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A234967
Smallest zeroless number such that a(n)^n has at least one zero.
0
32, 16, 7, 4, 4, 8, 5, 6, 2, 2, 2, 4, 5, 3, 3, 2, 3, 4, 2, 2, 2, 2, 4, 3, 2, 4, 4, 2, 2, 4, 3, 3, 4, 3, 3, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2
OFFSET
2,1
COMMENTS
It is very probable that a(n) = 2 for n > 87.
EXAMPLE
7 is the smallest number with nonzero digits such that 7^4 has at least one zero, so a(4) = 7.
MATHEMATICA
m[n_] := Min@ IntegerDigits@n; a[1]=0; a[n_] := Block[{k=2}, While[m[k] == 0 || m[k^n] > 0, k++]; k]; Array[a, 70] (* Giovanni Resta, Jan 11 2014 *)
PROG
(Python)
def f(x):
..for n in range(10**7):
....if str(n).find("0") == -1:
......if str(n**x).find("0") > -1:
........return n
x = 1
while x < 75:
..if f(x) == None:
....print(0)
..else:
....print(f(x))
..x += 1
CROSSREFS
Cf. A052382.
Sequence in context: A097614 A069892 A070621 * A033352 A364904 A140387
KEYWORD
nonn,base
AUTHOR
Derek Orr, Jan 02 2014
STATUS
approved