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%I #13 Oct 05 2021 04:50:50
%S 3,5,6,7,10,12,14,20,24,27,28,39,40,41,45,48,51,54,56,63,65,75,78,80,
%T 82,85,90,91,96,102,105,108,112,119,125,126,130,147,150,156,160,164,
%U 170,175,180,182,192,204,210,216,224,238,243,245,250,252,260,291,294,300
%N Possible bases for Pepin's primality test for Fermat numbers.
%C Prime elements of this sequence are given by A102742.
%H Arkadiusz Wesolowski, <a href="/A129802/b129802.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PepinsTest.html">Pepin's Test</a>
%F A positive integer 2^k*m, where m is odd and k>=0, belongs to this sequence iff the Jacobi symbol (F_n/m)=1 only for a finite number of Fermat numbers F_n=A000215(n).
%o (PARI) { isPepin(n) = local(s,S=Set(),t); n\=2^valuation(n,2); s=Mod(3,n); while( !setsearch(S,s), S=setunion(S,[s]); s=(s-1)^2+1); t=s; until( t==s, if( kronecker(lift(t),n)==1, return(0)); t=(t-1)^2+1);1 }
%o for(n=2,1000,if(isPepin(n),print1(n,", ")))
%o (PARI) for(b=2, 300, k=b/2^valuation(b, 2); if(k>1, i=logint(k, 2); m=Mod(2, k); z=znorder(m); e=znorder(Mod(2, z/2^valuation(z, 2))); t=0; for(c=1, e, if(kronecker(lift(m^2^(i+c))+1, k)==-1, t++, break)); if(t==e, print1(b, ", ")))); \\ _Arkadiusz Wesolowski_, Sep 22 2021
%Y Cf. A000215, A019434, A060377, A102742.
%K nonn
%O 1,1
%A _Max Alekseyev_, Jun 14 2007, corrected Dec 29 2007. Thanks to _Ant King_ for pointing out an error in the earlier version of this sequence.
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