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%I #25 Mar 24 2019 06:42:09
%S 4,8,32,128,168,224,756,8192,131072,524288,2147483648
%N Let 's' denote the sum of the deficient numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) is equal to x.
%C All numbers of the form 2^A000043(n) belong to the sequence.
%C Terms that are not of this form begin: 168, 224, 756, ... - _Amiram Eldar_, Mar 24 2019
%e Aliquot parts of 8 are 1, 2, 4 that are all deficient numbers: sigma(1 + 2 + 4) = sigma(7) = 8.
%e Aliquot parts of 168 are 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84 and the deficient numbers are 1, 2, 3, 4, 7, 8, 14, 21: sigma(1 + 2 + 3 + 4 + 7 + 8 + 14 + 21) = sigma(60) = 168.
%p with(numtheory); P:=proc(q) local a,b,c,k,n;
%p for n from 1 to q do a:=sort([op(divisors(n))]); b:=0; c:=0;
%p for k from 1 to nops(a)-1 do if sigma(a[k])<2*a[k] then b:=b+a[k]; fi; od;
%p if sigma(b)=n then print(n); fi; od; end: P(10^9);
%t seqQ[n_] := Module[{s = DivisorSum[n, # &, #<n && DivisorSigma[1,#] < 2# &]}, s>0 && DivisorSigma[1,s] == n]; Select[Range[10000], seqQ] (* _Amiram Eldar_, Mar 24 2019 *)
%o (PARI) isok(n) = my (s = sumdiv(n, d, d*((d!=n) && (sigma(d)/d < 2)))); s && (sigma(s) == n); \\ _Michel Marcus_, Feb 19 2015
%Y Cf. A000043, A000203, A005100, A254879, A254880.
%K nonn,more
%O 1,1
%A _Paolo P. Lava_, Feb 10 2015
%E a(11) from _Amiram Eldar_, Mar 24 2019
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