A292422 Numbers of the form x = 2^k*p*q, p,q odd primes, such that sigma(x)/phi(x) = 4.

418, 3596, 3956, 5396, 8636, 41656, 56536, 393104, 2072608, 2316448, 6543008, 17434528, 135394432, 217023616, 1264918784, 1490909824, 2710540544, 11444858368, 17669583104, 34797058304, 37698861568, 70572901376, 132968907776, 226965472256, 233356030976, 552070776832, 596357220352, 601188468736

Offset: 1

Author

Jud McCranie, Sep 16 2017

Keywords

• nonn

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

Examples

			418 = 2*11*19; sigma(418) = 720 = 4*phi(418).
		

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
• David A. Corneth, Some more terms ordered by their exponent of 2, specifying whether it is known that they are all such terms.

Crossrefs

Cf. A068390, A020492.

Subsequence of A293391 (sigma(x)/phi(x) square).

Programs

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