Keep Decoding This Number!
Python 3, 379 358 339 337 327 310 304 bytes

Conjecture: Is 13 the only number that will lead to a cycle? (There are no exceptions smaller than 106.)

Fixed a bug and -7 bytes thanks to Sherlock9.-3 byte thanks to Jonathan Frech.-16 bytes thanks to ovs.-6 bytes thanks to Mr. Xcoder.

If there is a cycle, it will repeat the whole cycle.



def p(a,x=0,n=1):
 while a:x+=1;a-=n%x;n*=x*x
 return x
def g(a):
 c=i=0
 for v in a:
  c+=int(v)*2-1
  if c&lt;0:return 0,0
  if c&lt;1:break
  i+=1
 if a:x=p(g(a[1:i])[0]);b,c=g(a[i+1:]);return(x&gt;c&gt;0)*(0,0)or(x*b,x)
 return 1,0
def f(a):
 x=a,
 while(x.count(a)&lt;3)*a:a,t=g(bin(a-~a)[2:]);x+=a,
 return x[:-1]


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Explanation:



def p(a,x=0,n=1):     # return the a-th prime
 while a:             # magical way to enumerate primes
  x+=1
  a-=n%x
  n*=x*x
 return x

def g(a):             # decode a 0/1 string
 c=i=0
 for v in a:
  c+=int(v)*2-1       # +1 if 1, -1 if 0
  if c&lt;0: return(0,0) # c&lt;0: unbalanced parentheses
  if c&lt;1: break       # first outermost parentheses
  i+=1
 if a:
   x=p(g(a[1:i])[0])  # recursive solve those inside the parentheses ...
   b,c=g(a[i+1:])     # and the remaining
   if x&gt;c and c:      # if not ascending
    return (0,0)
   return (x*b,x)     # return (result, value of first closed parentheses)
 return (1,0)         # empty string

def f(a):
 x=a,
 while a and x.count(a)&lt;3: # detect non-decodable or cycle
  a,t=g(bin(a-~a)[2:])     # a-~a is a*2+1
  x+=a,
 return x[:-1]