cp's OEIS Frontend

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

%I A261011 #28 Sep 08 2022 08:46:13
%S A261011 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,
%T A261011 80,85,90,95,100,105,110,115,120,125,126,132,138,144,150,156,162,168,
%U A261011 174,180,186,192,198,204,210,216,217,224,231,238,245,252,259
%N A261011 Positive integers n such that ceiling(n^(1/3)) divides n.
%C A261011 Positive terms of A000578 (cubes) are in the sequence. - _Michel Marcus_, Aug 15 2015
%C A261011 Theorem: The sequence consists precisely of the numbers k^3+1+i*(k+1) for k >= 0, 0 <= i <= 3*k.
%C A261011 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
%H A261011 Chai Wah Wu, <a href="/A261011/b261011.txt">Table of n, a(n) for n = 1..1000</a>
%p A261011 p:=3; a:=[]; M:=200; Digits:=30;
%p A261011 for n from 1 to M do
%p A261011 # is n a p-th power?
%p A261011 t1:=round(evalf(n^(1/p)));
%p A261011 if t1^p = n then a:=[op(a),n];
%p A261011 else t2:=ceil(evalf(n^(1/p)));
%p A261011       if (n mod t2) = 0 then a:=[op(a),n]; fi;
%p A261011 fi;
%p A261011 od:
%p A261011 a;
%o A261011 (Python)
%o A261011 from gmpy2 import iroot
%o A261011 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
%o A261011 (Magma) [n: n in [1..400] | n mod Ceiling((n^(1/3))) eq 0 ]; // _Vincenzo Librandi_, Aug 15 2015
%o A261011 (PARI) is(n)=n%ceiling(n^(1/3))==0 \\ _Anders Hellström_, Aug 15 2015
%Y A261011 Suggested by A261205. Cf. A261417.
%K A261011 nonn
%O A261011 1,2
%A A261011 _N. J. A. Sloane_, Aug 14 2015