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A241490
Least zeroless number k such that k^3 contains n zeros.
0
16, 52, 126, 252, 3138, 5852, 58752, 71138, 493352, 1916568, 11696559, 58633193, 191293929, 464296543, 386826983, 5886958939, 46493141317, 115356679131, 79633784516, 2154578383152, 6694429222569
OFFSET
1,1
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 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 = 1
while n < 50:
..print(Cu(n))
..n += 1
CROSSREFS
Cf. A052382.
Sequence in context: A381109 A009939 A009935 * A009931 A009936 A217736
KEYWORD
nonn,more,base,hard
AUTHOR
Derek Orr, Apr 23 2014
EXTENSIONS
a(17)-a(21) from Giovanni Resta, Apr 27 2014
STATUS
approved