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
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 14 2015
STATUS
approved