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%I #31 Oct 18 2021 05:36:07
%S 2,6,28,120,496,672,8128,30240,32760,523776,2178540,23569920,33550336,
%T 45532800,142990848,459818240
%N Numbers k that divide (sum of proper divisors of k + product of proper divisors of k).
%C All perfect numbers belong to this sequence.
%C Every term of A007691 is in this sequence. - _T. D. Noe_, Sep 29 2005
%C 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
%p isA087948 := proc(n)
%p if modp( A001065(n)+A007956(n),n) = 0 then
%p true;
%p else
%p false;
%p end if;
%p end proc:
%p for n from 2 do
%p if isA087948(n) then
%p printf("%d\n",n) ;
%p end if;
%p end do: # _R. J. Mathar_, Oct 15 2021
%t 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
%o (Python)
%o from math import prod
%o from sympy import divisors
%o def ok(n): d = divisors(n)[:-1]; return n > 1 and (sum(d) + prod(d))%n == 0
%o print([k for k in range(10**5) if ok(k)]) # _Michael S. Branicky_, Oct 15 2021
%Y Cf. A001065, A007956, A007691, A080257 (k which divide A007691(k)).
%Y Cf. A219544.
%K nonn,more
%O 1,1
%A _Joseph L. Pe_, Jan 08 2004
%E a(11)-a(16) from _Michael S. Branicky_, Oct 16 2021
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