OFFSET
1,1
COMMENTS
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).
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, ...
LINKS
Jianing Song, Table of n, a(n) for n = 1..1000
EXAMPLE
1683683 is a term because 1683683 = 59*28537 is composite, 1683683 == 11 (mod 12), and 3^((1683683-1)/2) == 1 (mod 1683683).
MATHEMATICA
q[k_] := MemberQ[{1, 11}, Mod[k, 12]] && CompositeQ[k] && PowerMod[3, (k-1)/2, k] == 1; Select[Range[173000], q] (* Amiram Eldar, Mar 21 2026 *)
PROG
(PARI) isA375917(k) = (k>1) && !isprime(k) && (k%12==1 || k%12==11) && Mod(3, k)^((k-1)/2) == 1
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Jianing Song, Sep 02 2024
STATUS
approved
