

A256567


Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p3 whose product is 1 modulo p.


5



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, 457, 463, 467, 479, 503, 521, 523, 557, 563, 569, 571, 593, 599, 607, 613, 617, 619, 631, 641, 643, 659, 661, 677, 691, 701, 709, 719, 727, 733, 743, 751, 757, 769, 773, 787, 797, 809, 821, 827, 829, 839, 853, 877, 881, 883, 907, 911, 919, 937, 941, 953, 971, 977, 983, 991, 997
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OFFSET

1,1


LINKS

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


EXAMPLE

For p=7: 4*5*6=120==1 (mod 7); for p=11: 5*6*7=210==1 (mod 11); for p=13: there is no such triple with product equivalent to 1 (mod 13); for p=17: 4*5*6=120==1 (mod 17); and so on.
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.


PROG

(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[i1] < as.vector(mod(((i1)*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.


CROSSREFS

Cf. A256572, A254678, A256580.
Sequence in context: A191055 A078497 A274505 * A155048 A296926 A063639
Adjacent sequences: A256564 A256565 A256566 * A256568 A256569 A256570


KEYWORD

nonn


AUTHOR

Marian Kraus, Apr 02 2015


STATUS

approved



