OFFSET
1,1
COMMENTS
All terms are composite because for odd primes p we always have 2^((p-1)/2) == (2/p) = A091337(p) (mod p).
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 == 5 (mod 8) such that 2^((k-1)/2) == 1 (mod k)". [Comment rewritten by Jianing Song, Sep 07 2024]
Also numbers k in A001567 and congruent to 5 modulo 8 such that k - 1 divided by the multiplicative order of 2 modulo k is an even number.
Euler pseudoprimes (A006970) that are not Euler-Jacobi pseudoprimes (A047713). - Amiram Eldar, Oct 28 2019
LINKS
Amiram Eldar, 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
341 is a term because (2/341) = -1, and 2^((341-1)/2) == 1 (mod 341).
PROG
(PARI) isA306310(k)=(k%8==5) && Mod(2, k)^((k-1)/2)==1
(PARI) isok(k) = (k>1) && (k%2) && (Mod(2, k)^((k-1)/2) == Mod(-kronecker(2, k), k)); \\ Michel Marcus, Feb 07 2019
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, Feb 06 2019
STATUS
approved