OFFSET
1,2
COMMENTS
Harborth proved that this sequence is infinite. He showed that the terms are numbers n such that n|sigma(n)*2^(d(n) - 1), where d(n) is the number of divisors of n (A000005) and sigma(n) is their sum (A000203), and that the even terms, numbers of the form r*2^m where r is odd and m > 0, are those with m = ord_2(r/gcd(r, sigma(r)))*i with i = 1, 2, ... (ord_2(k) is the multiplicative order of 2 mod k, A002326). Thus this sequence includes all the powers of 2, all the numbers of the form n = 2^m*(2^(m + 1) - 1) which include the even perfect numbers.
Pollack and Pomerance call these numbers "H-perfect numbers". They prove that k is H-perfect if and only if denominator(sigma(k)/k) is a power of 2. - Amiram Eldar, Jun 02 2020
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..500
Heiko Harborth, Eine Bemerkung zu den vollkommenen Zahlen, Elemente der Mathematik, Vol. 31 (1976), pp. 119-121 (in German).
Paul Pollack and Carl Pomerance, Prime-Perfect Numbers, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 12a, Paper A14, 2012.
MATHEMATICA
divSumSubQ[n_] := Divisible[DivisorSigma[1, n] * 2^(DivisorSigma[0, n] - 1), n]; Select[Range[100000], divSumSubQ]
PROG
(PARI) isok(n) = (sigma(n)*2^(numdiv(n)-1) % n) == 0; \\ Michel Marcus, Dec 21 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Dec 21 2018
STATUS
approved