%I #11 May 13 2013 01:47:55
%S 3,5,7,15,11,13,0,17,19,25,23,35,0,29,31,51,0,37,0,41,43,69,47,65,0,
%T 53,81,87,59,61,0,85,67,0,71,73,0,0,79,123,83,129,0,89,0,141,0,97,0,
%U 101,103,159,107,109,121,113,0,177,0,143,0,0,127,255,131,161,0,137
%N Smallest k such that phi(k) = 2n, or 0 if there is no such k.
%C The zero values are easy to prove because of the bounds on the phi function.
%C phi(15)=2*4, so a(4)=15.
%H T. D. Noe, <a href="/A051445/b051445.txt">Table of n, a(n) for n = 1..10000</a>
%e sigma(4)=7, 4 is the smallest, so a(7)=4.
%o (PARI) a(n)=n+=n;for(k=n+1, solve(x=n,if(n<20,99,5*n*log(log(n))), x/(exp(Euler)*log(log(x))+3/log(log(x)))-n), if(eulerphi(k)==n,return(k))); 0 \\ _Charles R Greathouse IV_, Dec 19 2011
%Y Cf. A002181. For records see A132012, A132115.
%K nonn
%O 1,1
%A _Jud McCranie_