1,1

Given that prime p has only one coach, the corresponding value of k inA003558must 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 setA216371and 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 inA332439). 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

FromJianing Song,Dec 24 2022: (Start)

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 ofA053447.

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)

FromJianing Song,Aug 11 2023: (Start)

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

The n-th primeA000040(n) is a term iffA376010(n) = 2. -Max Alekseyev,Sep 05 2024

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.

Charles R Greathouse IV,Table of n, a(n) for n = 1..10000(first 1000 terms from T. D. Noe)

Gerold Brändli and Tim Beyne,Modified Congruence Modulo n with Half the Amount of Residues,arXiv:1504.02757 [math.NT], 2016.

Marcelo E. Coniglio, Francesc Esteva, Tommaso Flaminio, and Lluis Godo,On the expressive power of Lukasiewicz's square operator,arXiv:2103.07548 [math.LO], 2021.

a(n) = 2*A054639(n) + 1. -L. Edson Jeffery,Dec 18 2012

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 sinceA135303(15) shows 3 coaches, withA003558(15) = 5.

13 is in the set sinceA135303(6) = 1; but 17 isn't sinceA135303(8) = 2.

isA216371:= proc(n)

if isprime(n) then

ifA135303((n-1)/2) = 1 then

true;

else

false;

end if;

else

false;

end if;

end proc:

A216371:= proc(n)

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:

seq(A216371(n), n=1..40); #R. J. Mathar,Dec 01 2014

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 *)

(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

Union ofA001122andA105874.

A105876is the subsequence of terms congruent to 3 modulo 4.

Complement ofA268923in the set of odd primes.

Cf.A082654(order of 4 mod n-th prime),A000010,A000040,A003558,A005034,A053447,A054639,A135303,A364867,A376010.

Sequence in context:A350577A079733A179538*A095747A132779A192869

Adjacent sequences:A216368A216369A216370*A216372A216373A216374

nonn,easy,changed

Gary W. Adamson,Sep 05 2012

approved