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A372894 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).

%I A372894 #21 Oct 29 2024 15:41:14
%S A372894 1,2,4,8,9,13,15,16,17,18,21,25,26,30,32,34,35,36,42,49,50,52,60,64,
%T A372894 68,70,72,81,84,97,98,100,104,117,120,121,123,128,135,136,140,144,153,
%U A372894 162,168,169,189,193,194,195,196,200,205,208,221,225,234,240,241,242,246,255,256,257,270,272,273,280,287,288,289
%N A372894 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).
%C A372894 Can be seen as the opposite of A129802.
%C A372894 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.
%C A372894 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.
%C A372894 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 even. But a squarefree term can have factors that are neither elite nor anti-elite, the smallest being 341 = 11*31.
%C A372894 Contains divisors of Fermat numbers >= 17 (A307843 \ {3,5}) since they are products of elite primes.
%H A372894 Jianing Song, <a href="/A372894/b372894.txt">Table of n, a(n) for n = 1..10000</a>
%e A372894 For n >= 1, we have 2^2^n + 1 == 65, 5, 17, 257 (mod 341) respectively for n == 0, 1, 2, 3 (mod 4). As we have (65/341) = (5/341) = (17/341) = (257/341) = 1, 341 is a term.
%o A372894 (PARI) isA372894(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
%Y A372894 Cf. A000215, A102742, A129802, A307843.
%Y A372894 Prime elements of this sequence are given by A128852.
%K A372894 nonn
%O A372894 1,2
%A A372894 _Jianing Song_, May 15 2024