A256567 -id:A256567 - cp's OEIS frontend

A033217 Primes of form x^2 + 23*y^2.

23, 59, 101, 167, 173, 211, 223, 271, 307, 317, 347, 449, 463, 593, 599, 607, 691, 719, 809, 821, 829, 853, 877, 883, 991, 997, 1097, 1117, 1151, 1163, 1181, 1231, 1319, 1451, 1453, 1481, 1553, 1613, 1669, 1697, 1787, 1789, 1867, 1871, 1879, 1889, 1913, 2027, 2053, 2143, 2309, 2339, 2347, 2381, 2393, 2423, 2539, 2647, 2677, 2693, 2707, 2741, 2819

Offset: 1

N. J. A. Sloane

Discriminant -23.
Also primes of the form x^2 + x*y + 6*y^2. - N. J. A. Sloane, Jun 02 2014
Also primes of the form x^2 - x*y + 6*y^2 with x and y nonnegative. - T. D. Noe, May 07 2005
Primes p such that X^3-X+1 is split modulo p. E.g., X^3-X+1 = (X-33)*(X-40)*(X-94) modulo 167. - Julien Freslon (julien.freslon(AT)wanadoo.fr), Feb 24 2007
It appears that, if x > 0, then tau(p) = A000594(p) == 2 (mod 23). - Comment from Jud McCranie
In fact, this sequence appears to be the same as primes p such that RamanujanTau(p) == {1,2} (mod 23). - Ray Chandler, Dec 01 2016
Excluding the first term, this sequence is the intersection of A191021 and A256567. - Arkadiusz Wesolowski, Oct 03 2021
From Amiram Eldar, Jan 10 2025: (Start)
a(2)..a(10000) are the first terms of the sequence of primes p such that tau(p) == 2 (mod 23), where tau is Ramanujan's tau function (A000594).
Moree and Noubissie (2024) proved that the following 3 conditions for a prime p are equivalent:
  1. tau(p) == 2 (mod 23).
  2. p divides A000931(p+3) where A000931 is the Padovan sequence.
  3. The number of distinct roots modulo p of the polynomial x^3 - x - 1 is 3. (End)

David A. Cox, Primes of the Form x^2 + n y^2, Wiley, 1989.
Joe Roberts, Lure of the Integers, Math. Assoc. America, 1992. See pp. 158-160, "Integer 23 - the Tau function".

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Pieter Moree and Armand Noubissie, Higher reciprocity laws and ternary linear recurrence sequences, International Journal of Number Theory, 2024; arXiv preprint, arXiv:2205.06685 [math.NT], 2022.
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
John Raymond Wilton, Congruence properties of Ramanujan's function τ(n), Proceedings of the London Mathematical Society 2.1 (1930): 1-10. The primes are listed in Table II.
John Raymond Wilton, Congruence properties of Ramanujan's function τ(n), annotated copy of page 8 only.

Cf. A000594, A191021, A256567. Primes in A028958.

QuadPrimes2[1, 0, 23, 10000] (* see A106856 *)
Join[{23}, nn=23; pMax=5000; Union[Reap[Do[p=x^2 + nn y^2; If[p<=pMax&&PrimeQ[p], Sow[p]], {x, Sqrt[pMax]}, {y, Sqrt[pMax/nn]}]][[2, 1]]]] (* Vincenzo Librandi, Sep 05 2016 *)

isok(p) = isprime(p) && !(kronecker(-23, p)==-1) && !polisirreducible(Mod(1, p)*(x^3-x-1)); \\ Arkadiusz Wesolowski, Oct 03 2021

isok(p) = p==23 || (isprime(p) && #polrootsmod(x^3-x-1, p)==3); \\ Arkadiusz Wesolowski, Oct 09 2021

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).

Original entry on oeis.org

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

Marian Kraus, Apr 02 2015

nonn

Is 23 the only prime with two triples?

			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.

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

Cf. A256567.

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

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.

|T| where T={x|x*(x+1)*(x+2)==1 (mod p), p is prime, 1

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.

Original entry on oeis.org

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

Marian Kraus, Apr 02 2015

nonn

If "quadruples" is changed to "pairs" we get A086937 (for the counts) and A038872 (for the primes for which the count is nonzero).

			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.

Cf. A256567, A256572, A038872, A086937.

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

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

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

Showing 1-5 of 5 results.

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A033217 Primes of form x^2 + 23*y^2.

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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).

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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.

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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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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.