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A244606
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Least number k > 1 such that k^n contains the digit k k times, or 0 if no such digit exists.
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0
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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
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OFFSET
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1,18
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COMMENTS
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1 < a(n) < 10 if a(n) is not 0. Thus a(n) = 0 is definite for the n-values above.
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LINKS
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EXAMPLE
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2^18, 2^19 and 2^21 all contain the digit 2 twice. So a(18) = a(19) = a(21) = 2.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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