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A245697
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Least number k such that (n!+k)/n and (n!-k)/n are both prime.
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3
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0, 4, 25, 42, 133, 152, 279, 170, 121, 204, 1079, 938, 5295, 3632, 2771, 1062, 1159, 2260, 7413, 682, 33281, 13704, 9725, 4966, 9099, 24724, 2929, 54690, 20429, 22688, 5379, 46274, 15365, 11052, 40441, 65854, 97149, 42520, 44731, 83958, 61877, 4796, 123885, 27922, 122999, 12912, 5047
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OFFSET
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3,2
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COMMENTS
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a(n) < n! for all n > 2.
It is believed that a(n) exists for all n > 2.
a(n) = n times (least m such that (n-1)!+m and (n-1)!-m are both prime) = n*A075409(n-1). - Jens Kruse Andersen, Jul 30 2014 [Goldbach's conjecture would then imply that a(n) always exists.]
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LINKS
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EXAMPLE
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(4!+4)/4 = 7 is prime and (4!-4)/4 = 5 is prime. Thus a(4) = 4.
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PROG
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(PARI)
a(n)=for(k=0, 10^7, s1=(n!-k)/n; s2=(n!+k)/n; if(floor(s1)==s1&&floor(s2)==s2, if(ispseudoprime(s1)&&ispseudoprime(s2), return(k))))
n=3; while(n<100, print1(a(n), ", "); n++)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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