login
This site is supported by donations to The OEIS Foundation.

 

Logo

Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing.
Other ways to donate

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

All terms are composite bacause for odd primes p we always have 2^((p-1)/2) == (2/p) = A091337(p) (mod p).

All terms congruent to 5 modulo 8. So this sequence is equivalent to "Numbers k == 5 (mod 8) such that 2^((k-1)/2) == 1 (mod k)".

Proof: in the following proof let p be any prime factor of k. Note that 2 is a quadratic residue modulo p if and only if p == 1, 7 (mod 8).

(a) if k == 3 (mod 8), then 2^((k-1)/2) == 1 (mod p), so 2^((k+1)/2) == 2 (mod p). Let x = 2^((k+1)/4), then x is an integer such that x^2 == 2 (mod p), so 2 is a quadratic residue modulo p => p == 1, 7 (mod 8). Since any prime factor of k is congruent to 1, 7 modulo 8 we have k == 1, 7 (mod 8), a contradiction.

(b) if k == 7 (mod 8), then 2^((k-1)/2) == -1 (mod p). If p == 7 (mod 8), then 2 is a quadratic residue modulo p while -1 is not, a contradiction. If p == 5 (mod 8), let d be the multiplicative order of 2 modulo p. 2 is not a quadratic residue modulo p, so d divides p - 1 but d does not divide (p - 1)/2, so v2(d) = v2(p-1) = 2, where v2 = A007814 is the 2-adic valuation. On the other hand, 2^(k-1) == 1 (mod p), so d divides k - 1, but k - 1 == 2 (mod 4), a contradiction. So p == 1, 3 (mod 8). Since any prime factor of k is congruent to 1, 3 modulo 8 we have k == 1, 3 (mod 8), a contradiction.

(c) if k == 1 (mod 8), then 2^((k-1)/2) == -1 (mod p). Let x = 2^((k-1)/8), then x^4 == -1 (mod p), so -1 is a quartic residue modulo p => p == 1 (mod 8). Let d be the multiplicative order of 2 modulo p, then d divides k - 1 but d does not divide (k-1)/2, so v2(d) = v2(k-1). On the other hand, 2^((p-1)/2) == 1 (mod p), so d divides (p-1)/2. So v2(p-1) >= v2(d) + 1 = v2(k-1) + 1. Let t = v2(k-1), then k - 1 is not divisible by 2^(t+1), but any prime factor p of k should have p - 1 is divisible by 2^(t+1), a contradiction.

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

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

Cf. A001567, A006970, A007814, A047713, A244626.

Sequence in context: A143688 A086250 A285549 * A210454 A069309 A086806

Adjacent sequences:  A306307 A306308 A306309 * A306311 A306312 A306313

KEYWORD

nonn

AUTHOR

Jianing Song, Feb 06 2019

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified December 14 03:31 EST 2019. Contains 329978 sequences. (Running on oeis4.)