OFFSET
1,1
COMMENTS
Note that if k is odd and b^((k-1)/2) == -(b/k) (mod k), then taking Jacobi symbol modulo k (which depends only on the remainder modulo k) yields (b/k)^((k-1)/2) = -(b/k), or (b/k)^((k+1)/2) = -1. This implies that (k+1)/2 is odd, so k == 1 (mod 4). Moreover, if k > 1, then (b/k) = -1 (see the Math Stack Exchange link below), so b^((k-1)/2) == 1 (mod k). In particular, this sequence is equivalent to "numbers k == 13, 17 (mod 20) such that 5^((k-1)/2) == 1 (mod k)". [Comment rewritten by Jianing Song, Sep 07 2024]
LINKS
Jianing Song, Table of n, a(n) for n = 1..1000
Mathematics Stack Exchange, There are no a in Z and odd k > 1 such that (a/k) = 1 and a^((k-1)/2) == -1 (mod k)
EXAMPLE
217 is a term because (5/217) = -1, and 5^((217-1)/2) == 1 (mod 217).
PROG
(PARI) isA375816(k) = (k>1) && gcd(k, 10)==1 && Mod(5, k)^((k-1)/2)==-kronecker(5, k)
(PARI) isA375816(k) = (k%20==13 || k%20==17) && Mod(5, k)^((k-1)/2)==1
CROSSREFS
| b=2 | b=3 | b=5 |
-----------------------------------+-------------------+---------+----------+
-----------------------------------+-------------------+---------+----------+
-----------------------------------+-------------------+---------+----------+
(b/k)=-1, b^((k-1)/2)==1 (mod k) | | | |
-----------------------------------+-------------------+---------+----------+
(union of first two) | | | |
-----------------------------------+-------------------+---------+----------+
(union of all three) | | | |
KEYWORD
nonn
AUTHOR
Jianing Song, Sep 01 2024
STATUS
approved