OFFSET
1,1
COMMENTS
It is believed that a(n) will have the pattern 1000...0003, 1000...0002, 1000...0001 after a(6).
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
Derek Orr, Apr 23 2014
STATUS
approved