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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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Is this a finite sequence?

It is possible to generalize this class of sequences by taking some integer-valued 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

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Last modified September 21 08:38 EDT 2020. Contains 337268 sequences. (Running on oeis4.)