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A244606
Least number k > 1 such that k^n contains the digit k k times, or 0 if no such digit exists.
0
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 5, 3, 2, 7, 0, 0, 5, 3, 6, 4, 0, 3, 5, 3, 0, 2, 6, 5, 3, 0, 0, 0, 6, 2, 3, 4, 2, 4, 2, 2, 2, 3, 4, 3, 5, 2, 2, 4, 5, 2, 2, 0, 4, 0, 3, 7, 2, 5, 3, 4, 0, 4, 2, 4, 2, 7, 7, 7, 2, 0, 3, 2, 8, 6, 6, 2, 0, 3, 7, 2, 4, 0, 6, 0, 0, 0, 8, 5, 4, 3, 0, 0, 6, 5, 2, 5, 0, 8, 3
OFFSET
1,18
COMMENTS
1 < a(n) < 10 if a(n) is not 0. Thus a(n) = 0 is definite for the n-values above.
EXAMPLE
2^18, 2^19 and 2^21 all contain the digit 2 twice. So a(18) = a(19) = a(21) = 2.
PROG
(Python)
def tes(n):
..for k in range(2, 10):
....if str(k**n).count(str(k)) == k:
......return k
n = 1
while n < 200:
..if tes(n):
....print(tes(n), end=', ')
..else:
....print(0, end=', ')
..n += 1
CROSSREFS
Sequence in context: A141720 A353449 A248017 * A273127 A103272 A065710
KEYWORD
nonn,base,easy
AUTHOR
Derek Orr, Jul 01 2014
STATUS
approved