A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

7, 11, 17, 19, 23, 37, 43, 53, 59, 61, 67, 79, 83, 89, 97, 101, 103, 107, 109, 113, 137, 149, 157, 167, 173, 181, 191, 199, 211, 223, 227, 229, 241, 251, 263, 271, 281, 283, 293, 307, 313, 317, 337, 347, 359, 367, 373, 379, 383, 389, 401, 419, 421, 431, 433, 449

Offset: 1

Author

Marian Kraus, Apr 02 2015

Keywords

• nonn

Comments

There may be one or more such triples, but 23 is the only prime up to 100000 having precisely two such triples. For the number of triples for each prime, see A256572.

Together with 5, supersequence of A191065. - Arkadiusz Wesolowski, Nov 24 2021

Examples

			For p=7: 4*5*6=120==1 (mod 7), so 7 is a term.
For p=11: 5*6*7=210==1 (mod 11), so 11 is a term.
For p=17: 4*5*6=120==1 (mod 17), so 17 is a term.
13 is not a term because there is no such triple with product ==  1 (mod 13).
		

Links

• Marian Kraus, Table of n, a(n) for n = 1..6390

Crossrefs

Cf. A191065, A256572, A254678, A256580.

Programs

PARI

isok(p) = {if (isprime(p), for (x=1, p-3, if (Mod(x*(x+1)*(x+2), p) == 1, return (1));););} \\ Michel Marcus, Oct 05 2021

R

library(numbers)
IP <- vector()
t <- vector()
S <- vector()
IP <- c(Primes(1000)) # Build a vector of all primes < 1000.
for (j in 1:(length(IP))){
   for (i in 3:(IP[j]-2))
      t[i-1] <- as.vector(mod(((i-1)*i*(i+1)),IP[j]))
   S[j] <- length(which(t==1))
}
IP[S!=0]
#The loop checks for every triple for every prime, what it is modulo that prime. "IP[S!=0]" lists the primes that have at least one triple. For all p<10000 it takes a few minutes. For all p<100000 a few hours.