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A256572 Number of triples (x,x+1,x+2) with 1 < x <= p-3 of consecutive integers less than p whose product is 1 modulo p, where p = prime(n). 3
0, 0, 0, 1, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 3, 1, 1, 0, 0, 1, 1, 1, 1, 3, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 3, 3, 0, 1, 1, 0, 0, 1, 3, 3, 1, 1, 0, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 3, 0, 1, 3, 0, 1, 3, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 3, 1, 0, 3, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 3, 3, 0, 3, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 3, 1, 1, 3, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 3, 0, 3, 0, 1, 3, 1, 3, 0, 0, 0, 3, 1, 3, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 3, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,9

COMMENTS

Is 23 the only prime with two triples?

LINKS

Marian Kraus, Table of n, a(n) for n = 1..9592

FORMULA

|T| where T={x|x*(x+1)*(x+2)==1 (mod p), p is prime, 1<x<=p-3}.

EXAMPLE

For p=7: 4*5*6==1 (mod 7); T={4}, |T|=1.

For p=23: 2*3*4==1 (mod 23) and 9*10*11==1 (mod 23); T={2,9}, |T|=2.

For p=59: 3*4*5==1 (mod 59), 12*13*14==1 (mod 59), and 41*42*43==1 (mod 59); T={3,12,41}, |T|=3.

PROG

(R)

library(numbers); IP <- vector(); t <- vector(); S <- vector(); IP <- c(Primes(1000)); LIP <- length(IP); for (j in 1:LIP){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))}; S

#Needs a lot of memory. For Primes(100000), this takes a few hours.

(PARI) a(n) = {my(p = prime(n)); sum(x=2, p-3, (x*(x+1)*(x+2)) % p == 1); } \\ Michel Marcus, Apr 03 2015

CROSSREFS

Cf. A256567.

Sequence in context: A101669 A243067 A086965 * A025463 A327686 A274472

Adjacent sequences:  A256569 A256570 A256571 * A256573 A256574 A256575

KEYWORD

nonn

AUTHOR

Marian Kraus, Apr 02 2015

STATUS

approved

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Last modified September 30 18:12 EDT 2020. Contains 337440 sequences. (Running on oeis4.)