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.

5, 13, 37, 41, 101, 113, 157, 173, 181, 197, 229, 241, 281, 313, 337, 349, 353, 373, 409, 421, 433, 509, 541, 617, 677, 701, 757, 761, 769, 773, 821, 929, 941, 977, 997, 1013, 1093, 1097, 1109, 1181, 1193, 1237, 1409, 1433, 1481, 1489, 1669, 1693, 1721, 1741

Offset: 1

Author

Vladimir Shevelev, Feb 27 2014

Keywords

• nonn

Comments

a(n) is prime(k(n)) for which A238444(k(n)) = 2.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A238444.

Programs

Mathematica

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 *)

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

Formula

a(n) == 1 (mod 4).

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.

Extensions

More terms from Peter J. C. Moses, Feb 27 2014