A243911 Least number k such that n^k ends in two identical digits, or 0 if no such number exists.

18, 0, 9, 0, 0, 0, 6, 0, 2, 1, 2, 17, 3, 0, 0, 11, 0, 5, 2, 0, 1, 0, 0, 0, 0, 0, 14, 0, 2, 7, 0, 1, 7, 0, 0, 7, 2, 5, 2, 0, 3, 0, 1, 0, 0, 0, 9, 0, 2, 0, 18, 3, 9, 1, 0, 0, 6, 5, 2, 0, 2, 0, 3, 0, 1, 0, 0, 0, 2, 3, 7, 11, 0, 0, 0, 1, 14, 5, 2, 0, 0, 0, 7, 0, 0, 0, 1, 0, 2, 9, 3, 0, 11

Offset: 2

Author

Derek Orr, Jun 14 2014

Keywords

• nonn
• base

Comments

For all n > 1, the 2-digit ending of n^k repeats itself after a certain k-value. Thus a(n) = 0 is definite.

a(10*n) = 2 for all n > 0. Thus there are infinitely many nonzero entries. a(5^n) = 0 for all n > 0. Thus there are infinitely many zero entries.

Examples

			2^18 = 262144 ends in two of the same digit. Thus a(2) = 18.
		

Crossrefs

Cf. A005054, A216099.

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 < 100:
  if a(p):
    print(a(p),end=', ')
  else:
    print(0,end=', ')
  p += 1