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%I #16 Sep 24 2021 16:12:22
%S 30,140,2480,6200,40640,167751680,42949345280,687193456640,
%T 11529215040699760640,13292279957849158723273463079769210880,
%U 957809713041180536473966890421518190654986607740846080,65820182292848241686198767302293614551117361591934715588918640640
%N Numbers k for which sigma(k)/k = 12/5.
%C This sequence will contain terms of the form 5*P, where P is a perfect number (A000396) not divisible by 5. Proof: sigma(5*P)/(5*P) = sigma(5)*sigma(P)/(5*P) = 6*(2*P)/(5*P) = 12/5. QED
%C Terms ending in "30", "40", or "80" have this form. Example: a(n) = 5*A000396(n) for n = 1, 2, 3 and a(n) = 5*A000396(n-1) for n = 5..12.
%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 6200 is a term, since sigma(6200)/6200 = 14880/6200 = 12/5.
%t Select[Range[5*10^8], DivisorSigma[1, #]/# == 12/5 &]
%t Do[If[DivisorSigma[1, k]/k == 12/5, Print[k]], {k, 5*10^8}]
%Y Cf. A000203, A000396.
%Y Subsequence of A005101 and A218407.
%K nonn
%O 1,1
%A _Timothy L. Tiffin_, Aug 23 2021
%E a(9)-a(10) from _Michel Marcus_, Aug 24 2021
%E a(11)-a(12) from _David A. Corneth_, Aug 24 2021
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