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A261011 Positive integers n such that ceiling(n^(1/3)) divides n. 3
1, 2, 4, 6, 8, 9, 12, 15, 18, 21, 24, 27, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 217, 224, 231, 238, 245, 252, 259 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Positive terms of A000578 (cubes) are in the sequence. - Michel Marcus, Aug 15 2015

Theorem: The sequence consists precisely of the numbers k^3+1+i*(k+1) for k >= 0, 0 <= i <= 3*k.

Proof: k^3+1 <= n <= (k+1)^3 iff k+1 = ceil(n^(1/3)). So n must be of the form k^3+1+i*(k+1) with 0 <= i <= 3*k, and both endpoints work. QED - N. J. A. Sloane, Aug 27 2015

LINKS

Chai Wah Wu, Table of n, a(n) for n = 1..1000

MAPLE

p:=3; a:=[]; M:=200; Digits:=30;

for n from 1 to M do

# is n a p-th power?

t1:=round(evalf(n^(1/p)));

if t1^p = n then a:=[op(a), n];

else t2:=ceil(evalf(n^(1/p)));

      if (n mod t2) = 0 then a:=[op(a), n]; fi;

fi;

od:

a;

PROG

(Python)

from gmpy2 import iroot

A261011_list = [n for n in range(1, 10**5) if not n % (lambda x:x[0] + (0 if x[1] else 1))(iroot(n, 3))] # Chai Wah Wu, Aug 14 2015

(MAGMA) [n: n in [1..400] | n mod Ceiling((n^(1/3))) eq 0 ]; // Vincenzo Librandi, Aug 15 2015

(PARI) is(n)=n%ceiling(n^(1/3))==0 \\ Anders Hellström, Aug 15 2015

CROSSREFS

Suggested by A261205. Cf. A261417.

Sequence in context: A071562 A241562 A100345 * A118363 A146982 A298305

Adjacent sequences:  A261008 A261009 A261010 * A261012 A261013 A261014

KEYWORD

nonn

AUTHOR

N. J. A. Sloane, Aug 14 2015

STATUS

approved

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Last modified January 16 15:53 EST 2019. Contains 319195 sequences. (Running on oeis4.)