

A358412


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


6




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



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

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


KEYWORD

nonn,bref,hard


AUTHOR



STATUS

approved



