OFFSET
1,1
COMMENTS
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.
(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.
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.
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.
It is easy to see that a positive integer n with sigma(n) odd must be a square or twice a square.
See also A279894 for a similar sequence.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..433
EXAMPLE
a(1) = 6 since 6 = 1*6 with phi(1)*sigma(6) = 2*6.
a(3) = 84 since 84 = 7*12 with phi(7)*sigma(12) = 2*84.
MATHEMATICA
sigma[n_]:=sigma[n]=DivisorSigma[1, n];
phi[n_]:=phi[n]=EulerPhi[n];
Dv[m_]:=Dv[m]=Divisors[m];
Ld[m_]:=Ld[m]=Length[Dv[m]];
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}]
(* Second program *)
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 22 2016
STATUS
approved