A306310 Odd numbers k > 1 such that 2^((k-1)/2) == -(2/k) = -A091337(k) (mod k), where (2/k) is the Jacobi (or Kronecker) symbol.

341, 5461, 10261, 15709, 31621, 49981, 65077, 83333, 137149, 176149, 194221, 215749, 219781, 276013, 282133, 534061, 587861, 611701, 653333, 657901, 665333, 688213, 710533, 722261, 738541, 742813, 769757, 950797, 1064053, 1073021, 1109461, 1141141, 1357621, 1398101

Offset: 1

Author

Jianing Song, Feb 06 2019

Keywords

• nonn

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

Examples

			341 is a term because (2/341) = -1, and 2^((341-1)/2) == 1 (mod 341).
		

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)

Crossrefs

Cf. A091337, A001567, A007814.

| b=2 | b=3 | b=5 |

-----------------------------------+-------------------+---------+---------+

(b/k)=1, b^((k-1)/2)==1 (mod k) | A006971 | A375917 | A375915 |

-----------------------------------+-------------------+---------+---------+

(b/k)=-1, b^((k-1)/2)==-1 (mod k) | A244628 U A244626 | A375918 | A375916 |

-----------------------------------+-------------------+---------+---------+

b^((k-1)/2)==-(b/k) (mod k), also | this seq | A375490 | A375816 |

(b/k)=-1, b^((k-1)/2)==1 (mod k) | | | |

-----------------------------------+-------------------+---------+---------+

Euler-Jacobi pseudoprimes | A047713 | A048950 | A375914 |

(union of first two) | | | |

-----------------------------------+-------------------+---------+---------+

Euler pseudoprimes | A006970 | A262051 | A262052 |

(union of all three) | | | |

Programs

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