A342755 a(1) = 2; for n > 1, a(n) is the least positive integer not occurring earlier such that a(n) shares no digit with a(n-1) and a(n-1)*a(n) shares no digit with either a(n-1) or a(n).

2, 3, 4, 5, 6, 7, 8, 9, 42, 15, 22, 14, 55, 12, 37, 16, 25, 36, 29, 47, 23, 46, 13, 44, 18, 32, 17, 38, 19, 33, 26, 35, 174, 53, 76, 59, 34, 27, 43, 67, 49, 62, 87, 106, 493, 57, 24, 75, 48, 65, 122, 39, 54, 72, 88, 45, 66, 73, 56, 77, 52, 79, 84, 63, 78, 123, 69, 58, 64, 92, 74, 68, 114, 85, 314

Offset: 1

Author

Scott R. Shannon, Mar 20 2021

Keywords

• nonn
• base

Comments

No term can end in 0 or 1 as that would result in the last digit of a(n-1)*a(n) being the same as a(n)'s last digit. Currently the last known term is a(173) = 922989, the next being at least 5*10^10 if it exists. It is unknown if the sequence is infinite.

a(174) = 60060666070067077 and a(175) has 52 digits (see b-file). If a(176) exists, it is > 10^71. - Michael S. Branicky, Apr 10 2023

Examples

			a(2) = 3 as 3 shares no digit with a(1) = 2 and a(1)*3 = 2*3 = 6 shares no digit with a(1) = 2 or 3.
a(9) = 42 as 42 shares no digit with a(8) = 9 and a(8)*42 = 9*42 = 378 shares no digit with a(8) = 9 or 42.
a(10) = 15 as 15 shares no digit with a(9) = 42 and a(9)*15 = 42*15 = 630 shares no digit with a(9) = 42 or 15. This is the first term that differs from A342442.
a(173) = 922989 as 922989 shares no digit with a(172) = 7154 and a(172)*922989 = 7154*922989 = 6603063306 shares no digit with a(172) = 7154 or 922989. This is currently the last known term.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..175

Crossrefs

Cf. A342442, A342441, A276633, A010784, A043537, A043096, A338466, A336285.

Programs

Python

def aupton(terms):
  alst, aset = [2], {2}
  while len(alst) < terms:
    an, anm1_digs = 2, set(str(alst[-1]))
    while True:
      while an in aset: an += 1
      an_digs = set(str(an))
      if (an_digs & anm1_digs) == set():
        prod_digs = set(str(an*alst[-1]))
        if (anm1_digs | an_digs) & prod_digs  == set():
          alst.append(an); aset.add(an); break
      an += 1
  return alst
print(aupton(173)) # Michael S. Branicky, Mar 21 2021