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%I #18 Feb 10 2023 05:40:55
%S 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,
%T 106,108,112,126,130,136,69,74,150,156,81,83,86,178,36,95,96,49,66,5,
%U 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
%N Multiplicative order of 3/2 modulo n-th prime.
%C 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.
%C If p == 1, 5, 19, 23 (mod 24), then 3/2 is a quadratic residue modulo p, so a(n) divides (p - 1)/2.
%C 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
%H Robert Israel, <a href="/A320383/b320383.txt">Table of n, a(n) for n = 3..10000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiplicative_order">Multiplicative order</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem">Zsigmondy's theorem</a>
%e Let ord(n,p) be the multiplicative order of n modulo p.
%e 3/2 == 4 (mod 5), so a(3) = ord(4,5) = 2.
%e 3/2 == 5 (mod 7), so a(4) = ord(5,7) = 6.
%e 3/2 == 7 (mod 11), so a(5) = ord(7,11) = 10.
%e 3/2 == 8 (mod 13), so a(6) = ord(8,13) = 4.
%p f:= proc(n) local p; p:= ithprime(n); numtheory:-order(3/2 mod p,p) end proc:
%p map(f, [$3..100]); # _Robert Israel_, Apr 20 2021
%t a[n_] := With[{p = Prime[n]}, Do[If[Divisible[3^k - 2^k, p], Return[k]], {k, Rest@Divisors[p-1]}]];
%t Table[a[n], {n, 3, 100}] (* _Jean-François Alcover_, Feb 10 2023 *)
%o (PARI) forprime(p=5,10^3,print1(znorder(Mod(3/2,p)),", ")) \\ _Joerg Arndt_, Oct 13 2018
%Y Cf. A001047, A211242, A320384.
%K nonn,look
%O 3,1
%A _Jianing Song_, Oct 12 2018
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