I can do 41 bytes.
x=3 while 1 do x=math.log(C,x)print(x)end
Thanks! combine with your method, I finally have a 34 bytes solution:
::a::x=math.log(C,x)print(x)goto a
btw, why (ln(C)+x)/(ln(x)+1) can be simplify to math.log(C,x)?
--
regards,
Xavier Wang.