OFFSET
1,2
COMMENTS
Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(2) = 5^2*7*...*29 and a(3) = 5^4*7^3*11^2*13^2*17*...*157 ~ 5.16404*10^66. a(4) = 5^5*7^4*11^3*13^3*17^2*19^2*23^2*29^2*31^2*37^2*41*...*853 ~ 1.83947*10^370 is too large to display.
LINKS
Jianing Song, Table of n, a(n) for n = 1..4
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, author’s version, Research Commons.
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, Journal of Number Theory 128 (2008) 1566-1575.
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.
EXAMPLE
a(2) = A047802(2) = 5391411025 is the smallest abundant number coprime to 2 and 3.
Even if there is a number k coprime to 2 and 3 with sigma(k)/k = 3, we have that k is a square since sigma(k) is odd. If omega(k) = m, then 3 = sigma(k)/k < Product_{i=3..m+2} (prime(i)/(prime(i)-1)) => m >= 33, and we have k >= prime(3)^2*...*prime(35)^2 ~ 6.18502*10^112 > A358413(2) ~ 5.16403*10^66. So a(3) = A358413(2).
Even if there is a number k coprime to 2 and 3 with sigma(k)/k = 4, there can be at most 2 odd exponents in the prime factorization of k (see Theorem 2.1 of the Broughan and Zhou link). If omega(k) = m, then 4 = sigma(k)/k < Product_{i=3..m+2} (prime(i)/(prime(i)-1)) => m >= 140, and we have k >= prime(3)^2*...*prime(140)^2*prime(141)*prime(142) ~ 2.65585*10^669 > A358414(2) ~ 1.83947*10^370. So a(4) = A358414(2).
CROSSREFS
KEYWORD
nonn,bref,hard
AUTHOR
Jianing Song, Nov 14 2022
STATUS
approved