A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

7, 17, 23, 31, 41, 47, 73, 89, 97, 103, 127, 137, 151, 167, 199, 223, 233, 239, 241, 257, 271, 281, 311, 313, 353, 359, 367, 383, 409, 431, 433, 439, 449, 479, 487, 503, 521, 577, 593, 601, 607, 647, 673, 719, 727, 743, 751, 761, 769, 839, 857, 881, 887, 911, 929, 937, 953, 967, 977, 983

Offset: 1

Author

Marian Kraus, Apr 02 2015

Keywords

• nonn

Examples

			p=7: 2*3*4*5=120 == 1 mod 7;
p=17: 2*3*4*5=120 == 1 mod 17 AND 12*13*14*15=32760 == 1 mod 17; for p=13: no triple == 1 mod 13;
p=23: 5*6*7*8 == 1 mod 23 AND 15*16*17*18== 1 mod 23 AND 19*20*21*22 == 1 mod 23; and so on. For the number of quadruples for a prime, see A256580.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A256567, A256580.

Programs

Mathematica

fsiQ[n_]:=AnyTrue[Times@@@Partition[Range[n-1],4,1],Mod[#,n]==1&]; Select[ Prime[Range[200]],fsiQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 02 2019 *)

PARI

lista(nn) = forprime(p=2, nn, if (sum(x=1, p-4, ((x*(x+1)*(x+2)*(x+3)) % p) == 1) > 0, print1(p, ", "))); \\ Michel Marcus, Apr 03 2015

R

library(numbers)
IP <- vector()
t <- vector()
S <- vector()
IP <- c(Primes(1000))
for (j in 1:(length(IP))){
   for (i in 2:(IP[j]-4)){
       t[i-1] <- as.vector(mod((i*(i+1)*(i+2)*(i+3)),IP[j]))
       Z[j] <- sum(which(t==1))
       S[j] <- length(which(t==1))
   }
}
IP[S!=0]
#Carefully increase Primes(1000). It takes several hours for 100000.

Formula

x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 <= x <= p-4.