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
KEYWORD
nonn,look
AUTHOR
Jianing Song, Oct 12 2018
STATUS
approved