A Note On Perfect Totient Numbers

Moujie Deng^{1} Department of Applied Mathematics College of Information Science and Technology Hainan University Haikou 570228 P. R. China

                      dmj2002@hotmail.com

Abstract

In this note we prove that there are no perfect totient numbers of the form 3^{k}p, k ≥ 4, where s = 2^{a}3^{b} + 1, r = 2^{c}3^{d}s + 1, q = 2^{e}3^{f}r + 1, and p = 2^{g}3^{h}q + 1 are primes with a, c, e, g ≥ 1, and b, d, f, h ≥ 0.

1 Introduction

Let ϕ denote Euler’s totient function. Define ϕ^{1}(n) = ϕ(n) and ϕ^{k}(n) = ϕ(ϕ^{k−1}(n)) for all integers n > 2 , k ≥ 2. Let c be the smallest positive integer such that ϕ^{c}(n) = 1. Define the arithmetic function S by

S(n) = ∑_{k=1}^{c} ϕ^{k}(n).

We say that n is a perfect totient number (or PTN for short) if S(n) = n.

There are infinitely many PTNs, since it is easy to show that 3^{k} is a PTN for all positive integers k. Perez Cacho [6] proved that 3p, for an odd prime p, is a PTN if and only if p = 4n + 1, where n is a PTN. Mohan and Suryanarayana [5] proved that 3p, for an odd prime p, is not a PTN if p ≡ 3 (mod 4). Thus PTNs of the form 3p have been completely characterized. D. E. Iannucci, the author and G. L. Cohen [3] investigated PTNs of the form 3^{k}p in the following cases:

1. k ≥ 2, p = 2^{c}3^{d}q + 1 and q = 2^{a}3^{b} + 1 are primes with a,c ≥ 1 and b, d ≥ 0; 2. k ≥ 2, p = 2^{e}3^{f}q + 1, q = 2^{c}3^{d}r + 1 and r = 2^{a}3^{b} + 1 are all primes with a, c, e ≥ 1

1Supported by the Natural Science Foundation of Hainan province, Grant No. 808101.

and b, d, f ≥ 0;

3. k ≥ 3, p = 2^{g}3^{h}q + 1, q = 2^{e}3^{f}r + 1, r = 2^{c}3^{d}s + 1 and s = 2^{a}3^{b} + 1, are all primes with a, c, e, g ≥ 1, b, d, f, h ≥ 0.

In the first case, they determined all PTNs for k = 2, 3 and proved that there are no PTNs of the form 3^{k}p for k ≥ 4 by solving the related Diophantine equations. In the remaining cases, they only found several PTNs by computer searches. The author ([1, 2]) gave all solutions to the Diophantine equations 2^{x} − 2^{y}3^{z} − 2 · 3^{u} = 9^{k} + 1, and 2^{x} − 2^{y}3^{z} − 4 · 3^{w} = 3 · 9^{k} + 1, which shows that there are no PTNs of the form 3^{k}p for k ≥ 4 in the second case mentioned above.

In general, letM be the set of all perfect totients, I. E. Shparlinski [7] has shown that M is of asymptotic density zero, and F. Luca [4] showed that ∑_{m∈M} \frac{1}{m} converges.

The purpose of this note is to prove that, in the third case mentioned above, there are no PTNs of the form 3^{k}p for k ≥ 4.

2 Lemmas

We first deduce related Diophantine equations. Let k ≥ 3, n = 3^{k}p. Suppose all of s = 2^{a}3^{b} + 1, r = 2^{c}3^{d}s + 1, q = 2^{e}3^{f}r + 1, and p = 2^{g}3^{h}q + 1 are prime with a, c, e, g ≥ 1, b, d, f, h ≥ 0. If n is a PTN , then S(n) = n by definition, which implies the diophantine equation

2^{g}(2^{e}(2^{c}(2^{a} − 3^{d+f+h+k−3}) − 3^{f+h+k−2}) − 3^{h+k−1}) = 3^{k} + 1.  (1)

Apparently, g = 1 or 2 for k even or odd, respectively. Next, according to k = 2k_{1} or k = 2k_{1} + 1, we consider more general Diophantine equations

2^{x} − 2^{y}3^{z} − 2^{u}3^{v} − 2 · 3^{w} = 9^{k_{1}} + 1,  (2)

with x ≥ 4, y, u, w > 0, z, v ≥ 0, k_{1} ≥ 2, and

2^{x} − 2^{y}3^{z} − 2^{u}3^{v} − 4 · 3^{w} = 3 · 9^{k_{1}} + 1,  (3)

with x ≥ 4, y, u, w > 0, z, v ≥ 0, k_{1} ≥ 1, respectively. Since the terms 2^{y}3^{z} and 2^{u}3^{v} have symmetry in (2) and (3), we need only determine the solutions (x, y, z, u, v, w, k) to (2) and (3) such that y > u or y = u, z ≥ v.

Let (x, y, z, u, v, w, k_{1}) be any solution to the equation (2) (or (3)), and let

(x, y, z, u, v, w, k_{1}) ≡ (α, β, γ, δ, λ, μ, ν,) (mod 36, 36, 36, 36, 36, 36, 18)

denote x ≡ α (mod 36), y ≡ β (mod 36), z ≡ γ (mod 36), u ≡ δ (mod 36), v ≡ λ (mod 36), w ≡ μ (mod 36), and k_{1} ≡ ν (mod 18). In solving equation (2) and equation (3), we first determine all the α, β, γ, δ, λ, μ, ν.

Lemma 1. Let (x, y, z, u, v, w, k_{1}) be any solution to the equation (2), and let

(x, y, z, u, v, w, k_{1}) ≡ (α, β, γ, δ, λ, μ, ν) (mod 36, 36, 36, 36, 36, 36, 18).

Then all the possible α, β, γ, δ, λ, μ, ν with 36 ≥ α, β, δ, λ ≥ 1, 35 ≥ γ, λ ≥ 0, 19 ≥ ν ≥ 2, β > δ or β = δ and γ ≥ λ are listed in Table 1 and Table 1′.

Proof: Since

2^{36} ≡ 1 (mod 5 · 7 · 13 · 19 · 37 · 73), 3^{36} ≡ 1 (mod 5 · 7 · 13 · 19 · 37 · 73),

α, β, γ, δ, λ, μ, ν must satisfy

2^{α} − 2^{β}3^{γ} − 2^{δ}3^{λ} − 2 · 3^{μ} ≡ 9^{ν} + 1 (mod 5 · 7 · 13 · 19 · 37 · 73).  (4)

But note that 2^{x} ≡ 0 (mod 2^{4}), 9^{k_{1}} ≡ 0 (mod 3^{3}), 2^{36} ≡ 1 (mod 3^{3}), 3^{36} ≡ 1 (mod 2^{4}); M = 36l + m implies 2^{M} ≡ 0 or 2^{m} (mod 2^{4}) and 3^{M} ≡ 0 or 3^{m} (mod 3^{3}). Hence α, β, γ, δ, λ, μ, ν must satisfy one of the 4 congruences

−2^{β} · B · 3^{γ} − 2^{δ} · D · 3^{λ} − 2 · 3^{μ} ≡ 9^{ν} + 1 (mod 2^{4})  (5)

and one of the 8 congruences

2^{α} − 2^{β}3^{γ} · C − 2^{δ}3^{λ} · E − 2 · 3^{μ} · F ≡ 1 (mod 3^{3}),  (6)

where B, C, D, E, F take value 0, 1 independently. The congruences (4), (5) and (6) were tested on a computer with a program written in UBASIC. All the (α, β, γ, δ, λ, μ, ν) that satisfy (4), (5) and (6) are divided into two parts: those listed in Table 1 are in fact solutions to equation (2), and the remainder, listed in Table 1′, are not. □

Similarly, we have

Lemma 2. Let (x, y, z, u, v, w, k_{1}) be any solution to the equation (3), and let

(x, y, z, u, v, w, k) ≡ (α, β, γ, δ, λ, μ, ν) (mod 36, 36, 36, 36, 36, 36, 18).

Then all the possible α, β, γ, δ, λ, μ, ν with 36 ≥ α, β, δ, μ ≥ 1, 35 ≥ γ, λ ≥ 0, 18 ≥ ν ≥ 1, β > δ or β = δ and γ ≥ λ are listed in Table 2 and Table 2′.

Lemma 3. Let (α, β, γ, δ, λ, μ, ν) be any solution to equation (2) or (3) that is listed in Table 1 or Table 2, and suppose

1. α > β > δ ; or

2. α > β + 2, β = δ ;

holds. Then there is no other solution (x, y, γ, u, λ, μ, ν) to equation (2) or (3) that satisfies (x, y, u) ≡ (α, β, δ) (mod 36, 36, 36)

Proof: Let x = α + 36i, y = β + 36j, u = δ + 36l. We have

2^{α}(2^{36i} − 1) = 2^{β}3^{γ}(2^{36j} − 1) + 2^{δ}3^{λ}(2^{36l} − 1).  (7)

In case 1, consideration of (7), modulo 2^{β} and 2^{α} in turn gives l = 0 and j = 0. Hence we have i = 0. In case 2, since 3^{γ} + 3^{λ} ≡ 2, 4 (mod 8), consideration of (7), modulo 2^{α}, gives j = l = 0, and therefore i = 0.

3 Main Results

Theorem 1. All the solutions to equation (2) are given by (x, y, z, u, v, w, k_{1}) = (α, β, γ, δ, λ, μ, ν) with α, β, γ, δ, λ, μ, ν listed in Table 1.

Proof: Let (x, y, z, u, v, w, k_{1}) be any solution to equation (2), and let

(x, y, z, u, v, w, k_{1}) ≡ (α, β, γ, δ, λ, μ, ν) (mod 36, 36, 36, 36, 36, 36, 18).

By Lemma 1, all of α, β, γ, δ, λ, μ, ν are listed in Table 1 or Table 1′. Put x = α + 36i, y = β + 36j, z = γ + 36l, u = δ + 36m, v = λ + 36n, w = μ + 36t, k_{1} = ν + 18t_{1}. Then we must have

2^{α+36i}−2^{β+36j}·3^{γ+36l}−2^{δ+36m}·3^{λ+36n}−2·3^{μ+36t} ≡ 9^{ν+36t_{1}}+1 (mod 11·31·181·331·631). (8)

For α, β, γ, δ, λ, μ, ν appearing in Table 1, since 2^{180} ≡ 3^{360} ≡ 1 (mod 11·31·181·331·631), we first test (8) within

4 ≥ i ≥ 0, 4 ≥ j ≥ 0, 9 ≥ l ≥ 0, 4 ≥ m ≥ 0, 9 ≥ n ≥ 0, 9 ≥ t ≥ 0, 9 ≥ t_{1} ≥ 0.

With computer assistance, it follows that l = n = t = t_{1} = 0 in this case. Since any α, β, γ, δ, λ, μ, ν that listed in Table 1 satisfy the conditions of Lemma 3, we must have i = j = m = 0 by Lemma 3.

For α, β, γ, δ, λ, μ, ν appearing in Table 1′, since (α, β, γ, δ, λ, μ, ν) is not a solution to equation(2), we have i ≥ 1. The congruence (8) was then tested on a computer within the ranges

5 ≥ i ≥ 1, 4 ≥ j ≥ 0, 9 ≥ l ≥ 0, 4 ≥ m ≥ 0, 9 ≥ n ≥ 0, 9 ≥ t ≥ 0, 9 ≥ t_{1} ≥ 0

with no (i, j, l, m, n, t, t_{1}) being found, which shows that (x, y, z, u, v, w, k_{1}) cannot be a solution to equation (2). □

Theorem 2. All the solutions to equation (3) are given by (x, y, z, u, v, w, k_{1}) = (α, β, γ, δ, λ, μ, ν) with α, β, γ, δ, λ, μ, ν listed in Table 2 .

Proof: The proof is basically the same as that for theorem 1, with the only difference being that (α, β, γ, δ, λ, μ, ν) = (9, 7, 0, 7, 0, 1, 2), listed in Table 2, does not satisfy the conditions of Lemma 3. Suppose that x = 9 + 36i, y = 7 + 36j, z = 36l, u = 7 + 36m, v = 36n, w = 1 + 36t, k_{1} = 2+36t_{1} is a solution to equation (3). Then a computer test of the related congruence within 4 ≥ i ≥ 0, 4 ≥ j ≥ 0, 9 ≥ l ≥ 0, 4 ≥ m ≥ 0, 9 ≥ n ≥ 0, 9 ≥ t ≥ 0, 9 ≥ t_{1} ≥ 0 gives l = n = t = t_{1} = 0. Consideration of (7) with α, β, γ, δ, λ replaced by 9, 7, 0, 7, 0 , modulo 2^{7}, gives j = l = 0. Therefor i = 0.

Theorem 3. There are no PTNs of the form 3^{k}p, k ≥ 4, where all of s = 2^{a}3^{b} + 1, r = 2^{c}3^{d}s + 1, q = 2^{e}3^{f}r + 1, and p = 2^{g}3^{h}q + 1 are prime with a, c, e, g ≥ 1, b, d, f, h ≥ 0.

Proof: Suppose (a, c, d, e, f, g, h, k) is a solution to equation (1). Let x = a + c + e + g, y = c+e+g, z = d+f +k−3, u = e+g, v = f +h+k−2, w = h+k−1, and k_{1} = \frac{k}{2} or k_{1} = \frac{k−1}{2} for k even or odd, respectively. Then (x, y, z, u, v, w, k_{1}) must be a solution to equation (2) or equation (3). From the first two theorems it follows that the only solutions to equation(1) are (a, c, d, e, f, h, k) = (4, 1, 0, 1, 2, 1, 3),(1, 2, 0, 4, 0, 0, 3),(3, 1, 0, 4, 1, 0, 3),(2, 2, 1, 4, 0, 0, 3), (8, 1, 4, 1, 0, 1, 3),(5, 1, 2, 4, 1, 0, 3),(4, 2, 0, 4, 2, 0, 3). □

α β γ δ λ μ ν α β γ δ λ μ ν α β γ δ λ μ ν
7 2 2 2 0 1 2 10 5 3 3 1 3 2 12 8 2 6 2 5 3
7 3 1 2 0 2 2 10 5 3 3 2 1 2 12 10 1 5 2 1 3
7 4 0 2 1 2 2 10 6 1 4 1 3 3 13 6 0 2 3 6 4
7 4 0 3 1 1 2 10 6 1 5 1 1 3 13 8 3 6 0 5 3
7 5 0 3 0 1 2 10 7 0 2 0 4 3 13 10 1 4 0 7 3
8 1 1 1 1 4 2 10 7 1 3 2 5 2 14 6 5 4 1 3 3
8 1 4 1 1 1 2 10 8 0 5 0 1 3 14 6 5 5 1 1 3
8 2 3 2 1 3 2 10 8 1 2 1 4 2 14 10 1 2 3 8 2
8 3 0 2 0 4 2 11 3 5 2 0 2 2 15 10 3 4 0 7 3
8 4 1 2 3 2 2 11 4 0 3 5 1 2 16 3 6 3 4 1 5
8 4 1 3 2 3 2 11 4 4 2 0 2 3 16 5 4 4 5 1 5
8 4 2 2 1 2 2 11 4 4 4 0 1 3 16 6 4 4 4 1 5
8 4 2 3 1 1 2 11 7 2 2 0 4 3 16 7 1 6 3 7 5
8 5 1 3 1 3 2 11 8 0 6 2 5 3 16 8 5 5 4 1 3
8 5 1 3 2 1 2 11 8 1 6 0 5 3 16 9 1 6 2 7 5
9 8 0 2 1 4 2 11 10 0 5 2 1 3 16 9 4 3 7 1 4
10 3 1 2 3 4 3 12 4 3 2 6 2 3 16 9 4 5 6 1 3
10 3 3 3 1 3 3 12 4 5 2 3 2 2 16 11 0 6 0 7 5
10 3 3 3 2 1 3 12 4 5 3 2 3 2 16 11 1 2 4 2 5
10 4 2 4 2 1 3 12 5 2 2 6 4 3 16 11 1 5 2 3 5
10 4 3 3 1 5 2 12 5 4 5 2 5 3 16 13 1 5 3 9 3
10 5 1 2 2 4 3 12 7 2 6 3 5 3 18 10 5 2 3 8 2
10 5 1 4 2 3 3 12 7 3 3 2 5 2 18 13 3 5 3 9 3
10 5 2 3 4 1 2 12 8 1 5 4 1 3

Table 1

α β γ δ λ μ ν α β γ δ λ μ ν α β γ δ λ μ ν
5 1 0 1 0 2 19 6 3 1 2 1 2 19 8 3 14 2 1 16 13
5 2 1 2 0 1 19 6 3 1 2 2 36 18 8 3 15 2 1 14 13
5 3 0 2 0 2 18 6 3 12 2 2 14 12 8 6 1 2 2 2 19
5 3 1 2 0 36 18 6 3 13 2 2 12 12 10 3 1 2 5 2 19
5 3 12 2 0 14 12 6 4 0 2 2 36 19 10 9 0 3 1 5 18
5 4 0 20 0 36 19 6 5 0 2 0 2 19 12 7 2 2 6 2 19
5 3 13 2 0 12 12 7 5 1 2 0 2 19 18 34 22 21 16 33 11
6 2 2 2 1 1 19 8 3 2 2 1 4 19 30 23 2 1 13 1 15
6 3 0 2 2 2 18 8 3 3 2 1 2 19

Table 1´

\begin{matrix}α β γ δ λ μ ν & α β γ δ λ μ ν & α β γ δ λ μ ν \\ 6 1 2 1 1 1 1 & 10 6 2 5 1 4 1 & 12 7 3 7 1 1 2 \\ 6 2 1 2 1 1 1 & 10 6 2 6 1 1 2 & 12 8 1 7 1 6 1 \\ 6 4 0 3 0 1 1 & 10 7 1 3 2 4 2 & 12 8 2 6 2 5 2 \\ 7 4 0 3 2 1 1 & 10 7 1 5 2 3 2 & 12 8 2 6 3 2 1 \\ 7 4 1 4 0 2 1 & 10 7 1 5 2 4 1 & 12 9 1 4 1 4 3 \\ 7 5 0 5 0 2 1 & 10 7 1 6 2 2 1 & 12 9 1 8 2 1 2 \\ 7 6 0 3 1 1 1 & 10 7 1 7 1 1 2 & 12 10 0 7 0 6 1 \\ 8 1 4 1 3 1 1 & 10 8 1 3 3 1 1 & 12 10 1 3 1 5 1 \\ 8 2 3 2 1 3 1 & 10 8 1 6 1 2 1 & 12 10 1 8 1 1 2 \\ 8 2 3 2 3 1 1 & 10 9 0 8 0 1 2 & 13 6 0 3 6 3 3 \\ 8 4 1 3 2 3 1 & 11 6 0 3 5 1 1 & 13 6 0 5 5 3 2 \\ 8 4 2 3 2 1 1 & 11 6 3 6 0 1 2 & 13 6 0 5 5 4 1 \\ 8 4 2 4 1 2 1 & 11 8 0 6 2 5 2 & 13 6 4 6 0 6 1 \\ 8 5 1 3 1 3 1 & 11 8 0 6 3 2 1 & 13 8 3 6 0 5 2 \\ 8 5 1 5 1 2 1 & 11 8 1 6 0 5 2 & 13 10 0 8 3 1 2 \\ 8 6 1 3 1 1 1 & 11 9 1 8 0 1 2 & 13 10 1 4 0 6 3 \\ 8 7 0 6 0 2 1 & 11 10 0 3 1 5 1 & 13 12 0 7 2 6 1 \\ 9 4 0 3 3 2 2 & 11 10 0 8 1 1 2 & 14 6 5 6 2 1 2 \\ 9 4 2 4 0 3 2 & 12 4 5 3 2 3 1 & 14 8 1 6 5 2 1 \\ 9 4 2 4 0 4 1 & 12 4 5 4 2 2 1 & 14 9 3 4 1 4 3 \\ 9 4 3 4 0 2 1 & 12 5 2 3 4 5 3 & 14 9 3 8 2 1 2 \\ 9 6 0 5 1 3 2 & 12 5 2 3 4 6 2 & 14 10 1 7 4 6 1 \\ 9 6 0 5 1 4 1 & 12 5 2 4 4 4 3 & 14 10 2 8 3 1 2 \\ 9 6 1 6 0 1 2 & 12 5 3 3 2 5 3 & 14 12 1 7 2 6 1 \\ 9 7 0 5 0 3 2 & 12 5 3 3 2 6 2 & 15 10 3 4 0 6 3 \\ 9 7 0 5 0 4 1 & 12 5 3 5 2 6 1 & 16 8 5 7 1 6 1 \\ 9 7 0 7 0 1 2 & 12 5 4 5 2 5 2 & 16 10 1 8 5 1 2 \\ 9 7 1 6 0 2 1 & 12 6 2 4 4 2 3 & 18 4 3 3 6 9 5 \\ 9 8 0 3 3 1 1 & 12 6 2 6 2 6 1 & 18 4 3 3 6 10 4 \\ 9 8 0 6 1 2 1 & 12 6 3 3 2 3 3 & 18 5 6 4 3 10 3 \\ 10 1 2 1 1 5 1 & 12 6 3 4 2 2 3 & 18 5 6 5 4 10 1 \\ 10 2 1 2 1 5 1 & 12 7 2 3 4 3 3 & 18 7 4 6 5 10 1 \\ 10 2 5 2 1 1 1 & 12 7 2 3 5 5 1 & 18 8 4 6 4 10 1 \\ 10 3 4 3 1 3 2 & 12 7 2 4 3 4 3 & 18 10 3 7 5 8 5 \\ 10 3 4 3 1 4 1 & 12 7 2 5 4 3 2 & 18 10 4 3 5 3 5 \\ 10 4 0 3 0 5 1 & 12 7 2 5 4 4 1 & 18 10 4 6 3 4 5 \\ 10 4 3 3 1 4 2 & 12 7 2 6 3 5 2 & 18 10 5 7 4 6 1 \\ 10 5 1 3 4 2 2 & 12 7 3 3 2 4 2 & 18 11 3 7 3 8 5 \\ 10 5 3 3 1 3 1 & 12 7 3 5 2 3 2 & 18 11 4 3 7 9 1 \\ 10 5 3 5 1 2 1 & 12 7 3 5 2 4 1 & 18 11 4 5 7 8 1 \\ 10 6 2 5 1 3 2 & 12 7 3 6 2 2 1 & 18 11 4 7 6 6 1 \end{matrix} Table 2

6

\begin{matrix}α β γ δ λ μ ν & α β γ δ λ μ ν & α β γ δ λ μ ν \\ 6 1 2 1 1 2 18 & 7 6 0 3 1 2 18 & 18 18 18 2 9 1 14 \\ 6 2 1 2 1 2 18 & 8 3 3 3 1 1 18 & 18 18 20 1 14 31 18 \\ 6 2 2 2 1 1 18 & 8 5 1 4 2 1 18 & 18 20 33 14 20 36 1 \\ 6 3 1 3 1 1 18 & 8 6 1 3 1 2 18 & 18 23 12 17 3 32 1 \\ 6 3 35 3 1 2 17 & 8 6 1 4 1 1 18 & 18 27 30 3 34 9 7 \\ 6 4 0 3 0 2 18 & 8 9 35 3 2 1 17 & 18 28 18 7 20 3 2 \\ 6 5 35 4 0 2 17 & 10 3 1 3 1 5 18 & 18 28 18 7 20 4 1 \\ 6 6 35 3 35 2 18 & 12 7 2 3 1 6 18 & 18 29 2 3 29 31 6 \\ 6 6 35 4 35 2 17 & 18 11 0 10 4 36 5 & 18 29 5 23 21 34 3 \\ 6 9 33 3 0 2 16 & 18 18 18 2 1 9 14 & 18 33 35 19 24 12 15 \end{matrix}

Table 2´

4 Acknowledgments

The author thanks Professor G. L. Cohen and Professor Huishi Li for useful advice and the referee for helpful suggestions which improved the quality of this paper.

References

[1] Moujie Deng, On the Diophantine equation 2^{x} − 2^{y}3^{z} − 2 · 3^{u} = 9^{k} + 1, (in Chinese), Journal of Natural Science of Heilongjiang University, 23 (2006), 87–91.

[2] Moujie Deng, On the Diophantine equation 2^{x} − 2^{y}3^{z} − 4 · 3^{w} = 3 · 9^{k} + 1,(in Chinese), Journal of Inner Mongolia Normal University, 37 (2008), 45–49.

[3] D. E. Iannucci, D. Moujie and G. L. Cohen, On perfect totient numbers, J. Integer Sequences 6 (2003), Article 03.4.5.

[4] F. Luca, On the distribution of perfect totients, J. Integer Sequences 9 (2006), Article 06.4.4.

[5] A. L. Mohan and D. Suryanarayana, “Perfect totient numbers”, in: Number Theory (Proc. Third Matscience Conf., Mysore, 1981) Lect. Notes in Math. 938, Springer-Verlag, New York, 1982, 101–105.

[6] L. Perez Cacho, Sobre la suma de indicadores de ordenes sucesivos, Revista Matematica Hispano-Americana 5.3 (1939), 45–50.

[7] I. E. Shparlinski, On the sum of iterations of the Euler function, J. Integer Sequences 9 (2006), Article 06.1.6.

2000 Mathematics Subject Classification: Primary 11A25.

Keywords: totient, perfect totient number, diophantine equation. (Concerned with sequence \underline{A082897}.) Received April 14 2009; revised version received July 18 2009. Published in Journal of Integer Sequences, August 30 2009. Return to Journal of Integer Sequences home page.