A211450 -id:A211450 - cp's OEIS frontend

A211449 (p-1)/x, where p = prime(n) and x = ord(4,p), the smallest positive integer such that 4^x == 1 mod p.

0, 2, 2, 2, 2, 2, 4, 2, 2, 2, 6, 2, 4, 6, 2, 2, 2, 2, 2, 2, 8, 2, 2, 8, 4, 2, 2, 2, 6, 8, 18, 2, 4, 2, 2, 10, 6, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 6, 2, 6, 8, 2, 20, 10, 32, 2, 2, 2, 6, 8, 6, 2, 6, 2, 4, 2, 22, 16, 2, 2, 8, 2, 2, 2, 2, 2, 2, 18, 4, 4, 2, 2, 10, 12

Offset: 1

T. D. Noe, Apr 11 2012

nonn

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001917, A094593, A211450, A211451, A211452, A211453, A211454, A006556.

nn = 4; Table[If[Mod[nn, p] == 0, 0, (p-1)/MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

A211451 (p-1)/x, where p = prime(n) and x = ord(6,p), the smallest positive integer such that 6^x == 1 mod p.

Original entry on oeis.org

0, 0, 4, 3, 1, 1, 1, 2, 2, 2, 5, 9, 1, 14, 2, 2, 1, 1, 2, 2, 2, 1, 1, 1, 8, 10, 1, 1, 1, 1, 1, 1, 1, 6, 4, 1, 1, 6, 2, 4, 1, 3, 10, 2, 14, 1, 2, 1, 1, 1, 1, 14, 12, 1, 1, 2, 2, 1, 1, 5, 2, 2, 6, 62, 6, 2, 2, 6, 1, 3, 11, 2, 1, 1, 6, 2, 4, 1, 1, 24, 1, 15, 10

Offset: 1

T. D. Noe, Apr 11 2012

nonn

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001917, A094593, A211449, A211450, A211452, A211453, A211454, A006556.

nn = 6; Table[If[Mod[nn, p] == 0, 0, (p-1)/MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

A211452 (p-1)/x, where p = prime(n) and x = ord(7,p), the smallest positive integer such that 7^x == 1 mod p.

Original entry on oeis.org

1, 2, 1, 0, 1, 1, 1, 6, 1, 4, 2, 4, 1, 7, 2, 2, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 4, 8, 1, 2, 2, 2, 2, 1, 3, 1, 2, 1, 1, 15, 19, 8, 2, 2, 1, 6, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 2, 14, 2, 1, 2, 10, 3, 2, 3, 6, 1, 1, 11, 1, 6, 6, 1, 2, 4, 1, 2, 17, 22, 6, 1, 1, 6

Offset: 1

T. D. Noe, Apr 11 2012

nonn

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001917, A094593, A211449, A211450, A211451, A211453, A211454, A006556.

nn = 7; Table[If[Mod[nn, p] == 0, 0, (p-1)/MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

A211453 (p-1)/x, where p = prime(n) and x = ord(8,p), the smallest positive integer such that 8^x == 1 mod p.

Original entry on oeis.org

0, 1, 1, 6, 1, 3, 2, 3, 2, 1, 6, 3, 2, 3, 2, 1, 1, 3, 3, 2, 24, 6, 1, 8, 6, 1, 6, 1, 9, 4, 18, 1, 2, 3, 1, 30, 3, 3, 2, 1, 1, 3, 2, 6, 1, 6, 3, 6, 1, 3, 8, 2, 30, 5, 16, 2, 1, 6, 3, 4, 3, 1, 9, 2, 6, 1, 33, 48, 1, 3, 4, 2, 6, 3, 3, 2, 1, 9, 2, 6, 1, 3, 10, 18

Offset: 1

T. D. Noe, Apr 11 2012

nonn

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001917, A094593, A211449, A211450, A211451, A211452, A211454, A006556.

nn = 8; Table[If[Mod[nn, p] == 0, 0, (p-1)/MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

A211454 (p-1)/x, where p = prime(n) and x = ord(9,p), the smallest positive integer such that 9^x == 1 mod p.

Original entry on oeis.org

1, 0, 2, 2, 2, 4, 2, 2, 2, 2, 2, 4, 10, 2, 2, 2, 2, 12, 6, 2, 12, 2, 2, 2, 4, 2, 6, 2, 4, 2, 2, 2, 2, 2, 2, 6, 4, 2, 2, 2, 2, 4, 2, 24, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2, 2, 2, 18, 4, 2, 2, 2, 18, 2, 8, 2, 2, 4, 2, 4, 2, 2, 6, 4, 2, 2, 2, 4, 2, 4, 2, 4, 10, 16

Offset: 1

T. D. Noe, Apr 11 2012

nonn

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A001917, A094593, A211449, A211450, A211451, A211452, A211453, A006556.

nn = 9; Table[If[Mod[nn, p] == 0, 0, (p-1)/MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

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), ", ")))

cp's OEIS Frontend

A211449 (p-1)/x, where p = prime(n) and x = ord(4,p), the smallest positive integer such that 4^x == 1 mod p.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A211451 (p-1)/x, where p = prime(n) and x = ord(6,p), the smallest positive integer such that 6^x == 1 mod p.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A211452 (p-1)/x, where p = prime(n) and x = ord(7,p), the smallest positive integer such that 7^x == 1 mod p.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A211453 (p-1)/x, where p = prime(n) and x = ord(8,p), the smallest positive integer such that 8^x == 1 mod p.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A211454 (p-1)/x, where p = prime(n) and x = ord(9,p), the smallest positive integer such that 9^x == 1 mod p.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI