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

2,1

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.

Links

Table of n, a(n) for n=2..94.

Example

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

Prog

(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

Crossrefs

Cf. A005054, A216099.

Sequence in context: A214359 A198810 A347532 * A289660 A096306 A335187

Adjacent sequences: A243908 A243909 A243910 * A243912 A243913 A243914

Keyword

nonn,base

Author

Derek Orr, Jun 14 2014

Status

approved