cp's OEIS Frontend

A238460 Primes p for which x! + (p-1)!/x!==0 (mod p) has only two solutions 1<=x<=p-2 following from Wilson theorem: x = 1 and x = p-2.

%I A238460 #29 Feb 28 2014 09:10:23
%S A238460 5,13,37,41,101,113,157,173,181,197,229,241,281,313,337,349,353,373,
%T A238460 409,421,433,509,541,617,677,701,757,761,769,773,821,929,941,977,997,
%U A238460 1013,1093,1097,1109,1181,1193,1237,1409,1433,1481,1489,1669,1693,1721,1741
%N A238460 Primes p for which x! + (p-1)!/x!==0 (mod p)  has only two solutions 1<=x<=p-2 following from Wilson theorem: x = 1 and x = p-2.
%C A238460 a(n) is prime(k(n)) for which A238444(k(n)) = 2.
%H A238460 Charles R Greathouse IV, <a href="/A238460/b238460.txt">Table of n, a(n) for n = 1..10000</a>
%F A238460 a(n) == 1 (mod 4).
%F A238460 Proof. Using Wilson's theorem, for every p>3, p==3(mod 4) we have, at least, 3 solution in [1,p-2] of x! + (p-1)!/x!==0 (mod p): x = 1, x = (p-1)/2, x = p-2.
%t A238460 A238444[n_] := a[n] = Module[{p, r}, p = Prime[n]; r = Range[p-2]; Count[r!+(p-1)!/r!, k_ /; Divisible[k, p]]]; A238460 = Prime /@ (Position[Table[A238444[n], {n, 1, 300}], 2] // Flatten) (* _Jean-François Alcover_, Feb 27 2014 *)
%o A238460 (PARI) is(p)=if(!isprime(p),return(0)); my(X=Mod(1,p),P=Mod((p-1)!,p));for(x=2,p-3,X*=x;P/=x;if(X+P==0,return(0))); p>3 \\ _Charles R Greathouse IV_, Feb 28 2014
%Y A238460 Cf. A238444.
%K A238460 nonn
%O A238460 1,1
%A A238460 _Vladimir Shevelev_, Feb 27 2014
%E A238460 More terms from _Peter J. C. Moses_, Feb 27 2014