A241490 Least zeroless number k such that k^3 contains n zeros.

1, 16, 52, 126, 252, 3138, 5852, 58752, 71138, 493352, 1916568, 11696559, 58633193, 191293929, 464296543, 386826983, 5886958939, 46493141317, 115356679131, 79633784516, 2154578383152, 6694429222569

Offset

0,2

Links

Table of n, a(n) for n=0..21.

Example

16 does not have a 0 but 16^3 = 4096 has 1 zero. So, a(1) = 16.
52 does not have a 0 but 52^3 = 140608 has 2 zeros. So, a(2) = 52.

Prog

(Python)
def Cu(n):
  k = 0
  while k < 10**50:
    if str(k).count("0") > 0:
      c = []
      d = ''
      for i in list(str(k).partition("0")):
        if len(i) and int(i) == 0:
          c.append('1'*len(i))
        else:
          c.append(i)
      for j in c:
        d += j
      k = int(d)
    if str(k**3).count("0") == n:
      return k
    else:
      k += 1
n = 0
while n < 50:
  print(Cu(n))
  n += 1
(Python)
def zeroless():
    yield from range(1, 10)
    for z in zeroless():
        yield from range(10*z + 1, 10*z + 10)
def A241490(n):
    return next(z for z in zeroless() if str(z*z*z).count('0') == n)
# David Radcliffe, Jun 25 2025

Crossrefs

Cf. A000578, A052382.

Sequence in context: A381109 A009939 A009935 * A009931 A009936 A217736

Adjacent sequences: A241487 A241488 A241489 * A241491 A241492 A241493

Keyword

nonn,more,base,hard

Author

Derek Orr, Apr 23 2014

Extensions

a(17)-a(21) from Giovanni Resta, Apr 27 2014

a(0)=1 prepended by Alois P. Heinz, Jun 25 2025

Status

approved