Python program from Ron Reble, July 18 2017, who says: The MaxN value is the first value that the program doesn't check.
   If you want a 10000-line B file, good luck: after translating to
   C++ and tuning, it's a half-day run on BitTwiddler.
   But a(1000)=180055; the python program can probably get that far.


#!/usr/bin/python

# %I A110081
# %S  1   7  25  31  37  73  79  85  97 103
#   193 241 253 271 313 319 337 343 361 517
#   553 661 703 721 727 733 745 751
# %N Integers n such that the digit set D = (0,1,-n) used in base 3
#  expansions of the form Sum_{ -N < j < infty} d_j 3^{-j},
#  all d_j in D, can represent all real numbers.
# %A njas, based on correspondence from R. K. Guy and Jeff Lagarias,
#  Aug 31 2005

# For N in the sequence:
#  N is 1 mod 3. (Matula Lemma 2)
#  If N is 4 mod 6, you can't represent N/2, so N is 1 mod 6.
#  N isn't 3 mod 8 and isn't 0,3,9 mod 13. (Matula theorem 5)

# If N written in ternary begins "12", you can't represent 2.

# For N == 1 mod 6 but not in the sequence:
#  There's an unrepresentable value in [0,N/6]. (Matula eqn (10))
#  If 3A or 3A+1 is unrepresentable, then A is unrepresentable:
#   the least positive unrepresentable value is 2 mod 3.

# For N in the sequence, print:
#   N, a longest value, and its representation.
#   "Longest value" means that, for values between 2 and N/6, this
#   value has a longest representation.
# For N not in the sequence, but not eliminated by the above modular
# and "12" criteria, print:
#   "#", N, and the least positive 2-mod-3 unrepresentable value.

PrintNonElements = False
PrintLongest = False
MaxN = 180056

# return None if there's no representation of val
def getRep(nn, val):
   cycleset = {val}
   v2 = val
   repstr = ""
   while v2 != 0:
       v2m3 = v2 % 3
       if v2m3 == 0:
           repstr = "0" + repstr
       elif v2m3 == 1:
           repstr = "1" + repstr
           # v2 -= 1 # The "v2 /= 3" below takes care of this.
       else:
           repstr = "x" + repstr
           v2 += nn
       v2 /= 3
       if v2 in cycleset: return None # no representation
       cycleset.add(v2)
   return repstr

gapst = 1*3+2 # the next "12" gap
gapnd = 2*3
for NN in range(1, MaxN, 6):
   if NN >= gapst:
       if NN < gapnd: continue
       gapst *= 3
       gapnd *= 3

   # mod 8, 13 criteria
   if (NN % 8) == 3: continue
   m13 = NN % 13
   if (m13 == 0) or (m13 == 3) or (m13 == 9): continue

# check 2 mod 3 values up to NN/6
   longval = 1
   longrep = "1"
   good = True
   for val in range(2, NN/6 + 1, 3):
       repstr = getRep(NN, val)
       if repstr == None:
           good = False
           break
       elif len(longrep) < len(repstr):
           longval = val
           longrep = repstr

   if good:
       if PrintLongest:
           print NN, longval, longrep
       else:
           print NN
   else:
       if PrintNonElements: print "#", NN, val

--- end of program ---