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%I #17 Aug 22 2021 22:15:08
%S 42,3472,56896,544635,234852352,60129083392,962070839296,
%T 16140901056979664896,18609191940988822212582848311676895232
%N Numbers k for which sigma(k)/k = 16/7.
%C This sequence will contain terms of the form 7*P, where P is a perfect number (A000396) not divisible by 7. Proof: sigma(7*P)/(7*P) = sigma(7)*sigma(P)/(7*P) = 8*(2*P)/(7*P) = 16/7. QED
%C Terms ending in "2" or "96" have this form. Example: a(n) = 7*A000396(n) for n = 1, 5, 6, 7, 8, 9 and a(n) = 7*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 544635 is a term, since sigma(544635)/544635 = 1244880/544635 = 16/7.
%t Select[Range[5*10^8], DivisorSigma[1, #]/# == 16/7 &]
%t Do[If[DivisorSigma[1, k]/k == 16/7, Print[k]], {k, 5*10^8}]
%Y Cf. A000203, A000396.
%Y Subsequence of A005101 and A218409.
%K nonn,more
%O 1,1
%A _Timothy L. Tiffin_, Aug 20 2021
%E a(8)-a(9) from _Michel Marcus_, Aug 21 2021
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