How will check a no. is of the format 2^i - 2^j ......Single line code ..... ?

1st Way:- 

using Loop:-

let the given number is n;

while(n&(n%2==0?n/=2:0) || ! printf("%s", n&(n+1)?"no":"yes")) ; 

2nd Way:-

 Without Loop:-

‎(((n-1)|n)+1) &n ==0