Odd perfect numbers not divisible by 3 are divisible by at least ten distinct primes

Abstract

Hagis and McDaniel have shown that the largest prime factor of an odd perfect number TV is at least 100111, and Pomerance has shown that the second largest prime factor is at least 139. Using these facts together with the method we develop, we show that if 3 | N, N is divisible by at least ten distinct primes.

On the total number of prime factors of an odd perfect number

“…Chein [2] and Hagis [6] each showed that if n is an odd perfect number, then ω(n) ≥ 8. Furthermore, Hagis [7] and Kishore [9] each showed that if 3 n, then ω(n) ≥ 11.…”

Odd perfect numbers not divisible by 3. II

“…Then (a) All a's are even except for one a in which case a =p = 1(4). We write tr for p. The proof of the next lemma is similar to that of Lemma 6 in [4].…”

“…If 31 TV, then Kanold (1949) proved that co(TV) s= 9, and the author (1977, [4]) proved that to(TV) > 10.…”

Sketch of a proof that an odd perfect number relatively prime to 3 has at least eleven prime factors

“…Throughout this paper N will represent an odd perfect number, and o>(N) will denote the number of distinct prime factors of TV. It was shown in [1] that w(7V) > 8, while if 3 J TV it was proved by Kishore [6] that u(N)> 10. The purpose of the present paper is to sketch a proof of the following improvement of Kishore's result.…”

“…Likewise,p4 = 13 or 17 since A(5M7Mll0O190O23a>29°°31°o37°°10090O10012900) < 2, and similar arguments show that ps = 17, 19 or 23 and 19 =S/?6 < 31 and p-j < 79. From Lemma 1 in [6] we have (10) 7r=l(modl2) and (11) ifp = 1 (mod 3) and pa || N then a z 2 (mod 3).…”