OFFSET
3,2
COMMENTS
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.]
LINKS
Jens Kruse Andersen, Table of n, a(n) for n = 3..101
EXAMPLE
(4!+4)/4 = 7 is prime and (4!-4)/4 = 5 is prime. Thus a(4) = 4.
PROG
(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++)
CROSSREFS
KEYWORD
nonn
AUTHOR
Derek Orr, Jul 29 2014
STATUS
approved