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A256580
Number of quadruples (x, x+1, x+2, x+3) with 1 < x < p-3 of consecutive integers whose product is 1 mod p.
3
0, 0, 0, 1, 0, 0, 2, 0, 3, 0, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 2, 0, 0, 0, 4, 0, 4, 0, 0, 4, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 4, 4, 2, 0, 2, 0, 0, 2, 0, 4, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 2, 4, 2, 0, 0, 2, 0, 0, 0, 2, 0, 0, 4, 2, 2, 0, 4, 0, 0, 0, 0, 2, 4, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 2, 2, 0, 0, 4, 4, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 4, 0, 2, 2, 0, 0, 4, 4, 0, 4, 2, 0, 0
OFFSET
1,7
COMMENTS
If "quadruples" is changed to "pairs" we get A086937 (for the counts) and A038872 (for the primes for which the count is nonzero).
FORMULA
|T| where T = {x|x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 < x < p-3}.
EXAMPLE
p=7, x_1=2, 2*3*4*5 == 1 (mod 7), T={2}, |T|=1;
p=17, x_1=2, 2*3*4*5 == 1 (mod 17), x_2=12, 12*13*14*15 == 1 (mod 17), T={2,12}, |T|=2;
p=23, x_1=5, 5*6*7*8 == 1 (mod 23), x_2=15, 15*16*17*18 == 1 (mod 23), x_3=19, 19*20*21*22 == 1 (mod 23), T={5,15,19}, |T|=3.
PROG
(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))}}; S
CROSSREFS
KEYWORD
nonn
AUTHOR
Marian Kraus, Apr 02 2015
STATUS
approved