A385168 -id:A385168 - cp's OEIS frontend

A385167 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is even. Here ord(a,m) is the multiplicative order of a modulo m.

11, 79, 131, 199, 211, 239, 251, 331, 359, 439, 479, 491, 571, 599, 691, 719, 811, 839, 919, 971, 1039, 1051, 1091, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1531, 1559, 1571, 1759, 1811, 1879, 1931, 1999, 2011, 2039, 2131, 2239, 2251, 2371, 2399, 2411, 2531, 2719, 2731, 2851, 2879, 2971, 2999

Offset: 1

Jianing Song, Jun 20 2025

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).
List of p = A002145(k) such that A385166(k) is even.
Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).

			359 is a term since the multiplicative order of 2+-i modulo 359 is 6444, and (360*ord(5,359))/6444 = 10 is even.

Jianing Song, Table of n, a(n) for n = 1..10001

Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).

Subsequence of the intersection of A122869 and A385168. Contains A385180 as a subsequence.

quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
isA385167(p) = isprime(p) && p%4==3 && quot(p)%2==0

A385180 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is divisible by 4. Here ord(a,m) is the multiplicative order of a modulo m.

Original entry on oeis.org

331, 571, 599, 691, 839, 919, 971, 1039, 1051, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1571, 1759, 1879, 2131, 2411, 2879, 2971, 3079, 3251, 3331, 3491, 3571, 3691, 3851, 4051, 4079, 4091, 4211, 4519, 4639, 4651, 4679, 4691, 4759, 4919, 4931, 5051, 5119, 5171, 5279, 5479, 5519, 5531

Offset: 1

Jianing Song, Jun 20 2025

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).
List of p = A002145(k) such that A385166(k) is divisible by 4.
Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).
By definition, a term that is in neither A385169 nor A385179 must be congruent to 31 or 79 modulo 80. The smallest such term is p = 1759 (ord(2+-i,p) = ((p+1)/4) * ord(5,p) = 128920); even if 1039 == 79 (mod 80), we have ord(2+-i,p) =  ((p+1)/8) * ord(5,p) = 22490 == 2 (mod 4), which means that 1039 is in A385179.

			571 is a term since the multiplicative order of 2+-i modulo 571 is 40755, and (572*ord(5,571))/40755 = 4 is divisible by 4.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).

Subsequence of A385167, which is itself a subsequence of intersection of A122869 and A385168.

quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
isA385180(p) = isprime(p) && p%4==3 && quot(p)%4==0

A385166 Let p = A002145(n) be the n-th prime == 3 (mod 4); a(n) = (p+1) * ord(5,p) / ord(2+-i,p) = (p+1) * ord(5,p) / A385165(n). Here ord(a,m) is the multiplicative order of a modulo m.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 5, 1, 1, 1, 1, 3, 2, 2, 1, 1, 2, 2, 1, 1, 1, 11, 1, 4, 3, 10, 1, 1, 1, 3, 1, 2, 1, 1, 1, 6, 1, 6, 1, 3, 1, 1, 1, 4, 3, 24, 1, 1, 1, 1, 1, 3, 1, 4, 2, 1, 1, 1, 1, 1, 2, 1, 1, 8, 1, 27, 1, 1, 1, 1, 20, 3, 1, 4, 1, 1

Offset: 1

Jianing Song, Jun 20 2025

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).

			The multiplicative order of 2+-i modulo A002145(3) = 11 is A385165(3) = 30, so a(3) = (12*ord(5,11))/30 = 2.
The multiplicative order of 2+-i modulo A002145(13) = 83 is A385165(13) = 2296, so a(13) = (84*ord(5,83))/2296 = 3.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A002145, A385165. Primes corresponding to special terms: A385168 (>1), A385167 (even), A385180 (divisible by 4).

Cf. A211450.

quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
forprime(p=3, 1e3, if(p%4==3, print1(quot(p), ", ")))

A385169 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is odd.

Original entry on oeis.org

331, 571, 599, 691, 839, 971, 1051, 1171, 1291, 1451, 1571, 1879, 2131, 2411, 2971, 3251, 3331, 3491, 3571, 3691, 3851, 4051, 4091, 4211, 4651, 4679, 4691, 4919, 4931, 5051, 5171, 5479, 5531, 5651, 5839, 5851, 5879, 6011, 6599, 6679, 6691, 7079, 7211, 7331, 7691, 8011, 8039, 8171, 8731, 8839, 9011, 9371, 9811

Offset: 1

Jianing Song, Jun 20 2025

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
  - Of course this is true if ord(2+-i,p) is odd.
  - If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
  - If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) is odd.

			8731 is a term since (2+-i)^635253 == 1 (mod 8731), and 635253 is odd.
8839 is a term since (2+-i)^57447 == 1 (mod 8839), and 57447 is odd.
9011 is a term since (2+-i)^2029953 == 1 (mod 9011), and 2029953 is odd.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A385165, A385179, A385192, A385217 (the actual multiplicative orders).

A385188 < this sequence < A385180 < A385167 < intersection of A122869 and A385168, where Ax < Ay means that Ax is a subsequence of Ay.

ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
isA385169(p) = isprime(p) && p%4==3 && ord(p)%2

A385179 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is congruent to 2 modulo 4.

Original entry on oeis.org

11, 131, 211, 251, 491, 811, 919, 1039, 1091, 1319, 1399, 1531, 1811, 1931, 2011, 2251, 2371, 2531, 2731, 2851, 3011, 3079, 3371, 3931, 4079, 4451, 4519, 4759, 5011, 5639, 6091, 6131, 6211, 6359, 6451, 6491, 6571, 6971, 7411, 7451, 7559, 7639, 8291, 8719, 8971, 9091, 9491, 9719, 9839, 9851, 9931

Offset: 1

Jianing Song, Jun 20 2025

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
  - Of course this is true if ord(2+-i,p) is odd.
  - If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
  - If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) == 2 (mod 4).

			919 is a term since (2+-i)^21114 == 1 (mod 919), (2+-i)^(21114/2) !== 1 (mod 919), and we have 21114 == 2 (mod 4).

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A385165, A385169, A385188, A385218 (the actual multiplicative orders).

Subsequence of A385167, which itself lies in the intersection of A122869 and A385168.

ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
isA385179(p) = isprime(p) && p%4==3 && ord(p)%4==2

A385188 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is not divisible by 2 or 3.

Original entry on oeis.org

599, 691, 1291, 1451, 2411, 3851, 4919, 5051, 5479, 5531, 5879, 6599, 7079, 7691, 8011, 8039, 11491, 13291, 14011, 15091, 15971, 16651, 17359, 18731, 19211, 19531, 20731, 22651, 23971, 24611, 25639, 25679, 26251, 32051, 32359, 32531, 32771, 32971, 35879, 37039, 37571, 38011, 38371

Offset: 1

Jianing Song, Jun 20 2025

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.

5 is a quadratic residue of integers modulo p for p being a term of this sequence. (See A385165).

			5479 is a term since (2+-i)^125081 == 1 (mod 5479), and 125081 is divisible by neither 2 nor 3.

Jianing Song, Table of n, a(n) for n = 1..10000

Cf. A385165, A385179, A385219 (the actual multiplicative orders).
this sequence < A385169 < A385180 < A385167 < intersection of A122869 and A385168, where Ax < Ay means that Ax is a subsequence of Ay.
Also a subsequence of A385191.

ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
isA385188(p) = isprime(p) && p%4==3 && ord(p)%2 && ord(p)%3

A384948 Primes p == 3 (mod 4) such that 5 is a primitive root of integers modulo p, but 2+-i are not primitive roots of Gaussian integers modulo p.

Original entry on oeis.org

83, 307, 347, 503, 587, 863, 947, 1103, 1223, 1523, 1567, 1667, 1787, 1907, 2063, 2087, 2267, 2663, 2683, 2687, 2903, 2963, 3167, 3343, 3347, 3623, 3803, 3863, 4283, 4463, 4523, 4643, 4967, 5147, 5303, 5387, 5507, 5563, 5807, 5843, 6047, 6203, 6607, 6863, 6983, 7187, 7247, 7523, 7583

Offset: 1

Jianing Song, Jun 20 2025

For p = A002145(k), A385165(k) divides (p+1) * ord(5,p), since we have (2+-i)^(p+1) == 5 (mod p). Hence if 2+-i are primitive roots of Gaussian integers modulo p, then 5 is a primitive root of integers modulo p. This sequence lists p such that the converse does not hold.

			5 is a primitive root modulo 83, but the multiplicative order of 2+-i modulo 83 in Gaussian integers is not 83^2 - 1 = 6888; it is 2296 = 6888/3. In other words, 2+-i are not generators of (Z[i]/83Z[i])*.

Jianing Song, Table of n, a(n) for n = 1..10000

By definition, subsequence of A019335, A122870, and A385168.

isprim(p) = my(f = factor(p^2-1)[,1]~); for(i=1, #f, if(Mod([2, -1; 1, 2], p)^((p^2-1)/f[i]) == 1, return(0))); return(1) \\ for a prime p == 3 (mod 4), determines if 2+-i are primitive roots modulo p
isA384948(p) = isprime(p) && (p%4==3) && znorder(Mod(5,p))==p-1 && !isprim(p)

cp's OEIS Frontend

A385167 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is even. Here ord(a,m) is the multiplicative order of a modulo m.

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A385180 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is divisible by 4. Here ord(a,m) is the multiplicative order of a modulo m.

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A385166 Let p = A002145(n) be the n-th prime == 3 (mod 4); a(n) = (p+1) * ord(5,p) / ord(2+-i,p) = (p+1) * ord(5,p) / A385165(n). Here ord(a,m) is the multiplicative order of a modulo m.

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A385169 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is odd.

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A385179 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is congruent to 2 modulo 4.

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A385188 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is not divisible by 2 or 3.

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A384948 Primes p == 3 (mod 4) such that 5 is a primitive root of integers modulo p, but 2+-i are not primitive roots of Gaussian integers modulo p.

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