OFFSET
1,1
COMMENTS
All integers m contain at least one divisor d (number 1) such that sigma(d) divides m.
See A309253 for the smallest numbers m with n divisors d such that sigma(d) divides m for n >= 1.
Supersequence of A097603 (multiples of perfect numbers).
From Bernard Schott, Sep 04 2019: (Start)
If m = 6 * k with k >= 1, then 2 divides m and sigma(2) = 3 also divides m; so, the positive multiples of 6 belong to this sequence.
This sequence is generated by the primitive terms. A primitive term m is necessarily of the form d * sigma(d) where 1 < d < m is a divisor of m. The first few primitives are: 6, 28, 117, 182, ...
Some subsequences of such primitives, not exhaustive list:
1) d is prime p and m = p * sigma(p) = p * (p+1) is oblong.
For p = 2, 13, 19, 37, ..., we get 6, 182, 380, 1406, ...
2) d = p^2 with p prime, and m = p^2 * (p^2 + p + 1).
For p = 2, 3, 5, 7, ..., we get m = 28, 117, 775, 2793, ...
3) d = 2^(q-1) and m = 2^(q-1) * (2^q -1), with q prime in A000043 and 2^q - 1 is a Mersenne prime in A000668, then m is a perfect number in A000039.
For q prime = 2, 3, 5, 7, 13, ..., we get m = 6, 28, 496, 8128, 33550336, ... (End)
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
Divisors d of 12: 1, 2, 3, 4, 6, 12; corresponding sigma(d):1, 3, 4, 7, 12, 28; sigma(d) divides 12 for 4 divisors d > 1: 2, 3 and 6.
MAPLE
filter:= proc(n) local d;
uses numtheory;
ormap(t -> n mod sigma(t) = 0, divisors(n) minus {1})
end proc:
select(filter, [$2..1000]); # Robert Israel, Oct 07 2019
MATHEMATICA
aQ[n_] := AnyTrue[Rest @ Divisors[n], Divisible[n, DivisorSigma[1, #]] &]; Select[Range[282], aQ] (* Amiram Eldar, Aug 31 2019 *)
PROG
(Magma) [m: m in [1..10^5] | #[d: d in Divisors(m) | IsIntegral(m / SumOfDivisors(d) ) and d gt 1] gt 0];
(PARI) isok(m) = fordiv(m, d, if ((d>1) && (!(m % sigma(d))), return(1))); \\ Michel Marcus, Sep 03 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Aug 30 2019
STATUS
approved
