login
Minimal residues of Pepin's Test for Fermat Numbers using the base 3.
7

%I #13 Apr 03 2023 10:36:11

%S 0,-1,-1,-1,-1,10324303,-6586524273069171148,

%T 110780954395540516579111562860048860420,

%U 5864545399742183862578018016183410025465491904722516203269973267547486512819

%N Minimal residues of Pepin's Test for Fermat Numbers using the base 3.

%C For n>=1 the Fermat Number F(n) is prime if and only if 3^((F(n) - 1)/2) is congruent to -1 (mod F(n)).

%C Any positive integer k for which the Jacobi symbol (k|F(n)) is -1 can be used as the base instead.

%D M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001, pp. 42-43.

%H Dennis Martin, <a href="/A152155/b152155.txt">Table of n, a(n) for n = 0..11</a>

%H Chris Caldwell, The Prime Pages: <a href="https://t5k.org/glossary/page.php?sort=PepinsTest">Pepin's Test</a>.

%F a(n) = 3^((F(n) - 1)/2) (mod F(n)), where F(n) is the n-th Fermat Number, using the symmetry mod (so (-F(n)-1)/2 < a(n) < (F(n)-1)/2).

%e a(4) = 3^(32768) (mod 65537) = 65536 = -1 (mod F(4)), therefore F(4) is prime.

%e a(5) = 3^(2147483648) (mod 4294967297) = 10324303 (mod F(5)), therefore F(5) is composite.

%p f:= proc(n) local F;

%p F:= 2^(2^n) + 1;

%p `mods`(3 &^ ((F-1)/2), F)

%p end proc:

%p seq(f(n), n=0..10); # _Robert Israel_, Dec 19 2016

%o (PARI) a(n)=centerlift(Mod(3,2^(2^n)+1)^(2^(2^n-1))) \\ _Jeppe Stig Nielsen_, Dec 19 2016

%Y A000215, A019434, A152153, A152154, A152156.

%K sign

%O 0,6

%A Dennis Martin (dennis.martin(AT)dptechnology.com), Nov 27 2008