

A261206


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


4



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

1,2


COMMENTS

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


LINKS

Table of n, a(n) for n=1..19.


PROG

(PARI) 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


CROSSREFS

Cf. A261205, A261341, A261342.
Subsequence of all of A087811, A002620, A261011, A261417.
Sequence in context: A282193 A180213 A081457 * A127105 A196444 A307617
Adjacent sequences: A261203 A261204 A261205 * A261207 A261208 A261209


KEYWORD

nonn,more,nice


AUTHOR

Max Alekseyev, Aug 11 2015


STATUS

approved



