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%I #8 Apr 06 2024 07:53:54
%S 1,84,252,12717,177744,189264,143747328
%N Numbers k such that abundance(k) = abundance(sigma(k)).
%C a(8) > 3.4*10^10, if it exists. - _Amiram Eldar_, Apr 06 2024
%e abundance(84) = sigma(84) - 2(84) = 56 = abundance(224) = abundance(sigma(84)), so 84 is a term in the sequence.
%t abund[n_] := DivisorSigma[1,n]-2n; l = {}; For[i = 1, i <= 10^6, i++, If[abund[i] == abund[DivisorSigma[1, i]], l = Append[l, i]]]; l
%o (PARI) is(k) = {my(s = sigma(k)); s - 2*k == sigma(s) - 2*s;} \\ _Amiram Eldar_, Apr 06 2024
%Y Cf. A000203 (sigma), A033880 (abundance).
%K nonn,more
%O 1,2
%A _Joseph L. Pe_, Mar 05 2008
%E a(7) from _Donovan Johnson_, Apr 27 2008
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