A243912 Numbers n > 1 such that n^k never ends in 2 or more identical digits for any k.

3, 5, 6, 7, 9, 15, 16, 18, 21, 23, 24, 25, 26, 27, 29, 32, 35, 36, 41, 43, 45, 46, 47, 49, 51, 56, 57, 61, 63, 65, 67, 68, 69, 74, 75, 76, 81, 82, 83, 85, 86, 87, 89, 93, 95, 96, 101, 103, 105, 106, 107, 109, 115, 116, 118, 121, 123, 124, 125, 126, 127, 129, 132, 135, 136, 141

Offset: 1

Author

Derek Orr, Jun 14 2014

Keywords

• nonn
• base

Comments

If n = 5^k for some k > 0, then n is in this sequence.

If n is a repdigit, then n is not in this sequence.

For n > 1, n is in this sequence iff n == {1, 3, 5, 6, 7, 9, 15, 16, 18, 21, 23, 24, 25, 26, 27, 29, 32, 35, 36, 41, 43, 45, 46, 47, 49, 51, 56, 57, 61, 63, 65, 67, 68, 69, 74, 75, 76, 81, 82, 83, 85, 86, 87, 89, 93, 95, 96} mod 100

Examples

			5^k ends in 25 for all k > 1. Thus it will never end in any number of identical digits, and 5 is a member of this sequence.
		

Programs

Python

def b(n,p):
  lst = []
  count = 0
  lst1 = []
  for i in range(1,5**(n+2)):
    st = str(p**i)
    if len(st) >= n:
      if int(st[len(st)-n:len(st)]) not in lst:
        lst.append(int(st[len(st)-n:len(st)]))
        lst1.append(i)
      else:
        return len(lst)+min(lst1)
def a(p):
  for i in range(1,b(2,p)+2):
    st = str(p**i)
    if int(st[len(st)-2:len(st)])%11==0:
      return i
p = 2
while p < 200:
  if not a(p):
    print(p,end=', ')
  p += 1