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.

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

1,3

Comments

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).

Links

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

Example

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.

Prog

(PARI) 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), ", ")))

Crossrefs

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

Cf. A211450.

Sequence in context: A261794 A328929 A098744 * A337584 A273975 A025429

Adjacent sequences: A385163 A385164 A385165 * A385167 A385168 A385169

Keyword

nonn,easy

Author

Jianing Song, Jun 20 2025

Status

approved