

A089748


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


1



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

1,1


COMMENTS

All perfect numbers belong to this sequence.
There are two sets of candidates of k: (i) kA001065(k) and kA007956(k) individually, or (ii) neither kA001065(k) nor kA007956(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 kA007956(k). This means for all k in A080257 we have kA007956(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 k2 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


LINKS



MAPLE

isA087948 := proc(n)
true;
else
false;
end if;
end proc:
for n from 2 do
if isA087948(n) then
printf("%d\n", n) ;
end if;


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 *)


PROG

(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


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



