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%I #15 May 04 2014 18:09:09
%S 101,101,101,101,101,351,1244,194,4648,951,4357,3757,2169,2392,7399,
%T 7501,9723,8683,13867,6152,15204,18898,40141,54631,29647,35586,46564,
%U 67743,84789,119421,43055,43642,83055,44411,142553,94501,135852,52299,174062,121201,196205
%N Least number k not divisible by 10 such that k^n contains n zeros.
%e 101 is not divisible by 10 and 101^1 (101), 101^2 (10201), 101^3 (1030301), 101^4 (104060401), and 101^5 (10510100501) all have 1, 2, 3, 4, and 5 zeros, respectively. So, a(1) = a(2) = a(3) = a(4) = a(5) = 101.
%o (Python)
%o def Cu(n):
%o ..for k in range(10**10):
%o ....if k% 10 != 0:
%o ......if str(k**n).count("0") == n:
%o ........return k
%o n = 1
%o while n < 100:
%o ..print(Cu(n))
%o ..n += 1
%o (PARI) a(n) = {k = 1; while (((k % 10) == 0) || (d = digits(k^n)) && (sum(i=1, #d, d[i] == 0) != n), k++); k;} \\ _Michel Marcus_, Apr 30 2014
%Y Cf. A233821.
%K nonn,base
%O 1,1
%A _Derek Orr_, Apr 24 2014
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