A216371 -id:A216371 - cp's OEIS frontend

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

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.

A364867 Primes p such that the multiplicative order of 9 modulo p is (p-1)/2.

Original entry on oeis.org

5, 7, 11, 17, 19, 23, 29, 31, 43, 47, 53, 59, 71, 79, 83, 89, 101, 107, 113, 127, 131, 137, 139, 149, 163, 167, 173, 179, 191, 197, 199, 211, 223, 227, 233, 239, 251, 257, 263, 269, 281, 283, 293, 311, 317, 331, 347, 353, 359, 379, 383, 389, 401, 419, 443, 449, 461, 463, 467, 479, 487

Offset: 1

Jianing Song, Aug 11 2023

Primes p such that the multiplicative order of 9 modulo p is of the maximum possible value.
Primes p such that 3 or -3 (or both) is a primitive root modulo p. Proof of equivalence: let ord(a,k) be the multiplicative order of a modulo p. if ord(3,p) = p-1, then clearly ord(9,p) = (p-1)/2. If ord(-3,p) = p-1, then we also have ord(9,p) = (p-1)/2. Conversely, suppose that ord(9,p) = (p-1)/2, then ord(3,p) = p-1 or (p-1)/2, and ord(-3,p) = p-1 or (p-1)/2. If ord(3,p) = ord(-3,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) 3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of 3 modulo p is (p-1)/2 (in this case, we have p == 11 (mod 12) since 3 is a quadratic residue modulo p).
A prime p is a term if and only if one of the two following conditions holds: (a) -3 is a primitive root modulo p; (b) p == 3 (mod 4), and the multiplicative order of -3 modulo p is (p-1)/2 (in this case, we have p == 7 (mod 12) since -3 is a quadratic residue modulo p).
No terms are congruent to 1 modulo 12, since otherwise we would have 9^((p-1)/4) = (+-3)^((p-1)/2) == 1 (mod p). - Jianing Song, May 14 2024

			7 is a term since the multiplicative order of 9 modulo 7 is 3 = (7-1)/2.

Jianing Song, Table of n, a(n) for n = 1..10000

Union of A019334 and A105875.
A105881 is the subsequence of terms congruent to 3 modulo 4.
Cf. A211245 (order of 9 mod n-th prime), A216371.

okQ[p_] := MultiplicativeOrder[9, p] == (p - 1)/2;
Select[Prime[Range[100]], okQ] (* Jean-François Alcover, Nov 24 2024 *)

isA364867(p) = isprime(p) && (p!=3) && znorder(Mod(9, p)) == (p-1)/2

from sympy import n_order, nextprime
from itertools import islice
def A364867_gen(startvalue=4): # generator of terms >= startvalue
    p = max(startvalue-1,3)
    while (p:=nextprime(p)):
        if n_order(9,p) == p-1>>1:
            yield p
A364867_list = list(islice(A364867_gen(),20)) # Chai Wah Wu, Aug 11 2023

A372797 Smallest prime p such that the multiplicative order of 4 modulo p is 2*n, or 0 if no such prime exists.

Original entry on oeis.org

3, 17, 31, 73, 151, 433, 631, 337, 127, 241, 331, 601, 4421, 673, 3061, 257, 1429, 1657, 1103, 3121, 2143, 1321, 18539, 1777, 2351, 37441, 2971, 2857, 3191, 17401, 683, 15809, 17029, 9929, 38431, 1801, 11471, 63689, 49999, 13121, 17467, 21169, 83077, 25609, 5581, 5153, 26227

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 4. Also the first prime p such that {k/p : 1 <= k <= p-1} has 2*n different cycles when written out in base 4.

Since ord(a^m,k) = ord(a,k)/gcd(m,ord(a,k)) for gcd(a,k) = 1, we have that (p-1)/ord(4,p) = ((p-1)/ord(2,p)) * gcd(2,ord(2,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-4 expansion of 1/3 is 0.(1), so the reptend has length 1 = (3-1)/2. Also, the base-4 expansions of 1/3 = 0.(1) and 2/3 = 0.(2) have two cycles 1 and 2. 3 is the smallest such prime, so a(1) = 3.
The base-4 expansion of 1/17 is 0.(0033), so the reptend has length 4 = (17-1)/4. Also, the base-4 expansions of 1/17, 2/17, ..., 16/17 have four cycles 0033, 0132, 1023 and 1122. 17 is the smallest such prime, so a(2) = 17.
The base-4 expansion of 1/31 is 0.(00133), so the reptend has length 5 = (31-1)/6. Also, the base-4 expansions of 1/31, 2/31, ..., 30/31 have three cycles 00201, 01203, 02211, 03213, 11223 and 13233. 13 is the smallest such prime, so a(3) = 13.

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

Cf. A082654, A216371.

In other bases: A101208, A101209, A372798, A372799, A054471, A372800.

a[n_] := a[n] = For[p = 2, True, p = NextPrime[p], If[MultiplicativeOrder[4, 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=4}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base,p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))

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.

Original entry on oeis.org

2, 2, 2, 2, 2, 16, 2, 2, 2, 6, 2, 8, 6, 2, 2, 2, 2, 2, 2, 8, 2, 2, 8, 32, 2, 2, 2, 54, 16, 18, 2, 8, 2, 2, 50, 6, 2, 2, 2, 2, 2, 2, 64, 2, 2, 2, 6, 2, 6, 8, 2, 80, 250, 256, 2, 2, 2, 6, 8, 6, 2, 18, 2, 8, 2, 22, 16, 2, 2, 32, 2, 2, 2, 2, 2, 2, 18, 16, 8, 2, 2, 10, 432, 6, 2, 64, 24, 2, 2, 2, 2, 2, 2, 6, 2, 2, 8, 2, 2, 2, 2, 2, 8, 10, 64, 2, 16, 2, 24, 2, 2, 8, 2, 14, 640, 6

Offset: 2

Max Alekseyev, Sep 05 2024

nonn

a(n) does not depend on the choice of a primitive root r modulo prime(n).
a(n) = prime(n) - 1 iff prime(n) is in A376008.
a(n) = 2 iff prime(n) is in A216371.
a(n) > 2 iff prime(n) is in A268923.

Cf. A216371, A268923, A376008.

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

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A364867 Primes p such that the multiplicative order of 9 modulo p is (p-1)/2.

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A372797 Smallest prime p such that the multiplicative order of 4 modulo p is 2*n, or 0 if no such prime exists.

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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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