A261011 -id:A261011 - 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

A261417 Numbers n such that both ceiling(sqrt(n)) and ceiling(n^(1/3)) divide n.

Original entry on oeis.org

1, 2, 4, 6, 9, 12, 36, 56, 64, 90, 100, 110, 132, 144, 156, 210, 400, 576, 702, 729, 870, 900, 930, 1056, 1089, 1122, 1332, 1560, 2352, 2450, 2970, 3600, 4032, 4096, 4556, 4624, 4692, 5112, 5184, 5256, 5852, 7140, 8190, 9702, 9900, 12432, 14400, 15500, 15625, 16770, 16900, 17030, 18090, 18225, 18360, 19740

Offset: 1

N. J. A. Sloane, Aug 26 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..229

Intersection of A002620 and A261011. Contains A261206 as a subsequence.

[n: n in [1..2000] | n mod Ceiling((n^(1/2))) eq 0 and n mod Ceiling((n^(1/3))) eq 0 ];

Select[Range[200000], Mod[#, Ceiling[#^(1/2)]] == Mod[#, Ceiling[#^(1/3)]] == 0 &] (* Vincenzo Librandi, Aug 21 2016 *)

is(n) = Mod(n, ceil(sqrt(n)))==0 && Mod(n, ceil(n^(1/3)))==0 \\ Felix Fröhlich, Aug 21 2016

cp's OEIS Frontend

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

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