A364867 -id:A364867 - cp's OEIS frontend

A211245 Order of 9 mod n-th prime: least k such that prime(n) divides 9^k-1.

1, 0, 2, 3, 5, 3, 8, 9, 11, 14, 15, 9, 4, 21, 23, 26, 29, 5, 11, 35, 6, 39, 41, 44, 24, 50, 17, 53, 27, 56, 63, 65, 68, 69, 74, 25, 39, 81, 83, 86, 89, 45, 95, 8, 98, 99, 105, 111, 113, 57, 116, 119, 60, 125, 128, 131, 134, 15, 69, 140, 141, 146, 17, 155, 39

Offset: 1

T. D. Noe, Apr 11 2012

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A364867.

In other bases: A014664, A062117, A082654, A211241, A211242, A211243, A211244, A002371, A372801.

A000040:=Filtered([1..350],IsPrime);;
List([1..Length(A000040)],n->OrderMod(9,A000040[n])); # Muniru A Asiru, Feb 06 2019

nn = 9; Table[If[Mod[nn, p] == 0, 0, MultiplicativeOrder[nn, p]], {p, Prime[Range[100]]}]

a(n,{base=9}) = my(p=prime(n)); if(base%p, znorder(Mod(base,p)), 0) \\ Jianing Song, May 13 2024

From Jianing Song, May 13 2024: (Start)
a(n) = A062117(n)/gcd(2, A062117(n)).
a(n) <= (prime(n) - 1)/2. Those prime(n) for which a(n) = (prime(n) - 1)/2 are listed in A364867. (End)

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

A372799 Smallest prime p such that the multiplicative order of 9 modulo p is 2*n, or 0 if no such prime exists.

Original entry on oeis.org

5, 13, 67, 313, 41, 61, 883, 433, 271, 2161, 683, 193, 1223, 8317, 2131, 769, 2551, 1621, 8513, 2521, 8779, 4357, 5843, 3889, 7451, 16069, 3079, 19993, 14327, 661, 23747, 95617, 42703, 2857, 15401, 17209, 2887, 7297, 547, 13441, 4019, 757, 41453, 29833, 54631, 31741, 20399

Offset: 1

Jianing Song, May 13 2024

nonn

First prime p such that the expansion of 1/p has period (p-1)/(2*n) in base 9. Also the first prime p such that {k/p : 1 <= k <= p-1} has 2*n different cycles when written out in base 9.

Since ord(a^m,k) = ord(a,k)/gcd(m,ord(a,k)) for gcd(a,k) = 1, we have that (p-1)/ord(9,p) = ((p-1)/ord(3,p)) * gcd(2,ord(3,p)) is always even. Here ord(a,k) is the multiplicative order of a modulo k.

			In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-9 expansion of 1/5 is 0.(17), so the reptend has length 2 = (5-1)/2. Also, the base-9 expansions of 1/5 = 0.(17), 2/5 = (0.35), 3/5 = 0.(53) and 4/5 = 0.(71) have two cycles 17 and 35. 5 is the smallest such prime, so a(1) = 5.
The base-9 expansion of 1/13 is 0.(062), so the reptend has length 3 = (13-1)/4. Also, the base-9 expansions of 1/13, 2/13, ..., 12/13 have four cycles 062, 134, 268 and 475. 13 is the smallest such prime, so a(2) = 13.
The base-9 expansion of 1/67 is 0.(01178285332), so the reptend has length 11 = (67-1)/6. Also, the base-9 expansions of 1/67, 2/67, ..., 66/67 have six cycles 01178285332, 02367581664, 03556877106, 04746273438, 07224865213 and 08414261545. 67 is the smallest such prime, so a(3) = 67.

Jean-François Alcover, Table of n, a(n) for n = 1..1000

Cf. A211245, A364867.

In other bases: A101208, A101209, A372797, A372798, A054471, A372800.

a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[9, p] == (p-1)/(2n), Return[p]]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 24 2024 *)

a(n,{base=9}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base,p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))

cp's OEIS Frontend

A211245 Order of 9 mod n-th prime: least k such that prime(n) divides 9^k-1.

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A372799 Smallest prime p such that the multiplicative order of 9 modulo p is 2*n, or 0 if no such prime exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI