%I #15 Jan 18 2014 16:38:21
%S 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,
%T 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,
%U 2,2,2,3,2,2
%N Smallest zeroless number such that a(n)^n has at least one zero.
%C It is very probable that a(n) = 2 for n > 87.
%e 7 is the smallest number with nonzero digits such that 7^4 has at least one zero, so a(4) = 7.
%t 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 *)
%o (Python)
%o def f(x):
%o ..for n in range(10**7):
%o ....if str(n).find("0") == -1:
%o ......if str(n**x).find("0") > -1:
%o ........return n
%o x = 1
%o while x < 75:
%o ..if f(x) == None:
%o ....print(0)
%o ..else:
%o ....print(f(x))
%o ..x += 1
%Y Cf. A052382.
%K nonn,base
%O 2,1
%A _Derek Orr_, Jan 02 2014