

A242560


Least number k such that (n!k)/(nk) is an integer.


0



2, 1, 2, 2, 4, 3, 6, 4, 6, 5, 10, 6, 12, 7, 10, 8, 16, 9, 18, 10, 14, 11, 22, 12, 24, 13, 18, 14, 28, 15, 30, 16, 22, 17, 28, 18, 36, 19, 26, 20, 40, 21, 42, 22, 30, 23, 46, 24, 42, 25, 34, 26, 52, 27, 44, 28, 38, 29, 58, 30, 60, 31, 42, 32, 52, 33, 66, 34, 46, 35
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OFFSET

1,1


COMMENTS

It is conjectured that a(2n) = n.
The largest k such that (n!k)/(nk) is an integer is n!
a(2n+1) <= 2n for all n >= 1.


LINKS



EXAMPLE

61 is not divisible by 6!1, 62 is not divisible by 6!2, but 63 is divisible by 6!3. So a(6) = 3.


PROG

(PARI) a(n)=for(k=1, n!, if(k!=n, s = (n!k)/(nk); if(floor(s)==s, return(k)); ));
n=1; while(n<100, print(a(n)); n+=1) \\ corrected by Michel Marcus, Oct 24 2015


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



