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).
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
Offset: 1
Keywords
Examples
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.
Links
- Marian Kraus, Table of n, a(n) for n = 1..9592
Crossrefs
Cf. A256567.
Programs
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PARI
a(n) = {my(p = prime(n)); sum(x=2, p-3, (x*(x+1)*(x+2)) % p == 1);} \\ Michel Marcus, Apr 03 2015
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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.
Formula
|T| where T={x|x*(x+1)*(x+2)==1 (mod p), p is prime, 1
Comments