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%I #43 Jul 02 2019 17:11:58
%S 7,17,23,31,41,47,73,89,97,103,127,137,151,167,199,223,233,239,241,
%T 257,271,281,311,313,353,359,367,383,409,431,433,439,449,479,487,503,
%U 521,577,593,601,607,647,673,719,727,743,751,761,769,839,857,881,887,911,929,937,953,967,977,983
%N Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.
%H Harvey P. Dale, <a href="/A254678/b254678.txt">Table of n, a(n) for n = 1..1000</a>
%F x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 <= x <= p-4.
%e p=7: 2*3*4*5=120 == 1 mod 7;
%e p=17: 2*3*4*5=120 == 1 mod 17 AND 12*13*14*15=32760 == 1 mod 17; for p=13: no triple == 1 mod 13;
%e p=23: 5*6*7*8 == 1 mod 23 AND 15*16*17*18== 1 mod 23 AND 19*20*21*22 == 1 mod 23; and so on. For the number of quadruples for a prime, see A256580.
%t fsiQ[n_]:=AnyTrue[Times@@@Partition[Range[n-1],4,1],Mod[#,n]==1&]; Select[ Prime[Range[200]],fsiQ] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jul 02 2019 *)
%o (R)
%o library(numbers)
%o IP <- vector()
%o t <- vector()
%o S <- vector()
%o IP <- c(Primes(1000))
%o for (j in 1:(length(IP))){
%o for (i in 2:(IP[j]-4)){
%o t[i-1] <- as.vector(mod((i*(i+1)*(i+2)*(i+3)),IP[j]))
%o Z[j] <- sum(which(t==1))
%o S[j] <- length(which(t==1))
%o }
%o }
%o IP[S!=0]
%o #Carefully increase Primes(1000). It takes several hours for 100000.
%o (PARI) lista(nn) = forprime(p=2, nn, if (sum(x=1, p-4, ((x*(x+1)*(x+2)*(x+3)) % p) == 1) > 0, print1(p, ", "))); \\ _Michel Marcus_, Apr 03 2015
%Y Cf. A256567, A256580.
%K nonn
%O 1,1
%A _Marian Kraus_, Apr 02 2015
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