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%I #4 Jul 01 2014 23:04:31
%S 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,
%T 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,
%U 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
%N Least number k > 1 such that k^n contains the digit k k times, or 0 if no such digit exists.
%C 1 < a(n) < 10 if a(n) is not 0. Thus a(n) = 0 is definite for the n-values above.
%e 2^18, 2^19 and 2^21 all contain the digit 2 twice. So a(18) = a(19) = a(21) = 2.
%o (Python)
%o def tes(n):
%o ..for k in range(2,10):
%o ....if str(k**n).count(str(k)) == k:
%o ......return k
%o n = 1
%o while n < 200:
%o ..if tes(n):
%o ....print(tes(n),end=', ')
%o ..else:
%o ....print(0,end=', ')
%o ..n += 1
%Y Cf. A244604, A244603.
%K nonn,base,easy
%O 1,18
%A _Derek Orr_, Jul 01 2014
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