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A242567
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Least number k >= 0 such that (n!+k)/(n+k) is an integer.
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3
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1, 1, 0, 1, 18, 1, 712, 5031, 14, 1, 18, 1, 479001586, 1719, 87178291184, 1, 3024, 1, 40, 633, 124748, 1, 86, 51847, 625793187628, 123, 20404, 1, 210, 1, 265252859812191058636308479999968, 755, 263130836933693530167218012159999966
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OFFSET
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1,5
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COMMENTS
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a(n) = 1 iff n+1 is prime.
For n > 2, in order for (n!+k)/(n+k) to be an integer, the smallest integer possible is 2. Thus, a(n) <= n!-2n for all n > 2.
Let q = (n!+k)/(n+k). Then, k = (n!-n)/(q-1)-n. So, a(n) = d-n, where d is the smallest divisor (> n) of n!-n. (for all n >= 4) - Hiroaki Yamanouchi, Sep 29 2014
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LINKS
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EXAMPLE
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(6!+1)/(6+1) = 103 is an integer. Thus a(6) = 1.
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PROG
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a(n)=for(k=1, 10^5, s=(n!+k)/(n+k); if(floor(s)==s, return(k)));
n=1; while(n<100, print(a(n)); n+=1)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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