A320383 Multiplicative order of 3/2 modulo n-th prime.

2, 6, 10, 4, 16, 3, 11, 7, 30, 36, 40, 21, 23, 13, 58, 12, 33, 7, 36, 26, 82, 88, 8, 25, 102, 106, 108, 112, 126, 130, 136, 69, 74, 150, 156, 81, 83, 86, 178, 36, 95, 96, 49, 66, 5, 222, 226, 228, 232, 119, 30, 250, 256, 131, 67, 270, 276, 40, 141, 73, 51, 155, 156, 79, 11, 168, 346, 348, 352, 179, 366, 124

Offset

3,1

Comments

Let p = prime(n). a(n) is the smallest positive k such that p divides 3^k - 2^k. Obviously, a(n) divides p - 1. If a(n) = p - 1, then p is listed in A320384.

If p == 1, 5, 19, 23 (mod 24), then 3/2 is a quadratic residue modulo p, so a(n) divides (p - 1)/2.

By Zsigmondy's theorem, for each k >=2 there is a prime that divides 3^k-2^k but not 3^j-2^j for j < k. Therefore each integer >= 2 appears in the sequence at least once. - Robert Israel, Apr 20 2021

Links

Robert Israel, Table of n, a(n) for n = 3..10000

Wikipedia, Multiplicative order

Wikipedia, Zsigmondy's theorem

Example

Let ord(n,p) be the multiplicative order of n modulo p.
3/2 == 4 (mod 5), so a(3) = ord(4,5) = 2.
3/2 == 5 (mod 7), so a(4) = ord(5,7) = 6.
3/2 == 7 (mod 11), so a(5) = ord(7,11) = 10.
3/2 == 8 (mod 13), so a(6) = ord(8,13) = 4.

Maple

f:= proc(n) local p; p:= ithprime(n); numtheory:-order(3/2 mod p, p) end proc:
map(f, [$3..100]); # Robert Israel, Apr 20 2021

Mathematica

a[n_] := With[{p = Prime[n]}, Do[If[Divisible[3^k - 2^k, p], Return[k]], {k, Rest@Divisors[p-1]}]];
Table[a[n], {n, 3, 100}] (* Jean-François Alcover, Feb 10 2023 *)

Prog

(PARI) forprime(p=5, 10^3, print1(znorder(Mod(3/2, p)), ", ")) \\ Joerg Arndt, Oct 13 2018

Crossrefs

Cf. A001047, A211242, A320384.

Sequence in context: A095105 A220338 A052194 * A073662 A086553 A342068

Adjacent sequences: A320380 A320381 A320382 * A320384 A320385 A320386

Keyword

nonn,look

Author

Jianing Song, Oct 12 2018

Status

approved