A089748 Numbers k that divide (sum of proper divisors of k + product of proper divisors of k).

2, 6, 28, 120, 496, 672, 8128, 30240, 32760, 523776, 2178540, 23569920, 33550336, 45532800, 142990848, 459818240

Offset: 1

Author

Joseph L. Pe, Jan 08 2004

Keywords

• nonn
• more

Comments

All perfect numbers belong to this sequence.

Every term of A007691 is in this sequence. - T. D. Noe, Sep 29 2005

There are two sets of candidates of k: (i) k|A001065(k) and k|A007956(k) individually, or (ii) neither k|A001065(k) nor k|A007956(k) but the remainders of A001065(k)/k and A007956(k)/k sum up to k. If k has at least 4 divisors, the product of the second and penultimate divisor (in the sorted divisors list) is k, so k|A007956(k). This means for all k in A080257 we have k|A007956(k), and the k that do not divide A007956(k) are in A000430, which means k=p or k=p^2 for some prime p. If k=p, A001065(k)+A007956(k) = 1+1 =2, and the requirement here reduces to k|2 and only k=2 is left. If k=p^2, A001065(k) +A007956(k) = 1+p+p = 1+2*p, and the requirement here reduces to p^2 | (1+2*p), which has no solutions. This means case (ii) does not generate any solutions besides k=2. And this means all other solutions are from case (i), and therefore elements A007691 > 1 are the only remaining candidates. - R. J. Mathar, Oct 15 2021

Crossrefs

Cf. A001065, A007956, A007691, A080257 (k which divide A007691(k)).

Cf. A219544.

Programs

Maple

isA087948 := proc(n)
    if modp( A001065(n)+A007956(n),n) = 0 then
        true;
    else
        false;
    end if;
end proc:
for n from 2 do
    if isA087948(n) then
        printf("%d\n",n) ;
    end if;
end do: # R. J. Mathar, Oct 15 2021

Mathematica

l = {}; Do[d = Drop[Divisors[n], -1]; p = Apply[Plus, d]; t = Apply[Times, d]; m = Mod[p + t, n]; If[m == 0, l = Append[l, n]], {n, 2, 10^6}]; l
Select[Range[2,22*10^5],Mod[Total[Most[Divisors[#]]]+Times@@Most[Divisors[#]],#]==0&] (* The program generates the first 11 terms of the sequence. *) (* Harvey P. Dale, Jun 05 2024 *)

Python

from math import prod
from sympy import divisors
def ok(n): d = divisors(n)[:-1]; return n > 1 and (sum(d) + prod(d))%n == 0
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Oct 15 2021

Extensions

a(11)-a(16) from Michael S. Branicky, Oct 16 2021