A261011 Positive integers n such that ceiling(n^(1/3)) divides n.

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

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 = ceiling(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 * A354360 A333426 A118363

Adjacent sequences: A261008 A261009 A261010 * A261012 A261013 A261014

Keyword

nonn

Author

N. J. A. Sloane, Aug 14 2015

Status

approved