%I #13 Aug 26 2021 15:30:36

%S 924,2970,16368,18018,268224,1107161088,283465678848,4535476813824,

%T 76092819268618420224,87729047721804447573604856326476791808

%N Numbers k for which sigma(k)/k = 32/11.

%C This sequence will contain terms of the form 33*P, where P is a perfect number (A000396) not divisible by 3 or 11. Proof: sigma(33*P)/(33*P) = sigma(33)*sigma(P)/(33*P) = 48*(2*P)/(33*P) = 32/11. QED

%C Terms ending in "24" or "x8" (where x is an even digit) have this form. Example: a(1) = 33*A000396(2), a(3) = 33*A000396(3), and a(n) = 33*A000396(n-1) for 5, 6, 7, 8.

%C Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma(33*A341623(n))/(33*A341623(n)) = sigma(11)*sigma(3*A341623(n))/(33*A341623(n)) = 12*(8*A341623(n))/(33*A341623(n)) = 32/11. Does this sequence, though, contain any additional terms that are not generated by A341623?

%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 2970 is a term, since sigma(2970)/2970 = 8640/2970 = 32/11.

%e 18018 is a term, since sigma(18018)/18018 = 52416/18018 = 32/11.

%t Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/11 &]

%t Do[If[DivisorSigma[1, k]/k == 32/11, Print[k]], {k, 5*10^8}]

%Y Cf. A000203, A000396, A341623.

%Y Subsequence of A005101.

%K nonn,more

%O 1,1

%A _Timothy L. Tiffin_, Aug 24 2021

%E a(9)-a(10) from _Michel Marcus_, Aug 26 2021