%I #20 Mar 05 2014 03:51:07
%S 7,11,19,31,43,47,107,127,131,151,163,167,179,191,211,223,263,283,347,
%T 367,443,487,491,523,547,587,643,659,751,827,839,911,1039,1051,1087,
%U 1103,1123,1163,1171,1223,1259,1283,1291,1327,1427,1439,1447,1487,1523,1543
%N Primes p for which x! + (p-1)!/x!==0 (mod p) has only three solutions 1<=x<=p-2.
%C All terms are of the form 4*k+3.
%C 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.
%F a(n) is prime(k(n)) for which A238444(k(n)) = 3.
%t 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 *)
%Y Cf. A238444, A238460.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Feb 27 2014
%E More terms from _Peter J. C. Moses_, Feb 27 2014
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