A294912 -id:A294912 - cp's OEIS frontend

A007520 Primes == 3 (mod 8).

3, 11, 19, 43, 59, 67, 83, 107, 131, 139, 163, 179, 211, 227, 251, 283, 307, 331, 347, 379, 419, 443, 467, 491, 499, 523, 547, 563, 571, 587, 619, 643, 659, 683, 691, 739, 787, 811, 827, 859, 883, 907, 947, 971, 1019, 1051, 1091, 1123, 1163, 1171, 1187

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

Primes of the form 3x^2 + 2xy + 3y^2 with x and y in Z. - T. D. Noe, May 07 2005
Also, primes of the form X^2 + 2Y^2, X=|x-y|, Y=x+y. - Zak Seidov, Dec 06 2011
Each term is the sum of no fewer than three positive squares. - T. D. Noe, Nov 15 2010
Smallest terms expressible as sum of three distinct positive squares: 59 = 1^2 + 3^2 + 7^2, 83 = 3^2 + 5^2 + 7^2, 107, 131, 139, 179, 211, 227, 251, 283, 307. - Zak Seidov, Dec 06 2011
Except for the first term it appears that the terms of the sequence are also primes of the form 2k+1 such that 3*(2k+1) divides 2^k+1. - Hilko Koning, Dec 06 2019
From Hilko Koning, Nov 24 2021: (Start)
Theorem (Legendre symbol): With p an odd prime and a an integer coprime to p the Legendre symbol L(a/p) = -1 if a is a quadratic non-residue (mod p) and L(2/p) = -1 if p == +-3 (mod 8).
Theorem (Euler's criterion): L(a/p) == a^((p-1)/2) (mod p) so with a = 2 and prime p = 2k + 1 then -1 == 2^k (mod (2k+1)). So prime numbers 2k+1 = +-3 (mod 8) are the prime numbers 2k+1 | 2^k+1.
If 2k+1 == -3 (mod 8) then k is even and 2^k+1 is not divisible by 3 and if 2k+1 == +3 (mod 8) then k is odd and 2^k+1 is divisible by 3.
Hence prime numbers 2k+1 == 3 (mod 8) are prime numbers such that 3*(2k+1) | 2^k+1. Or, including the first term of the sequence, prime numbers 2k+1 with k odd such that 2k+1 | 2^k+1.
(End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Alexander Kalmynin, On Novák numbers, arXiv:1611.00417 [math.NT], 2016. See P0 in Theorem 7 p. 11.

Cf. A294912, A045339 (for n >= 2).

[p: p in PrimesUpTo(2000) | p mod 8 eq 3]; // Vincenzo Librandi, Aug 07 2012

A007520 := proc(n)
    option remember;
    local a;
    if n = 1 then
        return 3;
    end if;
    a := nextprime(procname(n-1)) ;
    while modp(a,8) <> 3 do
        a := nextprime(a) ;
    end do:
    a ;
end proc:
seq(A007520(n),n=1..30) ; # R. J. Mathar, Apr 07 2017

lst={};Do[p=8*n+3;If[PrimeQ[p], AppendTo[lst, p]], {n, 0, 10^3}];lst (* Vladimir Joseph Stephan Orlovsky, Aug 22 2008 *)
p=3;k=0;nn=1000;Reap[While[kZak Seidov, Dec 06 2011 *)
Select[Prime[Range[200]],Mod[#,8]==3&] (* Harvey P. Dale, Apr 05 2023 *)

forprime(p=2,97,if(p%8==3,print1(p", "))) \\ Charles R Greathouse IV, Aug 17 2011

A244628 Composite numbers n such that n == 3 (mod 8) and 2^((n-1)/2) == -1 (mod n).

Original entry on oeis.org

476971, 877099, 1302451, 1325843, 1397419, 1441091, 1507963, 1530787, 1907851, 2004403, 3090091, 3116107, 5256091, 5919187, 7883731, 9371251, 11081459, 11541307, 12263131, 13057787, 13338371, 15976747, 17134043, 18740971, 19404139, 20261251, 21623659, 22075579, 24214051

Offset: 1

Gary Detlefs, Jul 02 2014

nonn

This sequence contains the n mod 8 = 3 pseudoprimes to the following modified Fermat primality criterion:
Conjecture 1: if p is a prime congruent to {3,5} mod 8 then 2^((p-1)/2) mod p = p-1.
This conjecture has been tested to 10^8.
This modified primality test has far fewer pseudoprimes than the 2^(n-1) mod n = 1 test and thus has a much higher probability of success. The number of pseudoprimes up to 10^k for the two tests are:
   10^3 0     0
   10^4 0     2
   10^5 0     5
   10^6 2    14
   10^7 16   48
This sequence appears to be a subset of the composites in A175865.
The n mod 8 = 3 pseudoprimes are much rarer than the n mod 8 = 5 pseudoprimes. There are 16 terms < 10^7. If the additional constraints 3^(n-1) mod n = 1 and 5^(n-1) mod n = 1 are added, no terms remain.
Number of terms < 10^k: 0, 0, 0, 0, 0, 2, 16, 50, 132, ..., . - Robert G. Wilson v, Jul 21 2014
Number of terms < 10^k for k=5..15: 0, 2, 16, 50, 132, 341, 876, 2330, 6234, 16625, 44885. - Jens Kruse Andersen, Jul 27 2014
It appears that the terms of the sequence are also the composite numbers of A294912. - Hilko Koning, Dec 05 2019
Also composite numbers 2k+1 with k odd such that 2k+1 | 2^k+1. - Hilko Koning, Jan 27 2022
Conjecture 1 is true. With p = 2k+1 then 2^k mod (2k+1) == 2k. So 2k+1 | 2k-2^k . Prime numbers 2k+1 == +-3 (mod 8) are the prime numbers such that 2k+1 | 2^k+1 (Comments A007520). A reflection across the x-axis and +1 translation across the y-axis of the graph (2k-2^k) / (2k+1) gives the graph (2^k+1) / (2k+1). So the k values of both 2k+1 | 2k-2^k and 2k+1 | 2^k+1 are identical. - Hilko Koning, Feb 04 2022

Jens Kruse Andersen, Table of n, a(n) for n = 1..10000 (first 132 terms from Robert G. Wilson v)

Cf. A003629, A070179, A175865.

for n from 3 to 10^8 by 8 do if Power(2,(n-1)/2) mod n =  n -1 and not isprime(n) then print(n) fi od

fQ[n_] := !PrimeQ@ n && PowerMod[2, (n - 1)/2, n] == n - 1; k = 3; lst = {}; While[k < 10^8, If[fQ@ k, AppendTo[lst, k]]; k += 8]; lst (* Robert G. Wilson v, Jul 21 2014 *)

A295835 Numbers k == 3 (mod 4) such that 2^((k-1)/2), 3^((k-1)/2) and 5^((k-1)/2) are congruent to 1 (mod k).

Original entry on oeis.org

71, 191, 239, 311, 359, 431, 479, 599, 719, 839, 911, 1031, 1151, 1319, 1439, 1511, 1559, 1871, 2039, 2111, 2351, 2399, 2591, 2711, 2879, 2999, 3119, 3191, 3359, 3671, 3719, 3911, 4079, 4271, 4391, 4679, 4751, 4799, 4871, 4919, 5039, 5231, 5279, 5351, 5399, 5471

Offset: 1

Jonas Kaiser, Nov 28 2017

nonn

There are very few composite numbers in this sequence: The probability of catching a pseudoprime number (A001567) with this definition is estimated at 1 in 263 billion.
Composite numbers in the sequence include the Carmichael numbers 131314855918751, 23282264781147191, 70122000249565031, 104782993259720471, 583701149409931151, 870012810301712351. - Robert Israel, Nov 28 2017
With the exception of the pseudoprimes, it seems that this is a subsequence of A139035. Primes of this form (A139035) have two special properties. 1. There exists a smallest m of the form m = (A139035 - 1)/2 such that 2^m == 1 (mod A139035). 2. m is odd. The core of this definition is based on these two properties. The term 2^((k-1)/2) == 1 (mod n) is based on the first property, while the term k == 3 (mod 4) is based on the second property. The terms 3^((k-1)/2) == 1 (mod n) and 5^((k-1)/2) == 1 (mod n) I just tried freely to Fermat.
Prime terms are congruent to 71 or 119 modulo 120. - Jianing Song, Nov 22 2018 [This is because 2, 3, and 5 must be quadratic residues modulo every prime number in this sequence. - Jianing Song, Sep 01 2024]
From Jianing Song, Sep 03 2024: (Start)
Compare this sequence to the sequence of absolute Euler pseudoprimes (A033181): odd composite numbers k such that a^((k-1)/2) == +-1 (mod k) for every a coprime to k. Such numbers k satisfy 2*psi(k) | (k-1), where psi = A002322, so we must have k == 1 (mod 4).
All terms in this sequence are congruent to 7 modulo 8. In fact, taking the Jacobi symbol modulo k (which only depends on the remainder modulo k) of both sides of 2^((k-1)/2) == 1 (mod k) yields (2/k)^((k-1)/2) = 1. Since k == 3 (mod 4), we have that (k-1)/2 is odd, so (2/k) = 1, which means that k == 7 (mod 8). (End)
Those numbers given above by Robert Israel are all congruent to 71 modulo 120. There are no known composite terms congruent to 119 modulo 120; cf. A294092. - Bill McEachen and Jianing Song, Sep 05 2024

Robert Israel, Table of n, a(n) for n = 1..10000
Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv:1608.00862 [math.GM], 2016.

Cf. A001567, A293394, A294717, A294993, A294919, A294912.

A294092 is a subsequence.

filter:= proc(n) [2&^((n-1)/2),3&^((n-1)/2), 5&^((n-1)/2)] mod n = [1,1,1]  end proc:
select(filter, [seq(i,i=3..10000,4)]); # Robert Israel, Nov 28 2017, corrected Feb 26 2018

fQ[n_] := PowerMod[{2, 3, 5}, (n - 1)/2, n] == {1, 1, 1}; Select[3 + 4Range@ 1500, fQ] (* Michael De Vlieger, Nov 28 2017 and slightly modified by Robert G. Wilson v, Feb 26 2018 based on the renaming *)

is(n) = n%4==3 && Mod(2, n)^(n\2)==1 && Mod(3, n)^(n\2)==1 && Mod(5, n)^(n\2)==1 && Mod(2, n)^logint(n+1,2)!=1 \\ Charles R Greathouse IV, Nov 28 2017

Definition corrected by Jonas Kaiser, Feb 05 2018

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A007520 Primes == 3 (mod 8).

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A244628 Composite numbers n such that n == 3 (mod 8) and 2^((n-1)/2) == -1 (mod n).

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A295835 Numbers k == 3 (mod 4) such that 2^((k-1)/2), 3^((k-1)/2) and 5^((k-1)/2) are congruent to 1 (mod k).

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