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%I #12 Apr 27 2014 16:34:47
%S 16,52,126,252,3138,5852,58752,71138,493352,1916568,11696559,58633193,
%T 191293929,464296543,386826983,5886958939,46493141317,115356679131,
%U 79633784516,2154578383152,6694429222569
%N Least zeroless number k such that k^3 contains n zeros.
%e 16 does not have a 0 but 16^3 = 4096 has 1 zero. So, a(1) = 16.
%e 52 does not have a 0 but 52^3 = 140608 has 2 zeros. So, a(2) = 52.
%o (Python)
%o def Cu(n):
%o ..k = 0
%o ..while k < 10**50:
%o ....if str(k).count("0") > 0:
%o ......c = []
%o ......d = ''
%o ......for i in list(str(k).partition("0")):
%o ........if int(i) == 0:
%o ..........c.append('1'*len(i))
%o ........else:
%o ..........c.append(i)
%o ......for j in c:
%o ........d += j
%o ......k = int(d)
%o ....if str(k**3).count("0") == n:
%o ......return k
%o ....else:
%o ......k += 1
%o n = 1
%o while n < 50:
%o ..print(Cu(n))
%o ..n += 1
%Y Cf. A052382.
%K nonn,more,base,hard
%O 1,1
%A _Derek Orr_, Apr 23 2014
%E a(17)-a(21) from _Giovanni Resta_, Apr 27 2014
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