A358418 Least number k coprime to 2, 3, and 5 such that sigma(k)/k >= n.

1, 20169691981106018776756331

Offset: 1

Author

Jianing Song, Nov 14 2022

Keywords

• nonn
• bref
• hard

Comments

Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(2) = 7^2*11^2*13*...*67 ~ 2.01697*10^25. a(3) = 7^3*11^3*13^2*17^2*19^2*23^2*29^2*31*...*569 ~ 2.54562*10^239 and a(4) = 7^5*11^3*13^3*17^3*19^3*23^2*...*97^2*101*...*4561 ~ 1.11116*10^1986 are too large to display.

Examples

			a(2) = A047802(3) = 20169691981106018776756331 is the smallest abundant number coprime to 2, 3, and 5.
Even if there is a number k coprime to 2, 3, and 5 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=4..m+3} (prime(i)/(prime(i)-1)) => m >= 97, and we have k >= prime(4)^2*...*prime(100)^2 ~ 2.46692*10^436 > A358413(3) ~ 2.54562*10^239. So a(3) = A358413(3).
Even if there is a number k coprime to 2, 3, and 5 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=4..m+3} (prime(i)/(prime(i)-1)) => m >= 606, and we have k >= prime(4)^2*...*prime(607)^2*prime(608)*prime(609) ~ 6.54355*10^3814 > A358414(3) ~ 1.11116*10^1986. So a(4) = A358414(3).
		

Links

• Jianing Song, Table of n, a(n) for n = 1..3
• 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.

Crossrefs

Smallest k-abundant number which is not divisible by any of the first n primes: A047802 (k=2), A358413 (k=3), A358414 (k=4).

Least p-rough number k such that sigma(k)/k >= n: A023199 (p=2), A119240 (p=3), A358412 (p=5), this sequence (p=7), A358419 (p=11).