A003147 -id:A003147 - cp's OEIS frontend

A308794 Primes p such that A001175(p) = (p-1)/9.

199, 919, 6679, 7489, 12979, 16921, 17011, 17659, 20089, 20431, 23059, 23599, 24391, 24859, 25309, 28081, 29629, 33301, 36901, 39079, 39439, 41761, 42589, 43399, 43669, 45361, 46261, 48619, 51481, 53479, 54091, 62011, 62191, 67411, 69499, 72019, 72091, 77419, 78301

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/9, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p not dividing k^2 + 4, the Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the Pisano period of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 9.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      2 |            78
  4 |      4 |           609
  5 |     49 |          4777
  6 |    405 |         39210
  7 |   3489 |        332136
  8 |  30132 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), this sequence (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
Reap[For[p = 2, p < 50000, p = NextPrime[p], If[Mod[p, 9] == 1, If[pn[p] == (p - 1)/9, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

Pisano_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4,p)==1, my(v=divisors(p-1)); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 80000, if(Pisano_for_decomposing_prime(p)==(p-1)/9, print1(p, ", ")))

A293200 Primes p with a primitive root g such that g^3 = g + 1 mod p.

Original entry on oeis.org

5, 7, 11, 17, 23, 37, 59, 67, 83, 101, 113, 167, 173, 199, 211, 227, 241, 251, 271, 283, 307, 317, 367, 373, 401, 433, 457, 479, 569, 571, 593, 599, 607, 613, 643, 659, 691, 701, 719, 727, 743, 757, 769, 809, 821, 829, 839, 853, 877, 883, 919, 941, 977, 991, 997, 1019, 1031, 1049

Offset: 1

Joerg Arndt, Oct 02 2017

nonn

Since g^3 = g + 1, we have g^4 = g^2 + g, g^5 = g^3 + g^2, g^6 = g^4 + g^3, ..., g^(k+3) = g^(k+1) + g^k. Hence using g and g^2 we can compute all powers of the primitive root similar to A003147, where we have g^(k+2) = g^(k+1) + g^k (see the Shanks reference).

Robert Israel, Table of n, a(n) for n = 1..10000
D. Shanks, Fibonacci primitive roots, end of article, Fib. Quart., 10 (1972), 163-168, 181.

Cf. A003147 (primitive root g such that g^2 = g + 1 mod p).
Cf. A293201 (primitive root g such that g^3 = g^2 + g + 1 mod p).
Cf. A104217.

filter:= proc(p) local x,r;
  if not isprime(p) then return false fi;
  for r in map(t -> rhs(op(t)), [msolve(x^3-x-1,p)]) do
    if numtheory:-order(r,p) = p-1 then return true fi
  od;
  false
end proc:
select(filter, [seq(i,i=3..2000,2)]); # Robert Israel, Oct 02 2017

selQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^3 - # - 1, p] == 0&];
Select[Prime[Range[200]], selQ] (* Jean-François Alcover, Jul 29 2020 *)

Z(r,p)=znorder(Mod(r,p))==p-1;  \\ whether r is a primitive root mod p
Y(p)=for(r=2,p-2,if( Z(r,p) && Mod(r^3-r-1,p)==0 , return(1))); 0; \\ test p
forprime(p=2,10^3,if(Y(p),print1(p,", ")) );

A293201 Primes p with a primitive root g such that g^3 = g^2 + g + 1 mod p.

Original entry on oeis.org

7, 11, 13, 17, 19, 41, 47, 53, 73, 107, 131, 139, 149, 163, 167, 199, 227, 257, 263, 269, 271, 293, 311, 347, 349, 359, 373, 401, 419, 421, 431, 479, 523, 557, 599, 617, 683, 701, 757, 761, 769, 809, 827, 863, 877, 907, 911, 929, 937, 953, 991, 1031, 1033

Offset: 1

Joerg Arndt, Oct 02 2017

nonn

Since g^3 = g^2 + g + 1, we have g^4 = g^3 + g^2 + g, g^5 = g^4 + g^3 + g^2, g^6 = g^5 + g^4 + g^3, ..., g^(k+3) = g^(k+2) + g^(k+1) + g^k. Hence using g and g^2 we can compute all powers of the primitive root similar to A003147, where we have g^(k+2) = g^(k+1) + g^k (see the Shanks reference).

Robert Israel, Table of n, a(n) for n = 1..10000
D. Shanks, Fibonacci primitive roots, end of article, Fib. Quart., 10 (1972), 163-168, 181.

Cf. A003147 (primitive root g such that g^2 = g + 1 mod p).

Cf. A293200 (primitive root g such that g^3 = g + 1 mod p).

filter:= proc(p) local x,r;
  if not isprime(p) then return false fi;
  for r in map(t -> rhs(op(t)), [msolve(x^3-x^2-x-1,p)]) do
    if numtheory:-order(r,p) = p-1 then return true fi
  od;
  false
end proc:
select(filter, [seq(i,i=3..2000,2)]); # Robert Israel, Oct 02 2017

selQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^3 - #^2 - # - 1, p] == 0&];
Select[Prime[Range[2, 200]], selQ] (* Jean-François Alcover, Jul 29 2020 *)

Z(r, p)=znorder(Mod(r, p))==p-1;  \\ whether r is a primitive root mod p
Y(p)=for(r=2,p-2,if( Z(r,p) && Mod(r^3-r^2-r-1,p)==0 , return(1))); 0; \\ test p
forprime(p=2,10^3,if(Y(p),print1(p,", ")) );

A268271 Primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all quadratic residues of p in the first (p-1)/2 iterations (for some b).

Original entry on oeis.org

11, 19, 29, 31, 59, 71, 79, 89, 101, 131, 179, 181, 191, 229, 239, 251, 271, 311, 349, 359, 379, 401, 419, 431, 439, 479, 491, 499, 509, 571, 599, 631, 659, 719, 739, 751, 761, 839, 941, 971, 1019, 1021, 1039, 1051, 1061, 1091, 1109, 1171, 1229, 1249, 1259, 1319, 1361, 1399

Offset: 1

Michel Marcus, Mar 02 2016

nonn

			p=11 is a term since, modulo 11, the sequence 1, 4, 5, 9, 3 satisfies 5=4+1, 9=5+4, 3=9+5, 1=9+3, ..., with a period of (11-1)/2 = 5.

Alexandru Gica, Quadratic Residues in Fibonacci Sequences, Fibonacci Quart. 46/47 (2008/2009), no. 1, 68-72.

Subsequence of A045468.
Cf. A003147 (similar sequence for a different period).
Cf. A168429, A070373 (examples of such Fibonacci-type sequences).

findr(p) = {for (k=1, (p-1)/2, if ((k^2 % p) == 5, return(k)););}
isok(p) = {if ((p % 2) && isprime(p), pm = p % 5; if ((pm == 1) || (pm == 4), rf = findr(p);(znorder(Mod((1+rf)/2, p)) == (p-1)/2) || (znorder(Mod((1-rf)/2, p)) == (p-1)/2);););}

A366951 a(n) = 2(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.

Original entry on oeis.org

2, 1, 0, 1, 2, 1, 1, 2, 1, 4, 2, 1, 2, 1, 3, 1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 4, 1, 3, 2, 3, 1, 2, 1, 6, 2, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 18, 10, 1, 1, 4, 9, 2, 2, 2, 1, 3, 2, 2, 1, 10, 1, 1, 7, 2, 1, 1, 6, 1, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 4, 2, 2, 10, 2, 1, 2, 1

Offset: 1

A.H.M. Smeets, Oct 29 2023

nonn

Cf. A000040, A003147, A060305, A047650, A047652, A071774, A124096, A124253, A296240, A308784 - A308794.

a(n) == 0 (mod 2) for prime(n) == +/- 1 (mod 5) and n > 2.
a(n) == 1 (mod 2) for Prime(n) == +/- 2 (mod 5) and n > 2.
a(n) = 1 iff prime(n) in A071774.
a(n) = 2 iff prime(n) in ({2} union A003147)/{5}.
a(n) = 3 iff prime(n) in A308784.
a(n) = 4 iff prime(n) in A308787.
a(n) = 6 iff prime(n) in A308788.
a(n) = 7 iff prime(n) in A308785.
a(n) = 8 iff prime(n) in A308789.
a(n) = 9 iff prime(n) in A308786.
a(n) = 10 iff prime(n) in A308790.
a(n) = 12 iff prime(n) in A308791.
a(n) = 14 iff prime(n) in A308792.
a(n) = 16 iff prime(n) in A308793.
a(n) = 18 iff prime(n) in A308794.
a(n) = A296240(n) iff prime(n) == +/- 2 (mod 5) and n > 3.
a(n) = 2*A296240(n) iff prime(n) == +/- 1 (mod 5) and n > 3.
a(n) in {2^k: k > 1} iff prime(n) in {A047650}.
a(n) == 3 (mod 6) iff prime(n) in {A124096}.
a(n) == 6 (mod 12) iff prime(n) in {A046652}.
a(n) == 0 (mod 14) iff prime(n) in {A125252}.

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A308794 Primes p such that A001175(p) = (p-1)/9.

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A293200 Primes p with a primitive root g such that g^3 = g + 1 mod p.

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A293201 Primes p with a primitive root g such that g^3 = g^2 + g + 1 mod p.

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A268271 Primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all quadratic residues of p in the first (p-1)/2 iterations (for some b).

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A366951 a(n) = 2*(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2*(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.

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A366951 a(n) = 2(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.