The 4-Nicol Numbers Having Five Different Prime Divisors

Qiao-Xiao Jin and Min Tang^{1} Department of Mathematics Anhui Normal University Wuhu 241000 P. R. China

                   tangmin@mail.ahnu.edu.cn

Abstract

A positive integer n is called a Nicol number if n | φ(n)+σ(n), and a t-Nicol number if φ(n) + σ(n) = tn. In this paper, we show that if n is a 4-Nicol number that has five different prime divisors, then n = 2^{α_{1}} · 3 · 5^{α_{3}} · p^{α_{4}} · q^{α_{5}}, or n = 2^{α_{1}} · 3 · 7^{α_{3}} · p^{α_{4}} · q^{α_{5}} with p ≤ 29.

1 Introduction

For any positive integer n, let ϕ(n), ω(n) and σ(n) be the Euler function of n, the number of prime divisors of n and the sum of divisors of n, respectively. We call n is a Nicol number if n | φ(n) + σ(n), and a t-Nicol number if φ(n) + σ(n) = tn. It is well-known that t ≥ 2, and n is prime if and only if φ(n) + σ(n) = 2n. In 1966, Nicol [4] conjectured that Nicol numbers are all even, and proved that if α is such that p = 2^{α−2} · 7 − 1 is prime, then n = 2^{α} · 3 · p is 3-Nicol number. In 1995, Ming-Zhi Zhang [6] showed that if n = p^{α}q then n cannot be a Nicol number, where p and q are distinct primes and α is a positive integer. In 1997, Lin and Zhang [2] showed that if ω(n) = 2, then n cannot be a Nicol number. In 2008, Luca and Sandor [3] showed that if n is a Nicol number and ω(n) = 3, then either n ∈ {560, 588, 1400} or n = 2^{α} · 3 · p with p = 2^{α−2} · 7 − 1 prime. In 2008, Wang [5] studied the Nicol numbers that have four different prime divisors. In 2009, Harris [1] showed that the Nicol numbers that have four different prime divisors must be one of the following forms:

1Corresponding author. This work was supported by the National Natural Science Foundation of China, Grant No. 10901002 and the SF of the Education Department of Anhui Province, Grant No. KJ2010A126.

1. n = 2^{3} · 3^{3} · 5^{2} · 11, 2^{4} · 3^{3} · 5 · 11, 2^{7} · 5 · 11 · 79, 2^{3} · 3^{3} · 5^{3} · 13^{2}, 2^{2} · 3^{2} · 17 · 241, 2^{2} · 3^{2} · 17^{2} · 2243;

2. n ∈ \{2^{a} · 3 · p_{3} · p_{4}|p_{4} = \frac{(7 · 2^{a−2} − 1)p_{3} + 9 · 2^{a−2} − 1}{p_{3} − (7 · 2^{a−2} − 1)}, where p_{3}, p_{4} are distinct

primes.}

Moreover, Harris [1] proved that all but finitely many Nicol numbers that have 5 different prime divisors are divisible by 6 and not 9.

In this paper, we study the 4-Nicol numbers that have five different prime divisors and obtain the following result:

Theorem 1. If n is a 4-Nicol number with ω(n) = 5, then either n = 2^{α_{1}}3^{α_{2}}5^{α_{3}}p^{α_{4}}q^{α_{5}}, or n = 2^{α_{1}}3^{α_{2}}7^{α_{3}}p^{α_{4}}q^{α_{5}} with p ≤ 29, where p, q are distinct primes, and α_{i}(i = 1, 2, ⋯, 5) are positive integers.

By the Harris result and Theorem 1, we have the following result:

Corollary 2. All but finitely many 4-Nicol numbers with 5 different prime divisors have the form n = 2^{α_{1}} · 3 · 5^{α_{3}} · p^{α_{4}} · q^{α_{5}}, or n = 2^{α_{1}} · 3 · 7^{α_{3}} · p^{α_{4}} · q^{α_{5}} with p ≤ 29.

Throughout this paper, let a and m be relatively prime positive integers, the least positive integer x such that a^{x} ≡ 1 (mod m) is called the order of a modulo m. We denote the order of a modulo m by ord_{m}(a). Let V_{p}(m) be the exponent of the highest power of p that divides m.

2 Lemmas

The following three lemmas are motivated by the work of Luca and Sándor [3]. Here we make some minor revisions.

Lemma 3. Let a, b be two natural numbers and p be an odd prime. If V_{p}(a − 1) ≥ 1, then

V_{p}(a^{b} − 1) = V_{p}(b) + V_{p}(a − 1).

Proof. Let V_{p}(b) = m and V_{p}(a − 1) = n. We may assume that b = p^{m}t with p ∤ t and a = 1 + p^{n}a_{0} with p ∤ a_{0}.

Since n ≥ 1, we have

a^{t} = (1 + p^{n}a_{0})^{t} = 1 + C_{t}^{1}p^{n}a_{0} + ⋯ + C_{t}^{t}(p^{n}a_{0})^{t} = 1 + p^{n}c, p ∤ c.

Thus

a^{tp} = (1 + p^{n}c)^{p} = 1 + C_{p}^{1}p^{n}c + ⋯ + C_{p}^{p}(p^{n}c)^{p} = 1 + p^{n+1}a_{1}, p ∤ a_{1}.

By induction on m, for all m ≥ 0 we have a^{b} = a^{tp^{m}} = 1 + p^{m+n}a_{m} with p ∤ a_{m}. Hence

V_{p}(a^{b} − 1) = m + n = V_{p}(b) + V_{p}(a − 1).

This completes the proof of Lemma 3.

Lemma 4. Let t be a natural number and p, q be two primes. We have

V_{p}(q^{t} − 1) ≤ V_{p}(q^{f} − 1) + V_{p}(t),

where f = ord_{p}(q), if p ≠ 2; and f = 2, if p = 2.

Proof. (i) p = 2. By [3, Lemma 1], we have

V_{2}(q^{t} − 1) ≤ V_{2}(q^{2} − 1) + V_{2}(t).

(ii) p > 2. Now consider the following two cases:

Case 1. q^{t} ≢ 1 (mod p). The above inequality is obvious.

Case 2. q^{t} ≡ 1 (mod p). Then ord_{p}(q) | t and ord_{p}(q) | p − 1. Let V_{p}(t) = m. We may assume that t = ord_{p}(q) · p^{m} · k with p ∤ k. Thus

V_{p}(q^{t} − 1) = V_{p}((q^{ord_{p}(q)})^{p^{m}·k} − 1)
= V_{p}(q^{ord_{p}(q)} − 1) + V_{p}(p^{m} · k)
= V_{p}(q^{ord_{p}(q)} − 1) + m
= V_{p}(q^{ord_{p}(q)} − 1) + V_{p}(t).

This completes the proof of Lemma 4.

Lemma 5. Let n = p_{1}^{α_{1}}p_{2}^{α_{2}} ⋯ p_{k}^{α_{k}} be the standard factorization of n and X = max{α_{j} | j = 1, 2, ⋯, k}. We fix i ∈ {1, ⋯, k} such that X = α_{i}. If n is a Nicol number, then we have

X − 1 ≤ ∑_{j=1,j≠i}^{k} V_{p_{i}}(p_{j}^{f_{j}} − 1) + \frac{k − 1}{log p_{i}} log(X + 1),

where f_{j} = ord_{p_{i}}(p_{j}), if p_{i} ≠ 2; and f_{j} = 2, if p_{i} = 2.

Proof. Since n | ϕ(n) + σ(n) and p_{i}^{X−1} | ϕ(n), we have

p_{i}^{X−1} | σ(n) = ∏_{j=1}^{k} (\frac{p_{j}^{α_{j}+1} − 1}{p_{j} − 1}).

Hence

p_{i}^{X−1}| ∏_{j=1}^{k}(p_{j}^{α_{j}+1} − 1).

The above relation implies that

X − 1 ≤ ∑_{j=1}^{k} V_{p_{i}}(p_{j}^{α_{j}+1} − 1) = ∑_{j=1,j≠i}^{k} V_{p_{i}}(p_{j}^{α_{j}+1} − 1).

By Lemma 4

X − 1 ≤ ∑_{j=1,j≠i}^{k} V_{p_{i}}(p_{j}^{f_{j}} − 1) + ∑_{j=1,j≠i}^{k} V_{p_{i}}(α_{j} + 1)

≤ ∑_{j=1,j≠i}^{k} V_{p_{i}}(p_{j}^{f_{j}} − 1) + ∑_{j=1,j≠i}^{k} \frac{log(α_{j} + 1)}{log p_{i}}

≤ ∑_{j=1,j≠i}^{k} V_{p_{i}}(p_{j}^{f_{j}} − 1) + \frac{k − 1}{log p_{i}} log(X + 1).

This completes the proof of Lemma 5.

Lemma 6. If n is a 4-Nicol number and ω(n) = 5, then n must be one of the following three forms:

1. n = 2^{α_{1}} · 3^{α_{2}} · 5^{α_{3}} · p^{α_{4}} · q^{α_{5}}, p, q are distinct primes and 7 ≤ p < q.

2. n = 2^{α_{1}} · 3^{α_{2}} · 7^{α_{3}} · p^{α_{4}} · q^{α_{5}}, p < q are distinct primes and p ≤ 29.

3. n = 2^{α_{1}} · 3^{α_{2}} · 11^{α_{3}} · 13^{α_{4}} · p^{α_{5}}, p ≤ 23 is prime.

Proof. Let n = p_{1}^{α_{1}}p_{2}^{α_{2}}p_{3}^{α_{3}}p_{4}^{α_{4}}p_{5}^{α_{5}} be the standard factorization of n. Put l = \frac{σ(n)}{n}.

Noting that n is a 4-Nicol number and φ(n)σ(n) < n^{2}, we have 4 = \frac{φ(n)}{n} + \frac{σ(n)}{n} < l+l^{−1},

hence l > 2 + \sqrt{3}. By \frac{n}{φ(n)} > \frac{σ(n)}{n} = l, we have

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} > l > 2 + \sqrt{3}.

If p_{2} ≥ 5 then

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} ≤ 2 · \frac{5}{4} · \frac{7}{6} · \frac{11}{10} · \frac{13}{12} < 2 + \sqrt{3},

a contradiction. Thus p_{2} = 3 and p_{1} = 2.

If p_{3} ≥ 13 then

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} ≤ 2 · \frac{3}{2} · \frac{13}{12} · \frac{17}{16} · \frac{19}{18} < 2 + \sqrt{3},

a contradiction, thus p_{3} ≤ 11.

Case 1. p_{3} = 7. Then p_{4} ≤ 29. In fact, if p_{4} ≥ 31 then

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} ≤ 2 · \frac{3}{2} · \frac{7}{6} · \frac{31}{30} · \frac{37}{36} < 2 + \sqrt{3},

a contradiction.

Case 2. p_{3} = 11. Then p_{4} = 13 and p_{5} ≤ 23. In fact, if p_{4} ≥ 17 then

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} ≤ 2 · \frac{3}{2} · \frac{11}{10} · \frac{17}{16} · \frac{19}{18} < 2 + \sqrt{3},

a contradiction.

If p_{5} ≥ 29 then

\frac{n}{φ(n)} = \frac{p_{1}}{p_{1} − 1} \frac{p_{2}}{p_{2} − 1} \frac{p_{3}}{p_{3} − 1} \frac{p_{4}}{p_{4} − 1} \frac{p_{5}}{p_{5} − 1} ≤ 2 · \frac{3}{2} · \frac{11}{10} · \frac{13}{12} · \frac{29}{28} < 2 + \sqrt{3},

a contradiction.

This completes the proof of Lemma 6.

3 Proof of Theorem 1

By Lemma 6, it is enough to show that there is no 4-Nicol numbers n = 2^{α_{1}}·3^{α_{2}}·11^{α_{3}}·13^{α_{4}}·p^{α_{5}} with p ≤ 23.

Assume that n = 2^{α_{1}} · 3^{α_{2}} · 11^{α_{3}} · 13^{α_{4}} · p^{α_{5}} with p ≤ 23 be a 4-Nicol number, then by

\frac{φ(n)}{n} + \frac{σ(n)}{n} = 4 we have:

\begin{matrix}2^{α_{1}+6} · 3^{α_{2}+1} · 5 · 11^{α_{3}−1} · 13^{α_{4}−1} · p^{α_{5}−1} · (133p + 10) · (p − 1) \\ = (2^{α_{1}+1} − 1) · (3^{α_{2}+1} − 1) · (11^{α_{3}+1} − 1) · (13^{α_{4}+1} − 1) · (p^{α_{5}+1} − 1). \end{matrix}

Case 1. p = 17, n = 2^{α_{1}} · 3^{α_{2}} · 11^{α_{3}} · 13^{α_{4}} · 17^{α_{5}}. Then

2^{α_{1}+10} · 3^{α_{2}+2} · 5 · 11^{α_{3}−1} · 13^{α_{4}−1} · 17^{α_{5}−1} · 757
= (2^{α_{1}+1} − 1)(3^{α_{2}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(17^{α_{5}+1} − 1).  (1)

By Lemma 5 we have X ≤ 35. Noting that

ord_{757}(2) = 756, ord_{757}(3) = 9, ord_{757}(11) = ord_{757}(13) = ord_{757}(17) = 189,

thus

757 ∤ 2^{α_{1}+1} − 1, 757 ∤ 11^{α_{3}+1} − 1, 757 ∤ 13^{α_{4}+1} − 1, 757 ∤ 17^{α_{5}+1} − 1.

By (1) we have 757 | 3^{α_{2}+1} − 1, thus α_{2} + 1 = 9k, k ∈ ℤ. By X ≤ 35, we have k = 1, 2, 3. Subcase 1: k = 1, α_{2} + 1 = 9. Then

\begin{matrix}2^{α_{1}+9} · 3^{10} · 5 · 11^{α_{3}−1} · 13^{α_{4}−2} · 17^{α_{5}−1} \\ = (2^{α_{1}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(17^{α_{5}+1} − 1). \end{matrix}

(i) α_{4} = 2. By ord_{61}(13) = 3 we have 61 | 13^{3} − 1, this is impossible.

(ii) α_{4} > 2. Then 13 | (2^{α_{1}+1} − 1)(11^{α_{3}+1} − 1)(17^{α_{5}+1} − 1). On the other hand, we have the following facts: If 13 | 2^{α_{1}+1} − 1, by ord_{13}(2) = 12, thus 12 | α_{1} + 1, and noting that ord_{7}(2) = 3 we have 7 | 2^{α_{1}+1} − 1, which is impossible. If 13 | 11^{α_{3}+1} − 1, by ord_{13}(11) = 12, thus 12 | α_{3} + 1, and noting that ord_{7}(11) = 3 we have 7 | 11^{α_{3}+1} − 1, which is impossible. If 13 | 17^{α_{5}+1} − 1, by ord_{13}(17) = 6, thus 6 | α_{5} + 1, and noting that ord_{7}(17) = 6 we have 7 | 17^{α_{5}+1} − 1, which is impossible. Thus 13 ∤ (2^{α_{1}+1} − 1)(11^{α_{3}+1} − 1)(17^{α_{5}+1} − 1), a contradiction.

Subcase 2: k = 2, α_{2} + 1 = 18. By ord_{7}(3) = 6, we have 7 | 3^{18} − 1, thus 7 | 3^{α_{2}+1} − 1, which contradicts (1).

Subcase 3: k = 3, α_{2} + 1 = 27. By ord_{757}(3) = 9, we have 757 | 3^{27} − 1, thus 757 | 3^{α_{2}+1} − 1, which contradicts (1).

Case 2. p = 19, n = 2^{α_{1}} · 3^{α_{2}} · 11^{α_{3}} · 13^{α_{4}} · 19^{α_{5}}. Then

2^{α_{1}+7} · 3^{α_{2}+3} · 5 · 11^{α_{3}−1} · 13^{α_{4}−1} · 19^{α_{5}−1} · 43 · 59
= (2^{α_{1}+1} − 1)(3^{α_{2}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(19^{α_{5}+1} − 1).  (2)

By Lemma 5 we have X ≤ 33. Noting that

ord_{59}(2) = ord_{59}(11) = ord_{59}(13) = 58, ord_{59}(3) = ord_{59}(19) = 29,

we have

59 ∤ 2^{α_{1}+1} − 1, 59 ∤ 11^{α_{3}+1} − 1, 59 ∤ 13^{α_{4}+1} − 1.

By (2) we have 59 | 3^{α_{2}+1} − 1 or 59 | 19^{α_{5}+1} − 1. If 59 | 3^{α_{2}+1} − 1, then 29 | α_{2} + 1. Since ord_{28537}(3) = 29, we have 28537 | 3^{α_{2}+1} − 1, which contradicts with (2), thus 59 ∤ 3^{α_{2}+1} − 1 . If 59 | 19^{α_{5}+1} − 1, then 29 | α_{5} + 1. Since ord_{233}(19) = 29, we have 233 | 19^{α_{5}+1} − 1, which contradicts with (2), thus 59 ∤ 19^{α_{5}+1} − 1.

Case 3. p = 23, n = 2^{α_{1}} · 3^{α_{2}} · 11^{α_{3}} · 13^{α_{4}} · 23^{α_{5}}. Then

2^{α_{1}+7} · 3^{α_{2}+3} · 5 · 11^{α_{3}+1} · 13^{α_{4}−1} · 23^{α_{5}−1} · 31
= (2^{α_{1}+1} − 1)(3^{α_{2}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(23^{α_{5}+1} − 1).  (3)

By Lemma 5 we have X ≤ 34. Noting that

ord_{31}(2) = 5, ord_{31}(3) = ord_{31}(11) = ord_{31}(13) = 30, ord_{31}(23) = 10,

by ord_{31}(3) = ord_{31}(11) = ord_{31}(13) = 30, we know that 30 | α_{i} + 1, i = 2, 3, 4. Noting that ord_{61}(3) = 10, ord_{19}(11) = ord_{61}(13) = 3, we have 61 | 3^{α_{2}+1}−1, 19 | 11^{α_{3}+1}−1, 61 | 13^{α_{4}+1}−1. Which contradicts with (3), then we have

31 ∤ 3^{α_{2}+1} − 1, 31 ∤ 11^{α_{3}+1} − 1, 31 ∤ 13^{α_{4}+1} − 1.

By (3) we know that 31 | 23^{α_{5}+1} − 1 or 31 | 2^{α_{1}+1} − 1.

If 31 | 23^{α_{5}+1} − 1, then by ord_{31}(23) = 10, we know that 10 | α_{5} + 1. Noting that ord_{41}(23) = 5 we have 41 | 23^{α_{5}+1} − 1, which contradicts (3).

If 31 | 2^{α_{1}+1} − 1, then α_{1} + 1 = 5k, k ∈ ℤ. By X ≤ 34, we have k = 1, 2, 3, 4, 5, 6.

Subcase 1: k = 1, α_{1}+1 = 5, n = 2^{4}·3^{α_{2}} ·11^{α_{3}} ·13^{α_{4}} ·23^{α_{5}}. Put m = 3^{α_{2}} ·11^{α_{3}} ·13^{α_{4}} ·23^{α_{5}}. Then

\frac{σ(m)}{m} < \frac{m}{φ(m)} = \frac{3}{2} · \frac{11}{10} · \frac{13}{12} · \frac{23}{22} = \frac{299}{160} = 1.86875.

On the other hand, noting that φ(n) + σ(n) = 4n, then 8φ(m) + 31σ(m) = 64m, thus

\frac{σ(m)}{m} > 1.9264 > 1.86875,

a contradiction.

Subcase 2: k = 2, α_{1} + 1 = 10. Thus

\begin{matrix}2^{16} · 3^{α_{2}+2} · 5 · 11^{α_{3}} · 13^{α_{4}−1} · 23^{α_{5}−1} \\ = (3^{α_{2}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(23^{α_{5}+1} − 1). \end{matrix}

Noting that the following facts:

(i) If 2^{α} | 3^{α_{2}+1} − 1, then α ≤ 4. In fact, if α ≥ 5, then by ord_{32}(3) = 8, we have α_{2} + 1 = 8s, s ∈ ℤ. Noting that ord_{41}(3) = 8, thus 41 | 3^{α_{2}+1} − 1, this is impossible.

(ii) If 2^{α} | 11^{α_{3}+1} − 1, then α ≤ 3. In fact, if α ≥ 4, then by ord_{16}(11) = 4, we have α_{3} + 1 = 4s, s ∈ ℤ. Noting that ord_{61}(11) = 4, thus 61 | 11^{α_{3}+1} − 1, this is impossible.

(iii) If 2^{α} | 13^{α_{4}+1} − 1, then α ≤ 3. In fact, if α ≥ 4, then by ord_{16}(13) = 4, thus α_{4} + 1 = 4s, s ∈ ℤ. Noting that ord_{7}(13) = 2, thus 7 | 13^{α_{4}+1} − 1, this is impossible.

(iv) If 2^{α} | 23^{α_{5}+1} − 1, then α ≤ 4. In fact, if α ≥ 5, then by ord_{32}(23) = 4, we have α_{5} + 1 = 4s, s ∈ ℤ. Noting that ord_{53}(23) = 4, thus 53 | 23^{α_{5}+1} − 1, this is impossible.

Let

\begin{aligned}A & = (3^{α_{2}+1} − 1)(11^{α_{3}+1} − 1)(13^{α_{4}+1} − 1)(23^{α_{5}+1} − 1), \\ B & = 2^{16} · 3^{α_{2}+2} · 5 · 11^{α_{3}} · 13^{α_{4}−1} · 23^{α_{5}−1}. \end{aligned}

We have V_{2}(A) ≤ 14 and V_{2}(B) = 16, this is impossible.

Subcase 3: k = 3, α_{1} + 1 = 15. By ord_{7}(2) = 3, we have 7 | 2^{α_{1}+1} − 1, which contradicts (3).

Subcase 4: k = 4, α_{1} + 1 = 20. By ord_{41}(2) = 20, we have 41 | 2^{α_{1}+1} − 1, which contradicts (3).

Subcase 5: k = 5, α_{1} + 1 = 25. By ord_{601}(2) = 25, we have 601 | 2^{α_{1}+1} − 1, which contradicts (3).

Subcase 6: k = 6, α_{1} + 1 = 30. By ord_{151}(2) = 15, we have 151 | 2^{α_{1}+1} − 1, which contradicts (3).

This completes the proof of Theorem 1.

References

[1] K. Harris, On the classification of integer n that divide φ(n) + σ(n), J. Number Theory 129 (2009), 2093–2110.

[2] Da-Zheng Lin and Ming-Zhi Zhang, On the divisibility relation n | φ(n) + σ(n), J. Sichuan Daxue Xuebao 34 (1997), 121–123.

[3] F. Luca and J. Sándor, On a problem of Nicol and Zhang, J. Number Theory 128 (2008), 1044–1059.

[4] C. A. Nicol, Some Diophantine equations involving arithmetic functions, J. Math. Anal. Appl. 15 (1966), 154–161.

[5] Wei Wang, The research on Nicol problems, Master Degree Theses of Nanjing Normal University, 2008.

[6] Ming-Zhi Zhang, A divisibility problem, J. Sichuan Daxue Xuebao 32 (1995), 240–242.

2010 Mathematics Subject Classification: Primary 11A25. Keywords: Nicol number; Euler’s totient function.

Received April 12 2011; revised version received July 7 2011. Published in Journal of Integer Sequences, September 4 2011. Revised, April 11 2012.

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