Remember we are dealing with k*2^n+1b) natural (due to some obvious implication)
The n's for a given k are all even or all odd so they differ by an even number.
n'=n+2*j for some positive interger j
k*2^n'+1=k*2^(n+2j)+1 = k*(2^n)*(2^2J)+1=k*(2^n)*4^j+1=(4^j)*k*2^n+1