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%I #13 Nov 20 2021 18:03:38
%S 546,45136,739648,5752422,3053080576,781678084096,12506920910848,
%T 209831713740735643648,241919495232854688763577028051799638016
%N Numbers k for which sigma(k)/k = 32/13.
%C This sequence will contain terms of the form 91*P, where P is a perfect number (A000396) not divisible by 7 or 13. Proof: sigma(91*P)/(91*P) = sigma(91)*sigma(P)/(91*P) = 112*(2*P)/(91*P) = 32/13. QED.
%C Terms ending in "6" or "48" have this form. Example: a(n) = 91*A000396(n) for n = 1, 5, 6, 7, 8, 9 and a(n) = 91*A000396(n+1) for n = 2, 3.
%H G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiperfect">Multiperfect Numbers and Hemiperfect Numbers</a>
%H Walter Nissen, <a href="http://upforthecount.com/math/abundance.html">Abundancy: Some Resources (preliminary version 4)</a>
%H Walter Nissen, <a href="http://upforthecount.com/math/ffp8.html">Primitive Friendly Pairs with friends < 2^34 with denom < 20000</a>
%e 5752422 is a term, since sigma(5752422)/5752422 = 14159808/5752422 = 32/13.
%t Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/13 &]
%t Do[If[DivisorSigma[1, k]/k == 32/13, Print[k]], {k, 5*10^8}]
%Y Cf. A000203, A000396.
%Y Subsequence of A005101.
%K nonn,more
%O 1,1
%A _Timothy L. Tiffin_, Aug 22 2021
%E a(8)-a(9) from _Michel Marcus_, Aug 23 2021
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