Search: a105876 -id:a105876
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A167800
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Numbers with primitive root -4.
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+10
19
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3, 7, 9, 11, 19, 23, 27, 47, 49, 59, 67, 71, 79, 81, 83, 103, 107, 121, 131, 139, 163, 167, 179, 191, 199, 211, 227, 239, 243, 263, 271, 311, 343, 347, 359, 361, 367, 379, 383, 419, 443, 463, 467, 479, 487, 491, 503, 523, 529, 547, 563, 587, 599, 607, 619, 647
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OFFSET
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1,1
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LINKS
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MATHEMATICA
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pr=-4; Select[Range[2, 2000], MultiplicativeOrder[pr, # ] == EulerPhi[ # ] &]
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CROSSREFS
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Cf. A105876 (primes with primitive root -4)
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A216371
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Odd primes with one coach: primes p such that A135303((p-1)/2) = 1.
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+10
13
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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
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OFFSET
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1,1
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COMMENTS
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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
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)
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
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REFERENCES
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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.
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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isA216371 := proc(n)
if isprime(n) then
true;
else
false;
end if;
else
false;
end if;
end proc:
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:
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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A105876 is the subsequence of terms congruent to 3 modulo 4.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A302141
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Multiplicative order of 16 mod 2n+1.
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+10
4
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1, 1, 1, 3, 3, 5, 3, 1, 2, 9, 3, 11, 5, 9, 7, 5, 5, 3, 9, 3, 5, 7, 3, 23, 21, 2, 13, 5, 9, 29, 15, 3, 3, 33, 11, 35, 9, 5, 15, 39, 27, 41, 2, 7, 11, 3, 5, 9, 12, 15, 25, 51, 3, 53, 9, 9, 7, 11, 3, 6, 55, 5, 25, 7, 7, 65, 9, 9, 17, 69, 23, 15, 7, 21, 37, 15, 6, 5, 13, 13, 33, 81, 5, 83, 39, 9, 43, 15, 29, 89, 45, 15, 9, 10, 9, 95, 24, 3, 49, 99, 33
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OFFSET
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0,4
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COMMENTS
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Reptend length of 1/(2n+1) in hexadecimal.
a(n) <= n; it appears that equality holds if and only if n=1 or is in A163778. - Robert Israel, Apr 02 2018
a(n) <= psi(2*n+1)/2 <= n. a(n) = psi(2*n+1)/2 if and only if the multiplicative order of 2 modulo 2*n+1 is psi(2*n+1) or psi(2*n+1)/2, and psi(2*n+1) == 2 (mod 4).
a(n) = n if and only if A053447(n) = n and A053447(n) is odd. As a result, a(n) = n if and only if 2*n+1 = p is a prime congruent to 3 modulo 4, and the multiplicative order of 2 modulo p is p-1 or (p-1)/2 (p-1 if p == 3 (mod 8), (p-1)/2 if p == 7 (mod 8)). Such primes p are listed in A105876. (End)
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LINKS
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FORMULA
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EXAMPLE
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The fraction 1/13 is equal to 0.13B13B... in hexadecimal, so a(6) = 3.
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MAPLE
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seq(numtheory:-order(16, 2*n+1), n=0..100); # Robert Israel, Apr 02 2018
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MATHEMATICA
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Table[MultiplicativeOrder[16, 2 n + 1], {n, 0, 150}] (* Vincenzo Librandi, Apr 03 2018 *)
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PROG
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(Magma) [1] cat [ Modorder(16, 2*n+1): n in [1..100]]; // Vincenzo Librandi, Apr 03 2018
(GAP) List([0..100], n->OrderMod(16, 2*n+1)); # Muniru A Asiru, Feb 25 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A114564
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Numbers of the form n=12s+7, where q=4s+3 is a prime for which the order of 2 is either q-1 or (q-1)/2.
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+10
1
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7, 19, 31, 55, 67, 139, 175, 199, 211, 235, 247, 307, 319, 391, 415, 487, 499, 535, 571, 595, 631, 679, 715, 787, 811, 931, 1039, 1075, 1099, 1135, 1147, 1255, 1327, 1387, 1399, 1435, 1459, 1471, 1507, 1567, 1639, 1687, 1759, 1795, 1819, 1855
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OFFSET
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1,1
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REFERENCES
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L. Heffter, Über das Problem der Nachbargebiete, Math. Ann., 38 (1891) 477-508.
G. Ringel, Map Color Theorem, Springer, 1974, p. 6.
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LINKS
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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