OFFSET
1,1
COMMENTS
A subset of A068390 and A020492 (balanced numbers). Conjectured to be infinite by Broughan and Zhou.
From David A. Corneth, Sep 21 2019: (Start)
Exactly 130 terms are of the form 2^35 * p * q.
We have phi and sigma are multiplicative and sigma(2^k) / phi(2^k) = 4 - 2/2^k, and sigma(p)/phi(p) = 1 + 2 / (p-1).
So we need (4 - 2/2^k) * (1 + 2 / (p-1)) <= 4 which gives a lower bound on p depending on k; p > nextprime(4*2^k).
We can then, given k and p, solve for q. Without loss of generality, p < q. Then search over the primes and stop for that value of k when p > q.
This method may be refined using insights from the article and/or given some k, solve the system (1 + 2 / (p-1)) * (1 + 2 / (q - 1)) = (a*m) / (b*m) for p and q where a/b is in lowest terms, m > 0. (End)
Furthermore, p < 8*2^k - 2. - David A. Corneth, Sep 26 2019
LINKS
David A. Corneth, Table of n, a(n) for n = 1..221
Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, vol 17 (2014) article 14.9.2.
David A. Corneth, Some more terms, including terms b-file
EXAMPLE
418 = 2*11*19; sigma(418) = 720 = 4*phi(418).
PROG
(PARI) is(n) = my(f = factor(n)); #f~ == 3 && f[2, 2] == 1 && f[3, 2] == 1 && f[1, 1] == 2 && sigma(f) / eulerphi(f) == 4 \\ David A. Corneth, Sep 21 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jud McCranie, Sep 16 2017
STATUS
approved