This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A216371 #110 Oct 31 2024 04:09:03
%S A216371 3,5,7,11,13,19,23,29,37,47,53,59,61,67,71,79,83,101,103,107,131,139,
%T A216371 149,163,167,173,179,181,191,197,199,211,227,239,263,269,271,293,311,
%U A216371 317,347,349,359,367,373,379,383,389,419,421,443,461,463,467,479,487
%N A216371 Odd primes with one coach: primes p such that A135303((p-1)/2) = 1.
%C A216371 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.
%C A216371 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
%C A216371 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
%C A216371 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
%C A216371 From _Jianing Song_, Dec 24 2022: (Start)
%C A216371 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.
%C A216371 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.
%C A216371 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.
%C A216371 {(a(n)-1)/2} is the sequence of indices of fixed points of A053447.
%C A216371 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)
%C A216371 From _Jianing Song_, Aug 11 2023: (Start)
%C A216371 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.
%C A216371 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)
%C A216371 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
%C A216371 The n-th prime A000040(n) is a term iff A376010(n) = 2. - _Max Alekseyev_, Sep 05 2024
%D A216371 P. Hilton and J. Pedersen, A Mathematical Tapestry, Demonstrating the Beautiful Unity of Mathematics, 2010, Cambridge University Press, pages 260-264.
%D A216371 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.
%H A216371 Charles R Greathouse IV, <a href="/A216371/b216371.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from T. D. Noe)
%H A216371 Gerold Brändli and Tim Beyne, <a href="https://arxiv.org/abs/1504.02757">Modified Congruence Modulo n with Half the Amount of Residues</a>, arXiv:1504.02757 [math.NT], 2016.
%H A216371 Marcelo E. Coniglio, Francesc Esteva, Tommaso Flaminio, and Lluis Godo, <a href="https://arxiv.org/abs/2103.07548">On the expressive power of Lukasiewicz's square operator</a>, arXiv:2103.07548 [math.LO], 2021.
%F A216371 a(n) = 2*A054639(n) + 1. - _L. Edson Jeffery_, Dec 18 2012
%e A216371 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.
%e A216371 13 is in the set since A135303(6) = 1; but 17 isn't since A135303(8) = 2.
%p A216371 isA216371 := proc(n)
%p A216371 if isprime(n) then
%p A216371 if A135303((n-1)/2) = 1 then
%p A216371 true;
%p A216371 else
%p A216371 false;
%p A216371 end if;
%p A216371 else
%p A216371 false;
%p A216371 end if;
%p A216371 end proc:
%p A216371 A216371 := proc(n)
%p A216371 local p;
%p A216371 if n = 1 then
%p A216371 3;
%p A216371 else
%p A216371 p := nextprime(procname(n-1)) ;
%p A216371 while true do
%p A216371 if isA216371(p) then
%p A216371 return p;
%p A216371 end if;
%p A216371 p := nextprime(p) ;
%p A216371 end do:
%p A216371 end if;
%p A216371 end proc:
%p A216371 seq(A216371(n),n=1..40) ; # _R. J. Mathar_, Dec 01 2014
%t A216371 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 *)
%t A216371 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 *)
%o A216371 (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
%o A216371 (PARI) is(p) = isprime(p) && (p>2) && znorder(Mod(4,p)) == (p-1)/2 \\ _Jianing Song_, Dec 24 2022
%Y A216371 Union of A001122 and A105874.
%Y A216371 A105876 is the subsequence of terms congruent to 3 modulo 4.
%Y A216371 Complement of A268923 in the set of odd primes.
%Y A216371 Cf. A082654 (order of 4 mod n-th prime), A000010, A000040, A003558, A005034, A053447, A054639, A135303, A364867, A376010.
%K A216371 nonn,easy
%O A216371 1,1
%A A216371 _Gary W. Adamson_, Sep 05 2012