OFFSET
1,2
COMMENTS
2^m is in the sequence iff m=0 or m+1 is prime (the proof is easy). Also all numbers of the form 3*2^(2^m-1) are in the sequence because d(phi(3*2^(2^m-1))) - phi(d(3*2^(2^m-1)))= d(2*2^(2^m-2)) - phi(2*2^m) = d(2^(2^m-1)) - phi(2^(m+1)) = 2^m - 2^m = 0. So this sequence is infinite. - Farideh Firoozbakht, Jan 25 2006
LINKS
T. D. Noe, Table of n, a(n) for n = 1..500
EXAMPLE
k = 24: d(24) = 8, phi(8) = 4, phi(24) = 8, d(8) = 4, so 24 is here.
MATHEMATICA
cm[x_] := DivisorSigma[0, EulerPhi[x]]-EulerPhi[DivisorSigma[0, x]] Do[s=cm[n]; If[Equal[s, 0], Print[n]], {n, 1, 100000}]
Select[Range[36000], DivisorSigma[0, EulerPhi[#]]==EulerPhi[ DivisorSigma[ 0, #]]&] (* Harvey P. Dale, Sep 02 2013 *)
PROG
(PARI) is(n)=numdiv(eulerphi(n))==eulerphi(numdiv(n)) \\ Charles R Greathouse IV, Feb 21 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Nov 26 2002
STATUS
approved