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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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..29.

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

Cf. A134845.

Sequence in context: A235660 A044118 A044499 * A009953 A009939 A009935

Adjacent sequences: A241486 A241487 A241488 * A241490 A241491 A241492

KEYWORD

nonn,base

AUTHOR

Derek Orr, Apr 23 2014

STATUS

approved

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Last modified February 6 23:12 EST 2023. Contains 360111 sequences. (Running on oeis4.)