Odd perfect numbers of a special form

Abstract

Abstract. We show that there is an effectively computable upper bound of odd perfect numbers whose Euler factors are powers of fixed exponent.

A note on the paper by Bugeaud and Laurent “Minoration effective de la distance p-adique entre puissances de nombres algébriques”

“…Hence N < ( i p i ) (p k −1)/2 (for other finiteness results, see, for example, [6,7,11]). We can improve this upper bound using Theorem 4.…”

An exponential diophantine equation related to odd perfect numbers

“…We have shown that, if N = p α (q 1 q 2 • • • q k ) 2β is an odd perfect number, then k ≤ 4β 2 + 2β + 2 in [19]. Recently, we have improved this upper bound by 2β 2 + 8β + 3 in [21], where the coefficient 8 of β can be replaced by 7 if 2β + 1 is not a prime or β ≥ 29.…”

On the divisibility of odd perfect numbers, quasiperfect numbers and amicable numbers by a high power of a prime

“…In their paper [22], Hagis and McDaniel conjecture that β 1 = • • • = β t = β does not occur. The author [26] proved that there are only finitely many odd perfect numbers for any given β. McDaniel [20] proved that we cannot have β 1 ≡ • • • ≡ β t ≡ 2 (mod 6), i.e., 3 cannot divide all of β 1 + 1, β 2 + 1, . .…”

“…Using the author's method, but with the aid of the large sieve instead of Selberg's sieve used by the author [26], Fletcher, Nielsen and Ochem [7] proved that if N = p α1 1 • • • p αs s q β1 1 • • • q βt t satisfies h(N ) = n/d and for each i, β i + 1 has a prime factor belonging to a finite set P of primes, then N has a prime divisor small than a effective constant C, depending only on n, s and P. Moreover, they proved that the smallest prime factor of an odd perfect number N satisfying the above condition with P = {3, 5} lies between 10 8 and 10 1000 , improving results in [3] and [27].…”

“…Hence, in the case z ≥ y, we can take B 0 = 2.01 in (19). Since z ≥ X 1 = x 1 ≥ x 101 0 , also in the case z < y, we can take B 0 = 2.01 in (26). In our setting of x 1 , we can take A 3 = 0.7 in (21) with k = l and k = P from (41) of Lemma 4.2, noting that x 1 ≥ x 0 ≥ exp P .…”