|α|^{(n−1)(k+1)}}{2}

< |α|^{(n−1)(k+1)−1},

and |U_{φ_{k}(n)}| ≥ |α|^{φ_{k}(n)−2}. Therefore, in order to prove φ_{k}(|U_{n}|) < |U_{φ_{k}(n)}|, it is sufficient to show that φ_{k}(n) − 2 ≥ (n − 1)(k + 1) − 1, i.e.,

φ_{k}(n) ≥ (n − 1)(k + 1) + 1.  (4)

Case II: |α| ≥ 2.

If |α| ≥ 2, again using Lemma 8 and Lemma 7, we obtain

φ_{k}(|U_{n}|) < \frac{|U_{n}|^{k}φ(|U_{n}|)}{2}

< \frac{|U_{n}|^{k}|U_{n}|}{2} = \frac{|U_{n}|^{k+1}}{2}

< \frac{|α|^{(n+1)(k+1)}}{2} ≤ |α|^{(n+1)(k+1)},

and |U_{φ_{k}(n)}| > |α|^{φ_{k}(n)−2}. Similar to Case I, in order to prove φ_{k}(|U_{n}|) < |U_{φ_{k}(n)}|, it suffices to show that φ_{k}(n) − 2 ≥ (n + 1)(k + 1), i.e.,

φ_{k}(n) ≥ (n + 1)(k + 1) + 2.  (5)

Since (n + 1)(k + 1) + 2 ≥ (n − 1)(k + 1) + 1 for every n ≥ 1 and k ≥ 2, therefore in order to prove (4) and (5), it is sufficient to prove (5), i.e., φ_{k}(n) ≥ (n + 1)(k + 1) + 2. Now we claim that φ(n) ≥ 4 for every n ≥ 7. For this, suppose that n ≥ 7. If there exists a prime p ≥ 5 dividing n, then by multiplicativity of φ, we obtain

φ(n) ≥ φ(p) ≥ p − 1 ≥ 4.

Suppose n = 2^{a}3^{b} for some a, b ∈ ℕ ∪ {0}. If a ≥ 3, then φ(n) ≥ φ(2^{3}) = 2^{3} − 2^{2} = 4. If a = 2, b ≥ 1, then φ(n) ≥ φ(2^{2})ϕ(3) = (2^{2} − 2)(3 − 1) = 4. If a ≤ 1, b ≥ 2, then φ(n) ≥ φ(3^{2}) = 6. Thus, φ(n) ≥ 4 for every n ≥ 7.

Now by Lemma 8, we have

φ_{k}(n) ≥ φ(n) (\frac{n}{2})^{k} ≥ 4 (\frac{n}{2})^{k} .

Hence, in order to prove (5), it is sufficient to prove that

4 (\frac{n}{2})^{k} ≥ (n + 1)(k + 1) + 2.  (6)

Consider the function f : [7, ∞) → ℝ defined by

f(x) = 4 (\frac{x}{2})^{k} − (x + 1)(k + 1) − 2.

If x ≥ 7, then f′(x) = 2k (\frac{x}{2})^{k−1} − (k + 1) > 6k − (k + 1) > 0, and so f is an

increasing function on [7, ∞). Therefore, f(x) ≥ f(7) = 4 (\frac{7}{2})^{k} − 8(k + 1) − 2 >

4(3)^{k} − (8)(k + 1) − 2 > 0, where k ≥ 2. This implies that

4 (\frac{x}{2})^{k} ≥ (x + 1)(k + 1) + 2 for x ≥ 7.

Hence, (6) holds and this proves that φ_{k}(|U_{n}|) < |U_{φ_{k}(n)}| for every n ≥ 7 and k ≥ 2. By Lemma 10, for 3 ≤ n ≤ 6 and k ≥ 2, (5) holds except for for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). This completes the proof of (iii).

3.2 Proof of Theorem 3

(i) The Fibonacci sequence (F_{n})_{n≥1} is the Lucas sequence of the first kind with r = 1 and

s = 1. The characteristic equation of (F_{n})_{n≥1} is x^{2} − x − 1 = 0 with roots α = \frac{1+\sqrt{5}}{2}

and β = \frac{1−\sqrt{5}}{2}. Since both α and β are real roots of the characteristic equation, all hypothesis of Theorem 1 holds. Hence, |U_{n}| = F_{n} and by using Theorem 1 (i), for every k ≥ 1, φ_{k}(F_{1}) = F_{φ_{k}(1)} and since F_{2} = 1, this implies that φ_{k}(F_{2}) = F_{φ_{k}(2)}. Now, using Theorem 1 (ii), φ_{1}(F_{n}) < F_{φ_{1}(n)} whenever n ≥ 16. Using definition of φ and a straightforward calculation for (n, 1) where 3 ≤ n ≤ 15, we obtain that φ_{1}(F_{n}) ≤ F_{φ_{1}(n)} except for (n, k) = (6, 1) for which the inequality is reversed, and equality holds for (n, k) = (4, 1). Using Theorem 1 (iii), φ_{k}(F_{n}) < F_{φ_{k}(n)} for every n ≥ 3 and k ≥ 2 except for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). Again using definition of φ and a straightforward calculation, we obtain φ_{k}(F_{n}) < F_{φ_{k}(n)} for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). This completes the proof of (i).

(ii) The Pell sequence (P_{n})_{n≥1} is the Lucas sequence of the first kind with r = 2 and s = 1. The characteristic equation of (P_{n})_{n≥1} is x^{2} − 2x − 1 = 0 with roots α = 1 + \sqrt{2} and β = 1 − \sqrt{2}. Since both α and β are real roots of the characteristic equation, all hypothesis of Theorem 1 holds. Hence, |U_{n}| = P_{n} and by using Theorem 1 (i), for every k ≥ 1, φ_{k}(P_{1}) = P_{φ_{k}(1)} and since P_{2} = 2, this implies that φ_{k}(P_{2}) = P_{φ_{k}(2)}. Now, using Theorem 1 (ii), φ_{1}(P_{n}) < P_{φ_{1}(n)} whenever n ≥ 16. Using definition of φ and a straightforward calculation for (n, 1) where 3 ≤ n ≤ 15, we obtain that φ_{1}(P_{n}) ≤ P_{φ_{1}(n)} except for (n, k) = (3, 1), (4, 1), (6, 1) for which the inequality is reversed. Using Theorem 1 (iii), we have φ_{k}(P_{n}) < P_{φ_{k}(n)} for every n ≥ 3 and k ≥ 2 except for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). Again using definition of φ and a straightforward calculation , we obtain φ_{k}(P_{n}) < P_{φ_{k}(n)} for (n, k) = (3, 3), (3, 4), (4, 2) and φ_{k}(P_{n}) > P_{φ_{k}(n)} for (n, k) = (3, 2). This completes the proof of (ii).

(iii) The balancing sequence (B_{n})_{n≥1} is the Lucas sequence of the first kind with r = 6 and s = −1. The characteristic equation of (B_{n})_{n≥1} is x^{2}−6x+1 = 0 with roots α = 3+\sqrt{2} and β = 3 − \sqrt{2}. Since both α and β are real roots of the characteristic equation, all hypothesis of Theorem 1 holds. Hence, |U_{n}| = B_{n} and by using Theorem 1 (i), for every k ≥ 1, φ_{k}(B_{1}) = B_{φ_{k}(1)} and since B_{2} = 6, this implies that φ_{k}(B_{2}) > B_{φ_{k}(2)}. Now, using Theorem 1 (ii), φ_{1}(B_{n}) < B_{φ_{1}(n)} whenever n ≥ 16. Using definition of φ and a straightforward calculation for (n, 1) where 3 ≤ n ≤ 15, we obtain that φ_{1}(B_{n}) ≤ B_{φ_{1}(n)} except for (n, k) = (3, 1), (4, 1), (6, 1), (12, 1) for which the inequality is reversed. Using Theorem 1 (iii), φ_{k}(B_{n}) < B_{φ_{k}(n)} for every n ≥ 3 and k ≥ 2 except for (n, k) = (3, 2), (3, 3), (3, 4), (4, 2). Again using definition of φ and a straightforward calculation, we obtain φ_{k}(F_{n}) < F_{φ_{k}(n)} for (n, k) = (3, 3), (3, 4), (4, 2) and φ_{k}(B_{n}) > B_{φ_{k}(n)} for (n, k) = (3, 2). This completes the proof of (iii).

4 Acknowledgments

The authors are thankful to the referee for the valuable suggestions. The first author is grateful to the University Grant Commission (U.G.C.), India for providing financial support (U.G.C. Ref. no. 191620159617) for this research work. The authors would also like to thank Dr. Anup Kumar Singh and Mr. Shubham Gupta for the fruitful discussion related with computational results.

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    at https://oeis.org/.

2020 Mathematics Subject Classification: Primary 11A25, Secondary 11B39. Keywords: Euler’s totient function, Lucas sequence, Fibonacci number, Pell number, balancing number.

(Concerned with sequences \underline{A000045}, \underline{A000129}, and \underline{A001109}.)

Received July 24 2023; revised versions received July 31 2023; September 29 2023; October 6 2023. Published in Journal of Integer Sequences, October 14 2023.

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