A335051 a(n) is the smallest decimal number > 1 such that when it is written in all bases from base 2 to base n those numbers all contain both 0 and 1.

2, 9, 19, 28, 145, 384, 1128, 2601, 2601, 101256, 103824, 382010, 572101, 971400, 1773017, 1773017, 22873201, 64041048, 64041048, 1193875201, 2496140640, 10729882801, 21660922801, 120068616277, 333679563001, 427313653201, 427313653201, 10436523921264, 10868368953601

Offset: 2

Author

Zach J. Shannon and Scott R. Shannon, May 21 2020

Keywords

• nonn
• base

Comments

The sequence is infinite since 1 + lcm(2,...,n)^2 is always a candidate for a(n). - Giovanni Resta, May 24 2020

Examples

			a(3) = 9 as 9_2 = 1001 and 9_3 = 100, both of which contain a 0 and 1.
a(6) = 145 as 145_2 = 10010001, 145_3 = 12101, 145_4 = 2101, 145_5 = 1040, 145_6 = 401, all of which contain a 0 and 1.
a(9) = 2601 as 2601_2 = 101000101001, 2601_3 = 10120100, 2601_4 = 220221, 2601_5 = 40401, 2602_6 = 20013, 2601_7 = 10404, 2601_8 = 5051, 2601_9 = 3510, all of which contain a 0 and 1. Note that, as 2601 also contains a 0 and 1, a(10) = 2601.
a(16) = 1773017 as 1773017_2 = 110110000110111011001, 1773017_3 = 10100002010022, 1773017_4 = 12300313121, 1773017_5 = 423214032, 1773017_6 = 102000225, 1773017_7 = 21033101, 1773017_8 = 6606731, 1773017_9 = 3302108, 1773017_10 = 1773017, 1773017_11 = 1001104, 1773017_12 = 716075, 1773017_13 = 4A102C, 1773017_14 = 342201, 1773017_15 = 250512, 1773017_16 = 1B0DD9, all of which contain a 0 and 1.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 2..32

Crossrefs

Cf. A335066, A258107, A145100, A317725, A181929, A331565, A153114.

Programs

Mathematica

a[n_] := Block[{k=2}, While[ AnyTrue[ Range[n, 2, -1], ! SubsetQ[ IntegerDigits[k, #], {0, 1}] &], k++]; k]; a /@ Range[2, 13] (* Giovanni Resta, May 24 2020 *)

Python

from numba import njit
@njit
def hasdigits01(n, b):
    has0, has1 = False, False
    while n >= b:
      n, r = divmod(n, b)
      if r == 0: has0 = True
      if r == 1: has1 = True
      if has0 and has1: return True
    return has0 and (has1 or n==1)
@njit
def a(n, start=2):
  k = start
  while True:
    for b in range(n, 1, -1):
      if not hasdigits01(k, b): break
    else: return k
    k += 1
anm1 = 2
for n in range(2, 21):
  an = a(n, start=anm1)
  print(an, end=", ")
  anm1 = an # Michael S. Branicky, Feb 09 2021

Extensions

a(29)-a(30) from Giovanni Resta, May 24 2020