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%I #13 Oct 07 2021 01:57:31
%S 94500,195300,1674000,27432000,56692800,325883250,735257250,
%T 113232384000,234013593600,28990808064000,59914336665600,
%U 463855583232000,559625737239000,958634872012800,1373356918809000,7782220152472338432000,16083254981776166092800,8972288971548182138209587578844217344000
%N Numbers k for which sigma(k)/k = 832/225.
%C This sequence contains terms of the form 3375*P and 6975*Q, where P is a perfect number (A000396) not divisible by 3 or 5, and Q is a perfect number not divisible by 3, 5, or 31. Proof: sigma(3375*P)/(3375*P) = sigma(3375)*sigma(P)/(3375*P) = 6240*(2*P)/(3375*P) = 832/225 and sigma(6975*Q)/(6975*Q) = sigma(6975)*sigma(Q)/(6975*Q) = 12896*(2*Q)/(6975*P) = 832/225. QED
%C Many terms ending in "00" will have one of these forms:
%C a( 1) = 94500 = 3375* 28 = 3375*A000396(2)
%C a( 2) = 195300 = 6975* 28 = 6975*A000396(2)
%C a( 3) = 1674000 = 3375* 496 = 3375*A000396(3)
%C a( 4) = 27432000 = 3375* 8128 = 3375*A000396(4)
%C a( 5) = 56692800 = 6975* 8128 = 6975*A000396(4)
%C a( 8) = 113232384000 = 3375* 33550336 = 3375*A000396(5)
%C a( 9) = 234013593600 = 6975* 33550336 = 6975*A000396(5)
%C a(10) = 28990808064000 = 3375* 8589869056 = 3375*A000396(6)
%C a(11) = 59914336665600 = 6975* 8589869056 = 6975*A000396(6)
%C a(12) = 463855583232000 = 3375* 137438691328 = 3375*A000396(7)
%C a(14) = 958634872012800 = 6975* 137438691328 = 6975*A000396(7)
%C a(16) = 7782220152472338432000 = 3375*2305843008139952128 = 3375*A000396(8)
%C a(17) = 16083254981776166092800 = 6975*2305843008139952128 = 6975*A000396(8).
%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 325883250 is a term, since sigma(325883250)/325883250 = 1205043840/325883250 = 832/225.
%t Select[Range[5*10^8], DivisorSigma[1, #]/# == 832/225 &]
%t Do[If[DivisorSigma[1, k]/k == 832/225, Print[k]], {k, 5*10^8}]
%Y Cf. A000203, A000396, A211680, A212610.
%Y Subsequence of A005101.
%K nonn
%O 1,1
%A _Timothy L. Tiffin_, Sep 24 2021
%E More terms from _Michel Marcus_, Oct 03 2021