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Search: a294912 -id:a294912
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Primes == 3 (mod 8).
(Formerly M2882)
+10
48
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
1,1
COMMENTS
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)
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
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.
MAPLE
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
MATHEMATICA
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[k<nn, If[PrimeQ[p], k++; Sow[p]]; p=p+8]][[2, 1]] (* Zak Seidov, Dec 06 2011 *)
Select[Prime[Range[200]], Mod[#, 8]==3&] (* Harvey P. Dale, Apr 05 2023 *)
PROG
(PARI) forprime(p=2, 97, if(p%8==3, print1(p", "))) \\ Charles R Greathouse IV, Aug 17 2011
(Magma) [p: p in PrimesUpTo(2000) | p mod 8 eq 3]; // Vincenzo Librandi, Aug 07 2012
CROSSREFS
Cf. A294912, A045339 (for n >= 2).
KEYWORD
nonn,easy
STATUS
approved
Composite numbers n such that n == 3 (mod 8) and 2^((n-1)/2) == -1 (mod n).
+10
15
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
1,1
COMMENTS
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
LINKS
Jens Kruse Andersen, Table of n, a(n) for n = 1..10000 (first 132 terms from Robert G. Wilson v)
MAPLE
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
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Gary Detlefs, Jul 02 2014
STATUS
approved
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).
+10
2
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
1,1
COMMENTS
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
MAPLE
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
MATHEMATICA
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 *)
PROG
(PARI) 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
CROSSREFS
A294092 is a subsequence.
KEYWORD
nonn
AUTHOR
Jonas Kaiser, Nov 28 2017
EXTENSIONS
Definition corrected by Jonas Kaiser, Feb 05 2018
STATUS
approved

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