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

1,1

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

Links

Table of n, a(n) for n=1..66.

Example

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.

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 < 200:
..if not a(p):
....print(p, end=', ')
..p += 1

Crossrefs

Sequence in context: A043817 A030731 A247454 * A242883 A163621 A124194

Adjacent sequences: A243909 A243910 A243911 * A243913 A243914 A243915

Keyword

nonn,base

Author

Derek Orr, Jun 14 2014

Status

approved