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

Author

N. J. A. Sloane, Aug 14 2015

Keywords

• nonn

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

Crossrefs

Suggested by A261205. Cf. A261417.

Programs

Magma

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

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;

PARI

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

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