Egyptian Fractions and Prime Power Divisors

John Machacek Department of Mathematics Michigan State University 619 Red Cedar Road C212 Wells Hall East Lansing, MI 48824 USA

                     machace5@math.msu.edu

Abstract From varying Egyptian fraction equations, we obtain generalizations of primary pseudoperfect numbers and Giuga numbers, which we call prime power pseudoperfect numbers and prime power Giuga numbers, respectively. We show that a sequence of Murthy in the On-line Encyclopedia of Integer Sequences is a subsequence of the sequence of prime power pseudoperfect numbers. We also provide prime factorization conditions sufficient to imply that a number is a prime power pseudoperfect number or a prime power Giuga number. The conditions on prime factorizations naturally give rise to a generalization of Fermat primes, which we call extended Fermat primes.

1 Introduction

We define and study two new types of integers, which we call prime power pseudoperfect numbers and prime power Giuga numbers. Each satisfies an Egyptian fraction equation that is a variation of a previously studied Egyptian fraction equation. Throughout we reference relevant sequences from the On-line Encyclopedia of Integer Sequences (OEIS) [7]. In Section 1 we review pseudoperfect numbers, primary pseudoperfect numbers, and Giuga numbers. In Section 2 we define prime power pseudoperfect numbers and show their relation to a sequence in the OEIS contributed by Murthy. Prime power Giuga numbers are defined in Section 3. We give some formulas that can produce more terms of our sequences in Section 4. In Section 5 we discuss some open problems and introduce extended Fermat primes. We have contributed each of our new sequences to the OEIS. The sequences of prime power pseudoperfect numbers, prime power Giuga numbers, and extended Fermat primes are \underline{A283423}, \underline{A286497}, and \underline{A286499}, respectively.

A pseudoperfect number is a positive integer n such that there exist integers 0 < d_{1} < ⋯ < d_{k} < n, where d_{i} | n for each i and n = d_{1} + ⋯ + d_{k}. For example, the number 20 is a pseudoperfect number since 20 = 1+4+5+10. Pseudoperfect numbers were first considered by Sierpiński [6] and form sequence \underline{A005835}. Open problems on pseudoperfect numbers can be found in the book [5, B2].

A primary pseudoperfect number is a positive integer n > 1 that satisfies the Egyptian fraction equation

∑_{p|n} \frac{1}{p} + \frac{1}{n} = 1,

where the sum is taken over all prime divisors on n. Primary pseudoperfect numbers were originally defined by Butske, Jaje, and Mayernik [3] and form sequence \underline{A054377}. When n > 1 is s primary pseudoperfect number, it follows that

∑_{p|n} \frac{n}{p} + 1 = n.

So, we see that, with the exception of 2, every primary pseudoperfect number is a pseudoperfect number.

A Giuga number is a positive composite integer n such that

∑_{p|n} \frac{1}{p} − \frac{1}{n} ∈ ℕ,

where the sum is taken over all prime divisors on n. Giuga numbers were defined by Borwein et al. [2], and form sequence \underline{A007850}. All known Giuga numbers satisfy the stronger Egyptian fraction equation

∑_{p|n} \frac{1}{p} − \frac{1}{n} = 1.

Giuga numbers are related to Giuga’s conjecture on primality [4]. Open problems relating to Giuga numbers can be found in the book [5, A17]

2 Prime power pseudoperfect numbers

A prime power pseudoperfect number is a positive integer n > 1 that satisfies the Egyptian fraction equation

∑_{p^{k}|n} \frac{1}{p^{k}} + \frac{1}{n} = 1,

where the sum is taken over all prime power divisors of n. As an example we can check that 20 is a prime power pseudoperfect number since

\frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{20} = 1.

Observe that, with the exception of powers of 2, all prime power pseudoperfect numbers are pseudoperfect. Also note that any primary pseudoperfect number is a prime power pseudoperfect number since primary pseudoperfect numbers must be squarefree. Prime power pseudoperfect numbers form OEIS sequence \underline{A283423}.

We now consider the sequences \underline{A073932} and \underline{A073935}, both of which were contributed to the OEIS by Murthy. We first define a function d on composite numbers by letting d(n) denote the largest nontrivial divisor of n. For example, d(15) = 5. Next we define a function f on positive integers greater than 1 by

f(n) := \begin{cases}n − 1, & if n is prime; \\ n − d(n), & otherwise. \end{cases}

As an example, we have f(15) = 10. Given any positive integer n > 1 we can iterate the function f until we reach 1. We let f^{(i)} to be the ith iterate of the function f. So f^{(i)} is the composition f ∘ f^{(i−1)} for i ≥ 1 and f^{(0)}(n) = n. In this way we obtain a triangle with nth row given by n, f(n), f^{(2)}(n), …, 1. The sequence \underline{A073932} is the sequence consisting of the entries of this triangle read by rows. For any positive integer n let D_{n} denote the set of divisors of n, and we define F_{n} := {n, f(n), f^{(2)}(n), …, 1}. The sequence consisting of all n such that the F_{n} = D_{n} is sequence \underline{A073935}.

If we consider n = 20 we obtain

f(20) = 10
f^{(2)}(20) = 5
f^{(3)}(20) = 4
f^{(4)}(20) = 2
f^{(5)}(20) = 1,

which are exactly the divisors of 20. We also notice that 20 is a prime power pseudoperfect number since

\frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{20} = 1.

We show in Theorem 3 that every number in the sequence \underline{A073935}, with the exception of 1, is a prime power pseudoperfect number. We first prove two lemmas.

Lemma 1. Let n > 1 be a positive integer with prime factorization n = p_{1}p_{2} ⋯ p_{ℓ}, where p_{1} ≤ p_{2} ≤ ⋯ ≤ p_{ℓ}. The function f is then given by

f(n) = (p_{1} − 1)p_{2} ⋯ p_{ℓ}.

Proof. Let n > 1 be an integer with prime factorization n = p_{1}p_{2} ⋯ p_{ℓ} where p_{1} ≤ p_{2} ≤ ⋯ ≤ p_{ℓ}. If n is prime, then ℓ = 1 and n = p_{1}. In this case f(n) = n − 1 = p_{1} − 1. When n is composite ℓ > 1, the largest nontrivial divisor is p_{2}p_{3} ⋯ p_{ℓ}. In this case

f(n) = n − d(n)
= p_{1}p_{2} ⋯ p_{ℓ} − p_{2}p_{3} ⋯ p_{ℓ}
= (p_{1} − 1)p_{2} ⋯ p_{ℓ}.

Hence the lemma holds.

Lemma 2. Let n > 1 be a positive integer with prime factorization

n = ∏_{i=1}^{ℓ} p_{i}^{a_{i}},

where p_{1} < p_{2} < ⋯ < p_{ℓ}. The integer n satisfies F_{n} = D_{n} if and only if

(p_{i} − 1) = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for 1 ≤ i ≤ ℓ.

Proof. Let n > 1 be an integer with prime factorization

n = ∏_{i=1}^{ℓ} p_{i}^{a_{i}},

where p_{1} < p_{2} < ⋯ < p_{ℓ}. The divisors of n are

D_{n} = {∏_{i=1}^{ℓ} p_{i}^{b_{i}} : 0 ≤ b_{i} ≤ a_{i}} .

We must show F_{n} = D_{n} if and only if

(p_{i} − 1) = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for 1 ≤ i ≤ ℓ.

First assume that

(p_{i} − 1) = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for 1 ≤ i ≤ ℓ. It follows that F_{n} = D_{n} as the divisors of n are obtained in lexicographic order of exponent vectors when we iterate f. To see this consider d ∈ D_{n} with

d = ∏_{i=1}^{ℓ} p_{i}^{b_{i}}

for some 0 ≤ b_{i} ≤ a_{i} and d ≠ 1. Let i′ be smallest index with b_{i′} > 0. Then

d = ∏_{i=i′}^{ℓ} p_{i}^{b_{i}}.

If i′ = 1, then

f(d) = p_{1}^{b_{1}−1} ∏_{i=2}^{ℓ} p_{i}^{b_{i}}.

Otherwise j > 1 and

f(d) = (∏_{i=1}^{i′−1} p_{i}^{a_{i}}) (p_{i′}^{b_{i′}−1}) ( ∏_{i=i′+1}^{ℓ} p_{i}^{b_{i}}) .

In either case we obtain the next divisor of n in lexicographic order of exponent vectors.

Next assume that F_{n} = D_{n}. We claim that p_{1} = 2. To see this, note that (p_{1} − 1) | f(n) by Lemma 1. Since F_{n} ⊆ D_{n} we have f(n) | n. Thus any prime divisor of p_{1} − 1 divides n, but a prime divisor of p_{1} − 1 is a prime strictly less than the smallest prime divisor of n, namely of p_{1}. Hence p_{1} − 1 can have no prime divisors, which implies that p_{1} = 2. Now assume that for j < i

(p_{j} − 1) = ∏_{j′=1}^{j−1} p_{j′}^{a_{j′}}.

By iterating f we initially obtain divisors of n of the form

(∏_{j=1}^{i−1} p_{j}^{b_{j}}) (∏_{j=i}^{ℓ} p_{j}^{a_{j}}) ,

where 0 ≤ b_{j} ≤ a_{j} for 1 ≤ j < i.

When we reach the divisor p_{i}^{a_{i}}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}, by Lemma 1 we get

f(p_{i}^{a_{i}}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}) = (p_{i} − 1)p_{i}^{a_{i}−1}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}.

Since F_{n} = D_{n}, (p_{i} − 1)p_{i}^{a_{i}−1}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}} divides n, which in turn implies that (p_{i} − 1) divides p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{i−1}^{a_{i−1}}. If

(p_{i} − 1) ≠ ∏_{j=1}^{i−1} p_{j}^{a_{j}},

we see (since f is a decreasing function) that

f(p_{i}^{a_{i}}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}) = (p_{i} − 1)p_{i}^{a_{i}−1}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}
< p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{i−1}^{a_{i−1}}p_{i}^{a_{i}−1}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}}.

Furthermore, the divisor p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{i−1}^{a_{i−1}}p_{i}^{a_{i}−1}p_{i+1}^{a_{i+1}} ⋯ p_{ℓ}^{a_{ℓ}} is not contained in F_{n}. Thus if F_{n} = D_{n} we must have

(p_{i} − 1) = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for all 1 ≤ i ≤ ℓ.

Algorithm 1 Nondeterministic algorithm to produce terms of sequence \underline{A073935}. n ← 2 loop

p ← largest prime divisor of n if n + 1 is prime then

nondeterministically choose n ← np or n ← n(n + 1) else

n ← np end if

end loop

Algorithm 1 is a nondeterministic algorithm that produces positive integers n with F_{n} = D_{n} (i.e., the terms of \underline{A073935}). Lemma 2 says that if n > 1 satisfies F_{n} = D_{n}, then F_{np} = D_{np}, where p is the largest prime divisor of n. If n + 1 is prime, then also F_{n(n+1)} = D_{n(n+1)}. Notice in the case that n + 1 is prime there are two ways to produce a new integer, and this causes the nondeterminism in Algorithm 1. Terms of the sequence coming from various branches of the algorithm are shown in Figure 1.

Theorem 3. Every integer n > 1 with F_{n} = D_{n} is a prime power pseudoperfect number.

Proof. Let n > 1 be such that F_{n} = D_{n}. Assume n has prime factorization

n = ∏_{i=1}^{ℓ} p_{i}^{a_{i}}.

By Lemma 2 we know that

(p_{i} − 1) = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for 1 ≤ i ≤ ℓ. We define

n_{i} := \frac{n}{p_{i}^{a_{i}}}
n′_{i} := \frac{n}{∏_{j=1}^{i} p_{j}^{a_{j}}}

for 1 ≤ i ≤ ℓ and define n′_{0} := n. We now compute

∑_{p^{k}|n} \frac{1}{p^{k}} + \frac{1}{n} = ∑_{i=1}^{ℓ} ∑_{j=1}^{a_{i}} \frac{1}{p_{i}^{j}} + \frac{1}{n}

= ∑_{i=1}^{ℓ} ∑_{j=0}^{a_{i}−1} \frac{p_{i}^{j}n_{i}}{n} + \frac{1}{n}

= ∑_{i=1}^{ℓ} \frac{(p_{i}^{a_{i}} − 1)n_{i}}{(p_{i} − 1)n} + \frac{1}{n}

= ∑_{i=1}^{ℓ} \frac{(p_{i}^{a_{i}} − 1)n′_{i}}{n} + \frac{1}{n}

= ∑_{i=1}^{ℓ} \frac{p_{i}^{a_{i}}n′_{i} − n′_{i}}{n} + \frac{1}{n}

= ∑_{i=1}^{ℓ} \frac{n′_{i−1} − n′_{i}}{n} + \frac{1}{n}

= \frac{n′_{0} − n′_{ℓ} + 1}{n}

= 1.

Therefore n is prime power pseudoperfect.

The converse of Theorem 3 is not true. For example, the number 23994 = 2 · 3^{2} · 31 · 41 is a prime power pseudoperfect number but F_{23994} ≠ D_{23994}.

2  6 42 1806 ^{77658 ⋮} 294 ^{2058 ⋮} 18 342 ^{6496 ⋮} 54 ^{162 ⋮} 4 20  ^{10010100 ⋮ 500 ⋮} 8  ^{16^{^{272 ⋮}^{32⋮}}}

Figure 1: Tree showing terms of sequence \underline{A073935} on various branches of Algorithm 1.

3 Prime power Giuga numbers

A prime power Giuga number is a positive composite integer n > 1 that satisfies the Egyptian fraction condition

∑_{p^{k}|n} \frac{1}{p^{k}} − \frac{1}{n} ∈ ℕ,

where the sum is taken over all prime power divisors of n. Since Giuga numbers are squarefree, it follows that all Giuga numbers are prime power Giuga numbers. All prime power Giuga numbers we have found obey the stricter Egyptian fraction equation

∑_{p^{k}|n} \frac{1}{p^{k}} − \frac{1}{n} = 1.

Prime power Giuga numbers form sequence \underline{A286497}.

We now prove a lemma analogous to Lemma 2.

Lemma 4. Let n > 1 be a positive integer with prime factorization

n = ∏_{i=1}^{ℓ} p_{i}^{a_{i}},

with p_{1} < p_{2} < ⋯ < p_{ℓ} and a_{ℓ} = 1. If

p_{i} − 1 = ∏_{j=1}^{i−1} p_{j}^{a_{j}}

for 1 ≤ i < ℓ and p_{ℓ} + 1 = \frac{n}{p_{ℓ}}, then the positive integer n is a prime power Giuga number.

Proof. Assume n > 1 is a positive integer satisfying the hypothesis of the lemma. Then by Lemma 2 and Theorem 3 we know the \frac{n}{p_{ℓ}} is a prime power pseudoperfect number. So,

∑_{p^{k}|n} \frac{1}{p^{k}} − \frac{1}{n} = ∑_{p^{k}| \frac{n}{p_{ℓ}}} \frac{1}{p^{k}} + \frac{1}{p_{ℓ}} − \frac{1}{n}

= \frac{\frac{n}{p_{ℓ}} − 1}{\frac{n}{p_{ℓ}}} + \frac{1}{p_{ℓ}} − \frac{1}{n}

= \frac{p_{ℓ}}{p_{ℓ} + 1} + \frac{1}{p_{ℓ}} − \frac{1}{(p_{ℓ} + 1)p_{ℓ}}

= 1.

Lemma 4 gives a sufficient but not necessary condition for being a prime power Giuga number. Table 1 shows prime power Giuga numbers less than 10^{7}. Notice some numbers in the table, such as 858 = 2 · 3 · 11 · 13, do not satisfy the condition in Lemma 4.

n prime factorization

12  2^{2} · 3
30  2 · 3 · 5
56  2^{3} · 7
306  2 · 3^{2} · 17
380  2^{2} · 5 · 19
858  2 · 3 · 11 · 13
992  2^{5} · 31
1722  2 · 3 · 7 · 41
2552  2^{3} · 11 · 29
2862  2 · 3^{3} · 53
16256  2^{7} · 127
30704  2^{4} · 19 · 101
66198 2 · 3 · 11 · 17 · 59
73712  2^{4} · 17 · 271
86142  2 · 3 · 7^{2} · 293
249500  2^{2} · 5^{3} · 499
629802 2 · 3^{3} · 107 · 109
1703872  2^{6} · 79 · 337
6127552  2^{6} · 67 · 1429

Table 1: Prime power Giuga numbers less than 10^{7}.

4 Producing more terms

In this section we give some formulas that can be helpful in finding solutions to our Egyptian fraction equations. Similar results for primary pseudoperfect numbers and Giuga numbers are given in the article [9, Theorem 8]. The article [8, Proposition 1] also contains conditions for primary pseudoperfect numbers. Results to help search for solutions of other related Egyptian fraction equations can be found in the articles [1, Proposition 12, Lemma 17] and [3, Lemma 4.1, Lemma 4.2].

Proposition 5. Let n > 1 be a positive integer.

(i) If n is in the sequence \underline{A073935} and p is largest prime divisor of n, then both \frac{n}{p} and np are in the sequence \underline{A073935}.

(ii) If n is in the sequence \underline{A073935} and n + 1 is prime, then n(n + 1)^{k} is in the sequence \underline{A073935} for any nonnegative integer k.

(iii) If n is a prime power pseudoperfect number and n + 1 is prime, then n(n + 1)^{k} is a prime power pseudoperfect number for any nonnegative integer k.

(iv) If n is a prime power pseudoperfect number and n − 1 is prime, then n(n − 1) is a prime power Giuga number.

Proof. Parts (i) and (ii) follow immediately from Lemma 2.

For part (iii) assume that n is a prime power pseudoperfect number and n + 1 is prime. So,

∑_{p^{k}|n(n+1)^{k}} \frac{1}{p^{k}} + \frac{1}{n(n + 1)^{k}} = ∑_{p^{k}|n} \frac{1}{p^{k}} + ∑_{j=1}^{k} \frac{1}{(n + 1)^{j}} + \frac{1}{n(n + 1)^{k}}

= \frac{n − 1}{n} + \frac{(n + 1)^{k} − 1}{n(n + 1)^{k}} + \frac{1}{n(n + 1)^{k}}

= \frac{(n − 1)(n + 1)^{k} + (n + 1)^{k}}{n(n + 1)^{k}}

= 1.

For part (iv) assume that n is a prime power pseudoperfect number and n − 1 is prime.

So,

∑_{p^{k}|n(n+1)} \frac{1}{p^{k}} − \frac{1}{n(n − 1)} = ∑_{p^{k}|n} \frac{1}{p^{k}} + \frac{1}{(n + 1)} − \frac{1}{n(n − 1)}

= \frac{n − 1}{n} + \frac{1}{(n + 1)} − \frac{1}{n(n − 1)}

= \frac{(n − 1)(n − 1) + n − 1}{n(n − 1)}

= 1.

Consider the number

n = 23994 = 2 · 3^{2} · 31 · 43,

which is a prime power pseudoperfect number. However, neither

\frac{n}{43} = 558 = 2 · 3^{2} · 31

nor

43n = 1031742 = 2 · 3^{2} · 31 · 43^{2}

are prime power pseudoperfect numbers. Hence, a version of Proposition 5 (i) does not hold from prime power pseudoperfect numbers. Also consider the number n = 18, which is a prime power pseudoperfect number, and the number n(n − 1) = 306, which is a prime power Giuga number, since n − 1 = 17 is prime. However, the number n(n − 1)^{2} = 5202 is not a prime power Giuga number. Thus a version of Proposition 5 (ii) or (iii) does not hold for prime power Giuga numbers.

5 Open questions

Proposition 5 immediately shows that there are infinitely many terms in both the sequence \underline{A073935} and the sequence of prime power pseudoperfect numbers \underline{A283423}. Proposition 5 does not give a way to produce infinitely many prime power Giuga numbers, but we conjecture there are infinitely many such numbers.

Conjecture 6. There are infinitely many prime power Giuga numbers.

A Mersenne prime is a prime number p such that p = 2^{k} −1 for some integer k. Mersenne primes form sequence \underline{A000668}. By Lemma 4, the number n = 2^{k}(2^{k} − 1) is a prime power Giuga number whenever 2^{k} − 1 is a Mersenne prime. Hence, Conjecture 6 would follow from an infinitude of Mersenne primes, and it is believed that there are infinitely many Mersenne primes.

\begin{matrix}p & p − 1 & level \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ 5 & 2^{2} & 1 \\ 7 & 2 · 3 & 2 \\ 17 & 2^{4} & 1 \\ 19 & 2 · 3^{2} & 2 \\ 43 & 2 · 3 · 7 & 3 \\ 101 & 2^{2} · 5^{2} & 2 \\ 163 & 2 · 3^{4} & 2 \\ 257 & 2^{8} & 1 \\ 487 & 2 · 3^{5} & 2 \\ 1459 & 2 · 3^{6} & 2 \\ 14407 & 2 · 3 · 7^{4} & 3 \\ 26407 & 2 · 3^{4} · 163 & 3 \\ 39367 & 2 · 3^{9} & 2 \\ 62501 & 2^{2} · 5^{6} & 2 \\ 65537 & 2^{16} & 1 \\ 77659 & 2 · 3 · 7 · 43^{2} & 4 \\ 1020101 & 2^{2} · 5^{2} · 101^{2} & 3 \end{matrix}

Table 2: Table of extended Fermat primes p along with factorizations of p − 1.

A Fermat prime is prime number p such that p = 2^{k}+1 for some positive integer k. Fermat primes form sequence \underline{A019434}. By Lemma 2 the number 2^{k} is in the sequence \underline{A073935} for any positive integer k, and the powers of 2 are the only numbers in the sequence that have a unique prime divisor. If a number with two distinct prime divisors is in sequence \underline{A073935} it must be of the form 2^{k}(2^{k} + 1)^{j} where 2^{k} + 1 is a Fermat prime and j is a positive integer.

The primes that occur as divisors of terms of the sequence \underline{A073935} are primes p such that

p − 1 = ∏_{i=1}^{ℓ} p_{i}^{a_{i}}

where for 1 ≤ i ≤ ℓ

p_{i} − 1 = ∏_{j=1}^{i−1} p_{j}^{a_{i}}.

If p is a prime such that p − 1 = ∏_{i=1}^{ℓ} p_{i}^{a_{i}} we say that p is a level-ℓ extended Fermat prime. A table of such primes is included in Table 2. By convention the prime 2 is the only level-0 extended Fermat prime. With this new definition usual Fermat primes are now level-1 extended Fermat primes. Extended Fermat primes form sequence \underline{A286499}. It is thought that there are only finitely many Fermat primes. However, we believe there are infinitely many extended Fermat primes and offer the following conjectures.

2^{⋯  ^{^{5^{⋯  62501 ⋮ 101 ⋮}}  ^{3^{⋯  ^{^{163 ⋮}  ^{^{19⋮}  ^{7^{⋯  ^{^{14407 ⋮}  ^{43^{⋯  ^{77659 p}}}}}}}}}}}}

Figure 2: A portion of the tree of extended Fermat primes.

Conjecture 7. There exists an extended Fermat prime p such that (p − 1)p^{k} + 1 is an extended Fermat prime for infinitely many values of k.

Conjecture 8. Given any positive integer ℓ there exists a level-ℓ extended Fermat prime.

Towards an answer to Conjecture 7, the prime 3 may give a example. Computation suggests there are many primes of the form 2 · 3^{k} + 1. The values of k for which 2 · 3^{k} + 1 is prime is sequence \underline{A003306}. In the direction of Conjecture 8, we have found a level-5 extended Fermat prime

p = 2 · 3 · 7 · 43^{2} · 77659^{197} + 1.

We can form a rooted tree of extended Fermat primes with root 2 as follows. Let p_{1} and p_{2} be two extended Fermat primes, then p_{2} is a descendant of p_{1} if and only if p_{1} | (p_{2} − 1). A portion of this tree, including a path to the level-5 extended Fermat prime p, is shown in Figure 2.

6 Acknowledgments

The author wishes to thank the anonymous referee for helpful comments that have improved this paper.

References

[1] L. Brenton, and R. Hill, On the Diophantine equation 1 = ∑ 1/n_{i} + 1/ ∏ n_{i} and a class of homologically trivial complex surface singularities. Pacific J. Math. 133 (1988), 41–67.

[2] D. Borwein, J. M. Borwein, P. Borwein, and R. Girgensohn, Giuga’s conjecture on primality, Amer. Math. Monthly 103 (1996), 40–50.

[3] W. Butske, L. M. Jaje, and D. R. Mayernik, On the equation ∑_{p|N} \frac{1}{p} + \frac{1}{N} =1, pseudoperfect numbers, and perfectly weighted graphs, Math. Comp. 69 (2000), 407 – 420.

[4] G. Giuga, Su una presumibile proprietá caratteristica dei numeri primi, Ist. Lombardo Sci. Lett. Rend. Cl. Sci. Mat. Nat. (3) 14(83) (1950), 511–528.

[5] R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, 2004.

[6] W. Sierpiński, Sur les nombres pseudoparfaits, Mat. Vesnik. 2 (17) (1965), 212–213.

[7] N. J. A. Sloane, editor, The On-Line Encyclopedia of Integer Sequences, published elec-

   tronically at http://oeis.org.

[8] J. Sondow and K. MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdős-Moser equation, Amer. Math. Monthly 124 (2017), 232–240.

[9] J. Sondow and E. Tsukerman, The p-adic order of power sums, the Erdős-Moser equation, and Bernoulli numbers, preprint, 2014. Available at http://arxiv.org/abs/1401.0322.

2010 Mathematics Subject Classification: Primary 11D68; Secondary 11A41, 11A51. Keywords: Egyptian fraction, pseudoperfect number, primary pseudoperfect number, Giuga number.

(Concerned with sequences \underline{A000668}, \underline{A003306}, \underline{A005835}, \underline{A007850}, \underline{A019434}, \underline{A054377}, \underline{A073932}, \underline{A073935}, \underline{A283423}, \underline{A286497}, and \underline{A286499}.)

Received June 7 2017; revised versions received September 14 2017; January 5 2018; January 22 2018; March 15 2018. Published in Journal of Integer Sequences, March 29 2018.

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