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A261206
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Numbers n such that ceiling(n^(1/k)) divides n for all integers k>=1.
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4
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1, 2, 4, 6, 12, 36, 132, 144, 156, 900, 3600, 4032, 7140, 18360, 44100, 46440, 4062240, 9147600, 999999000000
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,more,nice
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AUTHOR
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STATUS
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approved
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