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%I #16 Sep 05 2018 02:28:50
%S 16,52,101,252,1002,1001,10003,10002,10001,100003,100002,100001,
%T 1000003,1000002,1000001,10000003,10000002,10000001,100000003,
%U 100000002,100000001,1000000003,1000000002,1000000001,10000000003,10000000002,10000000001,100000000003,100000000002
%N Least number k not divisible by 10 such that k^3 contains n zeros.
%C It is believed that a(n) will have the pattern 1000...0003, 1000...0002, 1000...0001 after a(6).
%F For n > 6, a(n) = 10^(ceiling(n/3) + 1) + 3 - (n+2) mod 3.
%e 16 is not divisible by 10 and 16^3 = 4096, has 1 zero. So a(1) = 16.
%e 52 is not divisible by 10 and 52^3 = 140608, has 2 zeros. So a(2) = 52.
%o (Python)
%o def Cu(n):
%o ..for k in range(10**100):
%o ....if k % 10 != 0:
%o ......if str(k**3).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 ((d = digits(k^3)) && (((k % 10) == 0) || (sum(i=1, #d, d[i] == 0) != n)), k++); k;} \\ _Michel Marcus_, Apr 30 2014
%Y Cf. A134845.
%K nonn,base
%O 1,1
%A _Derek Orr_, Apr 23 2014
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