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A316225
Numbers k that divide the sum of sums of elements of all subsets of divisors of k (A229335).
1
1, 2, 4, 6, 8, 16, 24, 28, 32, 40, 64, 96, 120, 128, 224, 256, 288, 360, 384, 496, 512, 640, 672, 1024, 1536, 1792, 1920, 2016, 2048, 2176, 3744, 4096, 4320, 4680, 5632, 5760, 6144, 6528, 8128, 8192, 10240, 10880, 14336, 15872, 16384, 16896, 18432, 18688
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
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