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%I #32 Dec 23 2016 21:15:42
%S 6,28,84,120,234,360,496,588,600,1080,1638,2016,3000,3042,3240,3276,
%T 4116,4680,6048,7440,8128,9720,11466,14040,15000,18144,22320,22932,
%U 23400,28812,29160,30240,32640,32760,37200,39546,42120,42588,54432,55800,60480,60840,65520,66960
%N Numbers m that can be written as x*y with phi(x)*sigma(y) = 2*x*y, where x and y are positive integers, phi(.) is Euler's totient function and sigma(y) is the sum of all positive divisors of y.
%C Conjecture: (i) All the terms are even. Moreover, if x and y are positive integers with phi(x)*sigma(y) = 2*x*y, then y must be even.
%C (ii) If x and y are positive integers with phi(x)*sigma(y) = 2*x*y, then x = 1 or 3 | y. Thus, any term of the sequence is either a perfect number or a multiple of three.
%C As phi(1) = 1, the sequence contains all perfect numbers, and part (i) of the above conjecture implies the well-known conjecture that there are no odd perfect numbers.
%C We consider the terms of this sequence as natural extensions of perfect numbers. There are a total of 433 terms not exceeding 10^8, and they are all even.
%C It is easy to see that a positive integer n with sigma(n) odd must be a square or twice a square.
%C See also A279894 for a similar sequence.
%H Zhi-Wei Sun, <a href="/A279915/b279915.txt">Table of n, a(n) for n = 1..433</a>
%e a(1) = 6 since 6 = 1*6 with phi(1)*sigma(6) = 2*6.
%e a(3) = 84 since 84 = 7*12 with phi(7)*sigma(12) = 2*84.
%t sigma[n_]:=sigma[n]=DivisorSigma[1,n];
%t phi[n_]:=phi[n]=EulerPhi[n];
%t Dv[m_]:=Dv[m]=Divisors[m];
%t Ld[m_]:=Ld[m]=Length[Dv[m]];
%t n=0;Do[Do[If[sigma[Part[Dv[m],i]]phi[m/Part[Dv[m],i]]==2m,n=n+1;Print[n," ",m];Goto[aa]],{i,1,Ld[m]}];Label[aa];Continue,{m,1,70000}]
%t (* Second program *)
%t Select[Range[10^5], Function[n, Total@ Boole@ Map[EulerPhi[#1] DivisorSigma[1, #2] == 2 #1 #2 & @@ {#, n/#} &, Divisors@ n] > 0]] (* _Michael De Vlieger_, Dec 23 2016 *)
%Y Cf. A000010, A000203, A000396, A279894.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Dec 22 2016
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