A261417 -id:A261417 - cp's OEIS frontend

A261206 Numbers j such that ceiling(j^(1/k)) divides j for all integers k >= 1.

1, 2, 4, 6, 12, 36, 132, 144, 156, 900, 3600, 4032, 7140, 18360, 44100, 46440, 4062240, 9147600, 999999000000

Offset: 1

Max Alekseyev, Aug 11 2015

Is this a finite sequence?
It is possible to generalize this class of sequences by taking some integer-valued function f(j,k) decreasing in k such that f(j,1) = j and f(j,m) = c (for example, c=1 or c=2) for all sufficiently large m and considering those j that are divisible by all of f(j,1), f(j,2), ... If f(j,k) is slowly decreasing in k, then the set of corresponding j's is likely to have a very small number (if any) of terms, while if f(j,k) decreases rapidly, then there will be too many suitable j's. I believe the balance is achieved at functions like f(j,k) = floor(j^(1/k)) so that f(j,k) stabilizes to c at k ~= log(j). - Max Alekseyev, Aug 16 2015
If it exists, a(20) > 10^35. - Jon E. Schoenfield, Oct 17 2015

Cf. A261205, A261341, A261342.

Subsequence of all of A087811, A002620, A261011, A261417.

is(n) = my(k,t); if(n==1,return(1)); if(n%2,return(0)); k=2; while( (t=ceil((n-.5)^(1/k)))>2, if(n%t,return(0)); k++); 1
n=1;while(n<10^5,if(is(n),print1(n,", "));n++) /* Able to generate terms < 10^5 */ \\ Derek Orr, Aug 12 2015

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

Original entry on oeis.org

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

N. J. A. Sloane, Aug 14 2015

nonn

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

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

Suggested by A261205. Cf. A261417.

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

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;

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

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

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A261206 Numbers j such that ceiling(j^(1/k)) divides j for all integers k >= 1.

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

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