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Numbers with primitive root -2.
+10
19
5, 7, 13, 23, 25, 29, 37, 47, 49, 53, 61, 71, 79, 101, 103, 125, 149, 167, 169, 173, 181, 191, 197, 199, 239, 263, 269, 271, 293, 311, 317, 343, 349, 359, 367, 373, 383, 389, 421, 461, 463, 479, 487, 503, 509, 529, 541, 557, 599, 607, 613, 625, 647, 653, 661
MATHEMATICA
pr=-2; Select[Range[2, 2000], MultiplicativeOrder[pr, # ] == EulerPhi[ # ] &]
CROSSREFS
Cf. A105874 (primes with primitive root -2)
Odd primes with one coach: primes p such that A135303((p-1)/2) = 1.
+10
15
3, 5, 7, 11, 13, 19, 23, 29, 37, 47, 53, 59, 61, 67, 71, 79, 83, 101, 103, 107, 131, 139, 149, 163, 167, 173, 179, 181, 191, 197, 199, 211, 227, 239, 263, 269, 271, 293, 311, 317, 347, 349, 359, 367, 373, 379, 383, 389, 419, 421, 443, 461, 463, 467, 479, 487
COMMENTS
Given that prime p has only one coach, the corresponding value of k in A003558 must be (p-1)/2, and vice versa. Using the Coach theorem of Jean Pedersen et al., phi(b) = 2 * c * k, with b odd. Let b = p, prime. Then phi(p) = (p-1), and k must be (p-1)/2 iff c = 1. Or, phi(p) = (p-1) = 2 * 1 * (p-1)/2. Conjecture relating to odd integers: iff an integer is in the set A216371 and is either of the form 4q - 1 or 4q + 1, (q>0); then the top row of its coach (cf. A003558) is composed of a permutation of the first q odd integers. Examples: 11 is of the form 4q - 1, q = 3; with the top row of its coach [1, 5, 3]. 13 is of the form 4q + 1, q = 3; so has a coach of [1, 3, 5]. 37 is of the form 4q + 1, q = 9; so has a coach with the top row composed of a permutation of the first 9 odd integers: [1, 9, 7, 15, 11, 13, 3, 17, 5]. - Gary W. Adamson, Sep 08 2012 Odd primes p such that 2^m is not congruent to 1 or -1 (mod p) for 0 < m < (p-1)/2. - Charles R Greathouse IV, Sep 15 2012 These are also the odd primes a(n) for which there is only one periodic Schick sequence (see the reference, and also the Brändli and Beyne link, eq. (2) for the recurrence but using various inputs. See also a comment in A332439). This sequence has primitive period length (named pes in Schick's book) A003558((a(n)-1)/2) = A005034(a(n)) = A000010(a(n))/2 = (a(n) - 1)/2, for n >= 1. - Wolfdieter Lang, Apr 09 2020 Primes p such that the multiplicative order of 4 modulo p is (p-1)/2. Proof of equivalence: let ord(a,k) be the multiplicative of a modulo k.
If 2^m is not 1 or -1 (mod p) for 0 < m < (p-1)/2, then ord(2,p) is either p-1 or (p-1)/2. If ord(2,p) = p-1, then ord(4,p) = (p-1)/2. If ord(2,p) = (p-1)/2, then p == 3 (mod 4), otherwise 2^((p-1)/4) == -1 (mod p), so ord(4,p) = (p-1)/2.
Conversely, if ord(4,p) = (p-1)/2, then ord(2,p) = p-1, or ord(2,p) = (p-1)/2 and p == 3 (mod 4) (otherwise ord(4,p) = (p-1)/4). In the first case, (p-1)/2 is the smallest m > 0 such that 2^m == +-1 (mod p); in the second case, since (p-1)/2 is odd, 2^m == -1 (mod p) has no solution. In either case, so 2^m is not 1 or -1 (mod p) for 0 < m < (p-1)/2.
{(a(n)-1)/2} is the sequence of indices of fixed points of A053447. A prime p is a term if and only if one of the two following conditions holds: (a) 2 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 2 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 8) since 2 is a quadratic residue modulo p). (End)
Primes p such that 2 or -2 (or both) is a primitive root modulo p. Proof of equivalence: if ord(2,p) = p-1, then clearly ord(4,p) = (p-1)/2. If ord(-2,p) = p-1, then we also have ord(4,p) = (p-1)/2. Conversely, suppose that ord(4,p) = (p-1)/2, then ord(2,p) = p-1 or (p-1)/2, and ord(-2,p) = p-1 or (p-1)/2. If ord(2,p) = ord(-2,p) = (p-1)/2, then we have that (p-1)/2 is odd and (-1)^((p-1)/2) == 1 (mod p), a contradiction.
A prime p is a term if and only if one of the two following conditions holds: (a) -2 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of -2 modulo p is (p-1)/2 (in this case, we have p == 3 (mod 8) since -2 is a quadratic residue modulo p). (End)
No terms are congruent to 1 modulo 8, since otherwise we would have 4^((p-1)/4) = (+-2)^((p-1)/2) == 1 (mod p). - Jianing Song, May 14 2024
REFERENCES
P. Hilton and J. Pedersen, A Mathematical Tapestry, Demonstrating the Beautiful Unity of Mathematics, 2010, Cambridge University Press, pages 260-264.
Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Bokos Druck, Zürich, 2003 (ISBN 3-9522917-0-6). Tables 3.1 to 3.10, for odd p = 3..113 (with gaps), pp. 158-166.
EXAMPLE
Prime 23 has a k value of 11 = (23 - 1)/2 (Cf. A003558(11). It follows that 23 has only one coach ( A135303(11) = 1). 23 is thus in the set. On the other hand 31 is not in the set since A135303(15) shows 3 coaches, with A003558(15) = 5. 13 is in the set since A135303(6) = 1; but 17 isn't since A135303(8) = 2.
MAPLE
isA216371 := proc(n)
if isprime(n) then
true;
else
false;
end if;
else
false;
end if;
end proc:
local p;
if n = 1 then
3;
else
p := nextprime(procname(n-1)) ;
while true do
if isA216371(p) then
return p;
end if;
p := nextprime(p) ;
end do:
end if;
end proc:
MATHEMATICA
Suborder[a_, n_] := If[n > 1 && GCD[a, n] == 1, Min[MultiplicativeOrder[a, n, {-1, 1}]], 0]; nn = 150; Select[Prime[Range[2, nn]], EulerPhi[#]/(2*Suborder[2, #]) == 1 &] (* T. D. Noe, Sep 18 2012 *) f[p_] := Sum[Cos[2^n Pi/((2 p + 1))], {n, p}]; 1 + 2 * Select[Range[500], Reduce[f[#] == -1/2, Rationals] &]; (* Gerry Martens, May 01 2016 *)
PROG
(PARI) is(p)=for(m=1, p\2-1, if(abs(centerlift(Mod(2, p)^m))==1, return(0))); p>2 && isprime(p) \\ Charles R Greathouse IV, Sep 18 2012 (PARI) is(p) = isprime(p) && (p>2) && znorder(Mod(4, p)) == (p-1)/2 \\ Jianing Song, Dec 24 2022
CROSSREFS
A105876 is the subsequence of terms congruent to 3 modulo 4.
Complement of A268923 in the set of odd primes.
All odd primes a(n) such that for all odd primes q smaller than a(n) the order of 2 modulo a(n)*q is a proper divisor of phi(a(n)*q)/2. The totient function phi is given in A000010.
+10
6
17, 31, 41, 43, 73, 89, 97, 109, 113, 127, 137, 151, 157, 193, 223, 229, 233, 241, 251, 257, 277, 281, 283, 307, 313, 331, 337, 353, 397, 401, 409, 431, 433, 439, 449, 457, 499, 521, 569, 571, 577, 593, 601, 617, 631, 641, 643, 673, 683, 691, 727, 733, 739, 761, 769, 809, 811, 857, 881, 911, 919
COMMENTS
It seems that if for an odd prime p > 3 the order(2, p*3) < phi(p*3)/2 = p-1 then p is in this sequence.
Note that 2^(phi(p*q)/2) == 1 (mod p*q) for distinct odd primes p and q, due to Nagell's corollary on Theorem 64, p. 106. The products of distinct primes considered in the present sequence have order of 2 modulo p*q smaller than phi(p*q)/2.
Up to and including prime(100) = 541 the only odd primes p such that for all odd primes q smaller than p the order of 2 modulo p*q equals phi(p*q)/2 are 5, 7, and 11.
Complement of A216371 = A001122 U A105874 in the set of odd primes. Composed of the primes modulo which neither 2 nor -2 is a primitive root. Also, prime(n) is a term iff A376010(n) > 2. - Max Alekseyev, Sep 05 2024
EXAMPLE
n=1: Order(2, 17*3) = 8, and 8 is a proper divisor of phi(17*3)/2 = 16;
order(2, 17*5) = 8, and 8 is a proper divisor of phi(17*5)/2 = 32;
order(2, 17*7) = 24, and 24 is a proper divisor of phi(17*7)/2 = 48;
order(2, 17*11) = 40, and 40 is a proper divisor of phi(17*11)/2 = 80;
order(2, 17*13) = 24, and 24 is a proper divisor of phi(17*13)/2 = 96.
MATHEMATICA
Select[Prime@ Range[3, 157], Function[p, AllTrue[Prime@ Range[2, PrimePi@ p - 1], Function[q, With[{e = EulerPhi[p q]/2}, And[Divisible[e, #], # != e]] &@ MultiplicativeOrder[2, p q]]]]] (* Michael De Vlieger, Apr 01 2016, Version 10 *)
Primes for which -3 is a primitive root.
+10
4
2, 5, 11, 17, 23, 29, 47, 53, 59, 71, 83, 89, 101, 107, 113, 131, 137, 149, 167, 173, 179, 191, 197, 227, 233, 239, 251, 257, 263, 269, 281, 293, 311, 317, 347, 353, 359, 383, 389, 401, 419, 443, 449, 461, 467, 479, 503, 509, 521, 557, 563, 569, 587, 593, 599, 617, 641
COMMENTS
Also, primes for which -27 is a primitive root. Proof: -27 = (-3)^3, so -27 is a primitive root just when -3 is a primitive root and the prime is not 3k+1. Now if -3 is a primitive root, then -3 is not a quadratic residue and so the prime is not 3k+1. - Don Reble, Sep 15 2007
MATHEMATICA
pr=-3; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == #-1 &]
PROG
(Python)
from sympy import n_order, nextprime
from itertools import islice
def A105875_gen(startvalue=2): # generator of terms >= startvalue p = max(startvalue-1, 1)
while (p:=nextprime(p)):
if p!=3 and n_order(-3, p) == p-1:
yield p
Primes for which -8 is a primitive root.
+10
3
5, 23, 29, 47, 53, 71, 101, 149, 167, 173, 191, 197, 239, 263, 269, 293, 311, 317, 359, 383, 389, 461, 479, 503, 509, 557, 599, 647, 653, 677, 701, 719, 743, 773, 797, 821, 839, 863, 887, 941, 983, 1031, 1061, 1109, 1151, 1223, 1229, 1277, 1301, 1319, 1367, 1373, 1439
COMMENTS
Members of A105874 that are not congruent to 1 mod 3. Terms are congruent to 5 or 23 modulo 24. According to Artin's conjecture, the number of terms <= N is roughly ((3/5)*C)*PrimePi(N), where C is the Artin's constant = A005596, PrimePi = A000720. Compare: the number of terms of A001122 that are no greater than N is roughly C*PrimePi(N). (End)
FORMULA
Let a(p,q)=sum(n=1,2*p*q,2*cos(2^n*Pi/((2*q+1)*(2*p+1)))). Then 2*p+1 is a prime of this sequence when a(p,9)==1. - Gerry Martens , May 21 2015
MATHEMATICA
pr=-8; Select[Prime[Range[400]], MultiplicativeOrder[pr, # ] == #-1 &] (* N. J. A. Sloane, Jun 01 2010 *) a[p_, q_]:= Sum[2 Cos[2^n Pi/((2 q+1)(2 p+1))], {n, 1, 2 q p}]
2 Select[Range[800], Rationalize[N[a[#, 9], 20]] == 1 &] + 1
Numbers m such that all numbers {1...m} appear in the sequence {b(0) = m, b(n+1) = b(n)/2 if even, m-(b(n)+1)/2 otherwise}.
+10
1
1, 2, 3, 4, 6, 7, 10, 12, 15, 19, 24, 27, 30, 31, 34, 36, 40, 42, 51, 52, 54, 66, 70, 75, 82, 84, 87, 90, 91, 96, 99, 100, 106, 114, 120, 132, 135, 136, 147, 156, 159, 174, 175, 180, 184, 187, 190, 192, 195, 210, 211, 222, 231, 232, 234, 240, 244, 246, 252, 255, 262
COMMENTS
Lemma: A sequence {b(n)} defined as above with m>1 cannot have values outside [1,m]. (For m=1, b=(1,0,0,0....).)
Corollary: Such a sequence {b(n)} is periodic with period <= m (except maybe for some initial terms).
Lemma 2: For any m>1, b(1) = floor( m/2 ) and if b(n)=m-1, then b(n+1)= [ m/2 ].
Proposition: As soon as there is a term b(n)=2^k, the (b-)sequence continues b(n+1)=2^(k-1),...,b(n+k)=1, b(n+k+1)=m-1 and then starts over with b(n+k+2)=b(1).
Corollary 2: Numbers m=2^k, k>2 cannot appear in the present sequence.
Proposition: For any b(0)=m>1, sooner or later the value 1 is reached.
Generate a sequence b(n) by the following rule. If b(n-1) is divisible by 2 then b(n) = b(n-1)/2. If b(n-1) is not divisible by 2 then b(n) = b(0)-(b(n-1)+1)/2. When b(n)=1 it ends. Sequence gives all m such that all numbers k with 1<=k<=m-2 appear in b(n), b(0)=m.
Sequence contains 1 and numbers m>1 such that 2m-1 is prime and -2 or 2 is a primitive root modulo 2m-1. - Max Alekseyev, May 16 2008
EXAMPLE
6->3->4->2->1. 1,2,3,4=6-2 appear in b(n), b(0)=6. So 6 is a term of A137606.
MATHEMATICA
f[n_] := Block[{lst = {n}, a}, While[a = Last@ lst; a != 1, AppendTo[lst, If[ EvenQ@ a, a/2, lst[[1]] - (a + 1)/2]]]; Length@ lst - 1]; t = Array[f, 262]; Select[ Range @ 262], t[[ # ]] == # - 2 &] (* Robert G. Wilson v *)
PROG
(PARI) b137606(n)= n=[n]; for( i=1, n[1]-1, n=concat( n, if( n[i]%2, n[1]-(n[i]+1)/2, n[i]/2 )); n[i]>1 || break); n
A137606(Nmax) = for( n=1, Nmax, n==#b137606(n) && print1(n", "))
(PARI) forprime(p=3, 10^3, if(znorder(Mod(-2, p))==p-1||znorder(Mod(2, p))==p-1, print1((p+1)/2, ", ") )) \\ Max Alekseyev, May 16 2008
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