A046985 Multiply perfect numbers whose average divisor is an integer and divides the number itself.

1, 6, 672, 30240, 32760, 23569920, 45532800, 14182439040, 51001180160, 153003540480, 403031236608, 13661860101120, 154345556085770649600, 9186050031556349952000, 143573364313605309726720, 352338107624535891640320, 680489641226538823680000, 34384125938411324962897920

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Examples

			k = 45532800 is a term since, s0 = 384, s1 = 182131200, and the three quotients s1/k = 182131200/45532800 = 4, (k * s0)/s1 = (45532800 * 384)/182131200 = 96, and s1/s0 = 182131200/384 = 474300 are all integers.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..321

Crossrefs

Intersection of A003601, A007691 and A001599.

Cf. A000005, A000203, A046986, A046987.

Programs

Mathematica

q[n_] := Module[{d = DivisorSigma[0, n], s = DivisorSigma[1, n]}, Divisible[s, n] && Divisible[n * d, s] && Divisible[s, d]]; Select[Range[33000], q] (* Amiram Eldar, May 09 2024 *)

PARI

isok(n) = s1 = sigma(n); s0 = numdiv(n); !(s1 % n) && !(s1 % s0) && !((n*s0) % s1); \\ Michel Marcus, Dec 10 2013

PARI

is(k) = {my(f = factor(k), s = sigma(f), d = numdiv(f)); !(s % k) && !((k * d) % s) && !(s % d);} \\ Amiram Eldar, May 09 2024

Formula

Let s1 = sigma(k) = A000203(k) be the sum of divisors of k and s0 = d(k) = A000005(k) be the number of divisors of k. Then, k is a term if s1/k, (k * s0)/s1, and s1/s0 are all integers.

Extensions

a(10)-a(15) from Donovan Johnson, Nov 30 2008

Edited and a(16)-a(18) added by Amiram Eldar, May 09 2024