A256580 -id:A256580 - cp's OEIS frontend

A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

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

Offset: 1

Marian Kraus, Apr 02 2015

nonn

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.

Together with 5, supersequence of A191065. - Arkadiusz Wesolowski, Nov 24 2021

			For p=7: 4*5*6=120==1 (mod 7), so 7 is a term.
For p=11: 5*6*7=210==1 (mod 11), so 11 is a term.
For p=17: 4*5*6=120==1 (mod 17), so 17 is a term.
13 is not a term because there is no such triple with product ==  1 (mod 13).

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

Cf. A191065, A256572, A254678, A256580.

isok(p) = {if (isprime(p), for (x=1, p-3, if (Mod(x*(x+1)*(x+2), p) == 1, return (1));););} \\ Michel Marcus, Oct 05 2021

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[i-1] <- as.vector(mod(((i-1)*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.

A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

Original entry on oeis.org

7, 17, 23, 31, 41, 47, 73, 89, 97, 103, 127, 137, 151, 167, 199, 223, 233, 239, 241, 257, 271, 281, 311, 313, 353, 359, 367, 383, 409, 431, 433, 439, 449, 479, 487, 503, 521, 577, 593, 601, 607, 647, 673, 719, 727, 743, 751, 761, 769, 839, 857, 881, 887, 911, 929, 937, 953, 967, 977, 983

Offset: 1

Marian Kraus, Apr 02 2015

nonn

			p=7: 2*3*4*5=120 == 1 mod 7;
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;
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.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A256567, A256580.

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 *)

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

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))
   }
}
IP[S!=0]
#Carefully increase Primes(1000). It takes several hours for 100000.

x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 <= x <= p-4.

A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x(x+1)(x+2)...(x+i-1) == 1 mod p.

Original entry on oeis.org

0, 0, 1, 2, 6, 3, 8, 7, 13, 15, 13, 11, 13, 22, 18, 25, 36, 31, 34, 53, 42, 38, 38, 40, 55, 47, 41, 37, 77, 59, 62, 67, 66, 63, 55, 84, 74, 78, 90, 74, 90, 92, 85, 108, 100, 117, 98, 104, 136, 114, 118, 118, 141, 112, 118, 115, 122, 138, 132, 129, 115, 152

Offset: 1

Marian Kraus, Apr 03 2015

nonn

			prime(1)=2: There is no such product
=> a(1)=0;
prime(2)=3: There is no such product
=> a(2)=0;
prime(3)=5: 2*3=6==1 mod 5
=> i=1, x=2; a(3)=1;
prime(4)=7: 4*5*6==1 mod 7; 2*3*4*5==1 mod 7
=> a(3)=2;
prime(5)=11: 3*4==1 mod 11; 7*8==1 mod 11; 5*6*7==1 mod 11; 3*4*5*6*7==1 mod 11; 6*7*8*9*10==1 mod 11; 2*3*4*5*6*7*8*9==1 mod 11
=> x in {3,7,5,3,6,2}
=> a(5)=6.

Cf. A256567, A256580.

f[n_] := Block[{r = Range[2, Prime[n] - 1]}, Sum[Length@ Select[Times @@@ Partition[r, k, 1], Mod[#, Prime@ n] == 1 &], {k, 2, Prime@ n}]]; Array[f, 72] (* Michael De Vlieger, Apr 03 2015 *)

library(numbers)
p <- vector()
n <- vector()
NumTup <- vector()
p <- Primes(m)
n <- length(p)
m <- 17 #all primes will be checked up to this number
Piprod <- matrix(0,m,m) #Matrix with zeros
#loop: every ordered combination of products
for (i in 2:m)
for (j in 2:m)
  Piprod[j,i] <- ifelse(i

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A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

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A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

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A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x(x+1)(x+2)...(x+i-1) == 1 mod p.

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cp's OEIS Frontend

A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

Views

Author

Keywords

Comments

Examples

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Crossrefs

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PARI

R

A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

R

Formula

A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x*(x+1)*(x+2)*...*(x+i-1) == 1 mod p.

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Examples

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Mathematica

R

A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x(x+1)(x+2)...(x+i-1) == 1 mod p.