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%I #18 Sep 07 2024 16:03:28
%S 121,1729,2821,7381,8401,10585,15457,15841,18721,19345,23521,24661,
%T 28009,29341,31621,41041,46657,47197,49141,50881,52633,55969,63973,
%U 74593,75361,82513,87913,88573,93961,111361,112141,115921,125665,126217,138481,148417,172081
%N Composite numbers k == 1, 11 (mod 12) such that 3^((k-1)/2) == 1 (mod k).
%C Odd composite numbers k such that 3^((k-1)/2) == (3/k) = 1 (mod k), where (3/k) is the Jacobi symbol (or Kronecker symbol).
%C It seems that most terms are congruent to 1 modulo 12. The first terms congruent to 11 modulo 12 are 1683683, 1898999, 2586083, 2795519, 4042403, 4099439, 5087171, 8243111, ...
%H Jianing Song, <a href="/A375917/b375917.txt">Table of n, a(n) for n = 1..1000</a>
%e 1683683 is a term because 1683683 = 59*28537 is composite, 1683683 == 11 (mod 12), and 3^((1683683-1)/2) == 1 (mod 1683683).
%o (PARI) isA375917(k) = (k>1) && !isprime(k) && (k%12==1 || k%12==11) && Mod(3,k)^((k-1)/2) == 1
%Y | b=2 | b=3 | b=5 |
%Y -----------------------------------+-------------------+----------+---------+
%Y (b/k)=1, b^((k-1)/2)==1 (mod k) | A006971 | this seq | A375915 |
%Y -----------------------------------+-------------------+----------+---------+
%Y (b/k)=-1, b^((k-1)/2)==-1 (mod k) | A244628 U A244626 | A375918 | A375916 |
%Y -----------------------------------+-------------------+----------+---------+
%Y b^((k-1)/2)==-(b/k) (mod k), also | A306310 | A375490 | A375816 |
%Y (b/k)=-1, b^((k-1)/2)==1 (mod k) | | | |
%Y -----------------------------------+-------------------+----------+---------+
%Y Euler-Jacobi pseudoprimes | A047713 | A048950 | A375914 |
%Y (union of first two) | | | |
%Y -----------------------------------+-------------------+----------+---------+
%Y Euler pseudoprimes | A006970 | A262051 | A262052 |
%Y (union of all three) | | | |
%K nonn
%O 1,1
%A _Jianing Song_, Sep 02 2024