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A261206
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Numbers j such that ceiling(j^(1/k)) divides j 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(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
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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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