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%I A243152 #13 Jun 01 2014 00:10:11
%S A243152 1,1,6,3,9,4,1,1,6,0,8,5,4,8,7,1,4,1,6,0,6,3,8,9,3,4,9,6,7,6,7,4,6,0,
%T A243152 3,9,6,0,9,5,1,3,1,6,2,7,7,9,4,8,9,2,7,7,7,9,9,4,9,9,1,0,9,5,9,5,5,6,
%U A243152 4,3,2,6,3,7,4,7,2,0,1,2,8,1,9,6,9,9,1,9,3,0
%N A243152 a(n) = the digit that is repeated exactly n times in the string A243151(n)^n.
%C A243152 In the event of a tie, choose the smaller integer.
%e A243152 A243151(2) = 11. So A243151(2)^2 = 121. Since the digit 1 is the one repeated two times, a(2) = 1.
%o A243152 (Python)
%o A243152 def c(n):
%o A243152 ..for k in range(10**7):
%o A243152 ....if k % 10 == 0:
%o A243152 ......lst = []
%o A243152 ......count = 0
%o A243152 ......for i in range(10):
%o A243152 ........if str(k**n).count(str(i)) == n:
%o A243152 ..........return i
%o A243152 n = 1
%o A243152 while n < 100:
%o A243152 ..print(c(n),end=', ')
%o A243152 ..n+=1
%Y A243152 Cf. A243151.
%K A243152 nonn,base
%O A243152 1,3
%A A243152 _Derek Orr_, May 31 2014

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