

A238460


Primes p for which x! + (p1)!/x!==0 (mod p) has only two solutions 1<=x<=p2 following from Wilson theorem: x = 1 and x = p2.


3



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
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OFFSET

1,1


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


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,p2] of x! + (p1)!/x!==0 (mod p): x = 1, x = (p1)/2, x = p2.


MATHEMATICA

A238444[n_] := a[n] = Module[{p, r}, p = Prime[n]; r = Range[p2]; Count[r!+(p1)!/r!, k_ /; Divisible[k, p]]]; A238460 = Prime /@ (Position[Table[A238444[n], {n, 1, 300}], 2] // Flatten) (* JeanFrançois Alcover, Feb 27 2014 *)


PROG

(PARI) is(p)=if(!isprime(p), return(0)); my(X=Mod(1, p), P=Mod((p1)!, p)); for(x=2, p3, X*=x; P/=x; if(X+P==0, return(0))); p>3 \\ Charles R Greathouse IV, Feb 28 2014


CROSSREFS

Cf. A238444.
Sequence in context: A266102 A288180 A141408 * A107144 A137815 A089523
Adjacent sequences: A238457 A238458 A238459 * A238461 A238462 A238463


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Feb 27 2014


EXTENSIONS

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


STATUS

approved



