OFFSET
1,1
COMMENTS
All terms are of the form 4*k+3.
Using Wilson's theorem, for every p>3, p==3(mod 4) we have, at least, 3 solutions in [1,p-2] of x! + (p-1)!/x!==0 (mod p): x = 1, x = (p-1)/2, x = p-2.
FORMULA
a(n) is prime(k(n)) for which A238444(k(n)) = 3.
MATHEMATICA
kmax = 400; Select[Select[4*Range[kmax]+3, PrimeQ], (r = Range[#-2]; Count[r!+(#-1)!/r!, k_ /; Divisible[k, #]] == 3)&] (* Jean-François Alcover, Mar 05 2014 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Feb 27 2014
EXTENSIONS
More terms from Peter J. C. Moses, Feb 27 2014
STATUS
approved