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

Author

Max Alekseyev, Aug 11 2015

Keywords

• nonn
• more
• nice

Comments

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

Crossrefs

Cf. A261205, A261341, A261342.

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

Programs

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