

A241489


Least number k not divisible by 10 such that k^3 contains n zeros.


0



16, 52, 101, 252, 1002, 1001, 10003, 10002, 10001, 100003, 100002, 100001, 1000003, 1000002, 1000001, 10000003, 10000002, 10000001, 100000003, 100000002, 100000001, 1000000003, 1000000002, 1000000001, 10000000003, 10000000002, 10000000001, 100000000003, 100000000002
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OFFSET

1,1


COMMENTS

It is believed that a(n) will have the pattern 1000...0003, 1000...0002, 1000...0001 after a(6).


LINKS



FORMULA

For n > 6, a(n) = 10^(ceiling(n/3) + 1) + 3  (n+2) mod 3.


EXAMPLE

16 is not divisible by 10 and 16^3 = 4096, has 1 zero. So a(1) = 16.
52 is not divisible by 10 and 52^3 = 140608, has 2 zeros. So a(2) = 52.


PROG

(Python)
def Cu(n):
..for k in range(10**100):
....if k % 10 != 0:
......if str(k**3).count("0") == n:
........return k
n = 1
while n < 100:
..print(Cu(n))
..n += 1
(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


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



