A268923 -id:A268923 - cp's OEIS frontend

A345388 a(n) = 0, 1, or 2 according to whether A065091(n), the n-th odd prime, is in A001122, A139035, or A268923, respectively.

0, 0, 1, 0, 0, 2, 0, 1, 0, 2, 0, 2, 2, 1, 0, 0, 0, 0, 1, 2, 1, 0, 2, 2, 0, 1, 0, 2, 2, 2, 0, 2, 0, 0, 2, 2, 0, 1, 0, 0, 0, 1, 2, 0, 1, 0, 2, 0, 2, 2, 1, 2, 2, 2, 1, 0, 1, 2, 2, 2, 0, 2, 1, 2, 0, 2, 2, 0, 0, 2, 1, 1, 0, 0, 1, 0, 2, 2, 2, 0, 0, 2, 2, 2, 0, 2, 2

Offset: 1

Howard Givner, Jun 17 2021

nonn

The three OEIS sequences A001122, A139035, and A268923 are implicitly described in a Zoom lecture that was given May 14, 2021, by James Tanton. Here is a link to the video, followed by a description of how the sequences can be obtained by carrying out the procedure that the speaker described in his talk.
Description of the method:
James Tanton defined GOOD, HALF-GOOD, and BAD odd prime integers and a procedure for determining which of the three categories an odd prime integer belongs to.
Procedure for categorizing an odd prime integer P:
Step 1. Begin with an initial partition (1,P-1) of P.
Step 2. Generate a successor partition, derived from an existing partition.
When (x,y) is an existing partition and x is even, the successor partition is (s,t), where s=x/2 and t=P-s.
When (x,y) is an existing partition and x is odd, the successor partition is (s,t), where t=y/2 and s=P-t.
Step 3. Repeat step 2 until you return to (1,P-1).
He then classified P as either GOOD, HALF-GOOD, or BAD as follows:
P is GOOD when every integer from 1 to P-1 appears among the left parts of the set of generated partitions.
P is HALF-GOOD when P does not meet the requirements for GOOD, but every integer from 1 to P-1 appears somewhere in the set of generated partitions.
P is BAD when P does not meet the requirements for GOOD or HALF-GOOD.
The sequence of GOOD odd prime integers is identical to A001122.
The sequence of HALF-GOOD odd prime integers is identical to A139035.
The sequence of BAD odd prime integers is identical to A268923.

			For P=5, the generated partition set is:
  (1,4), (3,2), (4,1), (2,3), (1,4), and thus 5 is GOOD, so a(2)=0.
For P=7, the generated partition set is:
  (1,6), (4,3), (2,5), (1,6), and thus 7 is HALF-GOOD, so a(3)=1.
For P=17, the generated partition set is:
  (1,16), (9,8), (13,4), (15,2), (16,1), (8,9), (4,13), (2,15), (1,16),
  but 3, 5, 6, 7, 10, 11, 12, and 14 do not appear, and thus 17 is BAD, so a(6)=2.

James Tanton, How to fold things into thirds, sevenths, and thirty-sevenths!, video, May 14, 2021.

Cf. A001122, A065091, A139035, A268923.

Name edited by Felix Fröhlich, Jun 28 2021

A216371 Odd primes with one coach: primes p such that A135303((p-1)/2) = 1.

Original entry on oeis.org

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

Offset: 1

Gary W. Adamson, Sep 05 2012

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
From Jianing 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 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)
From Jianing 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 prime A000040(n) is a term iff A376010(n) = 2. - Max Alekseyev, Sep 05 2024

			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.

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.

Union of A001122 and A105874.
A105876 is the subsequence of terms congruent to 3 modulo 4.
Complement of A268923 in the set of odd primes.
Cf. A082654 (order of 4 mod n-th prime), A000010, A000040, A003558, A005034, A053447, A054639, A135303, A364867, A376010.

isA216371 := proc(n)
    if isprime(n) then
        if A135303((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 *)

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

is(p) = isprime(p) && (p>2) && znorder(Mod(4,p)) == (p-1)/2 \\ Jianing Song, Dec 24 2022

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

A333855 Numbers 2*k + 1 with A135303(k) >= 2, for k >= 1, sorted increasingly.

Original entry on oeis.org

17, 31, 33, 41, 43, 51, 57, 63, 65, 73, 85, 89, 91, 93, 97, 99, 105, 109, 113, 117, 119, 123, 127, 129, 133, 137, 145, 151, 153, 155, 157, 161, 165, 171, 177, 185, 187, 189, 193, 195, 201, 205, 209, 215, 217, 219, 221, 223, 229, 231, 233, 241, 247, 249, 251, 255

Offset: 1

Wolfdieter Lang, May 12 2020

nonn

These are the numbers a(n) for which there is more than one periodic Schick sequence. In his notation B(a(n)) >= 2, for n >= 1.
These are also the numbers a(n) for which there is more than one coach in the complete coach system Sigma(b = a(n)) of Hilton and Pedersen, for n >= 1
These are the numbers a(n) for which there is more than one cycle in the complete system MDS(a(n)) (Modified Doubling Sequence) proposed in the comment by Gary W. Adamson, Aug 20 2019, in A003558.
The complement relative to the odd numbers >= 3 is given in A333854.
The subsequence for odd primes is identical with A268923.

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, pp. 261-264.
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.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020

Cf. A003558, A135303, A216371, A268923, A333854.

1 + 2 Select[Range[2, 127], 2 <= EulerPhi[#2]/(2 If[#2 > 1 && GCD[#1, #2] == 1, Min[MultiplicativeOrder[#1, #2, {-1, 1}]], 0]) & @@ {2, 2 # + 1} &] (* Michael De Vlieger, Oct 15 2020 *)

isok8(m, n) = my(md = Mod(2, 2*n+1)^m); (md==1) || (md==-1);
A003558(n) = my(m=1); while(!isok8(m, n) , m++); m;
isok(m) = (m%2) && eulerphi(m)/(2*A003558((m-1)/2)) >= 2; \\ Michel Marcus, Jun 09 2020

Sequence {a(n)}_{n>=1} of numbers 2*k + 1 satisfying A135303(k) >= 2, for k >= 1, ordered increasingly.

A333854 Numbers 2*k + 1 with A135303(k) = 1, for k >= 1, sorted increasingly.

Original entry on oeis.org

3, 5, 7, 9, 11, 13, 15, 19, 21, 23, 25, 27, 29, 35, 37, 39, 45, 47, 49, 53, 55, 59, 61, 67, 69, 71, 75, 77, 79, 81, 83, 87, 95, 101, 103, 107, 111, 115, 121, 125, 131, 135, 139, 141, 143, 147, 149, 159, 163, 167, 169, 173, 175, 179, 181, 183, 191, 197, 199, 203

Offset: 1

Wolfdieter Lang, May 03 2020

nonn

These are the numbers a(n) for which there is only one periodic Schick sequence. In Schick's notation B(a(n)) = 1, for n >= 1.
These are the numbers a(n) for which there is only one coach in the complete coach system Sigma(b = a(n)) of Hilton and Pedersen, for n >= 1.
These are also the numbers a(n) for which there is only one cycle in the complete system MDS(a(n)) (Modified Doubling Sequence) proposed in the comment by Gary W. Adamson, Aug 20 2019, in A003558.
The subsequence of prime numbers is A216371.
The complement relative to the odd numbers >= 3 is given in A333855.

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, pp. 261-264.
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.

Cf. A003558, A135303, A216371, A268923, A333855.

isok8(m, n) = my(md = Mod(2, 2*n+1)^m); (md==1) || (md==-1);
A003558(n) = my(m=1); while(!isok8(m, n) , m++); m;
isok(m) = (m%2) && eulerphi(m)/(2*A003558((m-1)/2)) == 1; \\ Michel Marcus, Jun 10 2020

Sequence {a(n)}_{n >= 1} of numbers 2*k + 1 satisfying A135303(k) = 1, for k >= 1, ordered increasingly.

cp's OEIS Frontend

A345388 a(n) = 0, 1, or 2 according to whether A065091(n), the n-th odd prime, is in A001122, A139035, or A268923, respectively.

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A216371 Odd primes with one coach: primes p such that A135303((p-1)/2) = 1.

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A333855 Numbers 2*k + 1 with A135303(k) >= 2, for k >= 1, sorted increasingly.

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A333854 Numbers 2*k + 1 with A135303(k) = 1, for k >= 1, sorted increasingly.

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A376010 a(n) is the smallest distance between a pair of equal terms in the sequence s(0) = 1, s(1) = r, and s(k) = s(k-1)^2/(4*s(k-2)) mod p for k>=2, where p = prime(n) (=A000040(n)) and r is a primitive root modulo p.

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