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%I #10 Jul 30 2014 09:19:35
%S 0,4,25,42,133,152,279,170,121,204,1079,938,5295,3632,2771,1062,1159,
%T 2260,7413,682,33281,13704,9725,4966,9099,24724,2929,54690,20429,
%U 22688,5379,46274,15365,11052,40441,65854,97149,42520,44731,83958,61877,4796,123885,27922,122999,12912,5047
%N Least number k such that (n!+k)/n and (n!-k)/n are both prime.
%C a(n) < n! for all n > 2.
%C It is believed that a(n) exists for all n > 2.
%C 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.]
%H Jens Kruse Andersen, <a href="/A245697/b245697.txt">Table of n, a(n) for n = 3..101</a>
%e (4!+4)/4 = 7 is prime and (4!-4)/4 = 5 is prime. Thus a(4) = 4.
%o (PARI)
%o 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))))
%o n=3;while(n<100,print1(a(n),", ");n++)
%Y Cf. A075409, A245695, A245696.
%K nonn
%O 3,2
%A _Derek Orr_, Jul 29 2014
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