A241489 Least number k not divisible by 10 such that k^3 contains n zeros.

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

Offset: 1

Author

Derek Orr, Apr 23 2014

Keywords

• nonn
• base

Comments

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

Examples

			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.
		

Crossrefs

Cf. A134845.

Programs

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

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

Formula

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