A129802 Possible bases for Pepin's primality test for Fermat numbers.

3, 5, 6, 7, 10, 12, 14, 20, 24, 27, 28, 39, 40, 41, 45, 48, 51, 54, 56, 63, 65, 75, 78, 80, 82, 85, 90, 91, 96, 102, 105, 108, 112, 119, 125, 126, 130, 147, 150, 156, 160, 164, 170, 175, 180, 182, 192, 204, 210, 216, 224, 238, 243, 245, 250, 252, 260, 291, 294, 300

Offset: 1

Author

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

Keywords

• nonn

Comments

Prime elements of this sequence are given by A102742.

From Jianing Song, May 15 2024: (Start)

Let m be an odd number and ord(2,m) = 2^r*d be the multiplicative order of 2 modulo m, where d is odd, then 2^2^n + 1 is congruent to one of 2^2^r + 1, 2^2^(r+1) + 1, ..., 2^2^(r+ord(2,d)-1) + 1 modulo m, so it suffices to check these ord(2,d) numbers.

Note that if m > 1, then m does not divide 2^2^n + 1 for n >= r, otherwise we would have 2^(2^n*d) = (2^ord(2,m))^2^(n-r) == 1 (mod m) and 2^(2^n*d) = (2^2^n)^d == (-1)^d == -1 (mod m). As a result, m is a term if and only if the Jacobi symbol ((2^2^n + 1)/m) is equal to -1 for m = r, r+1, ..., r+ord(2,d)-1.

By definition, a squarefree number that is a product of elite primes (A102742) or anti-elite primes (A128852) is a term if and only if its number of elite factors is odd. But a squarefree term can have factors that are neither elite nor anti-elite, the smallest being 551 = 19*29. (End)

Examples

			For n >= 2, we have 2^2^n + 1 == 170, 461, 17, 257, 519, 539 (mod 551) respectively for n == 0, 1, 2, 3, 4, 5 (mod 6). As we have (170/551) = (461/551) = (17/551) = (257/551) = (519/551) = (539/551) = -1, 551 is a term. - _Jianing Song_, May 19 2024
		

Links

• Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Pepin's Test.

Crossrefs

Cf. A000215, A019434, A060377, A102742, A128852, A372894.

Programs

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 }
for(n=2,1000,if(isPepin(n),print1(n,", ")))

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

PARI

isA129802(n) = n = (n >> valuation(n,2)); my(d = znorder(Mod(2, n)), StartPoint = valuation(d, 2), LengthTest = znorder(Mod(2, d >> StartPoint))); for(i = StartPoint, StartPoint + LengthTest - 1, if(kronecker(lift(Mod(2, n)^2^i + 1), n) == 1, return(0))); 1 \\ Jianing Song, May 19 2024

Formula

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 for only a finite number of Fermat numbers F_n = A000215(n).