A308791 Primes p such that A001175(p) = (p-1)/6.

541, 709, 2389, 3121, 3529, 4561, 4861, 5869, 7069, 8821, 9001, 10789, 12421, 12781, 13309, 14341, 14869, 16981, 18289, 19249, 19309, 19429, 19501, 20389, 20809, 20929, 21649, 22741, 23629, 24181, 25189, 26821, 27109, 27409, 28669, 30181, 30469, 30781, 30949, 31189

Offset: 1

Author

Jianing Song, Jun 25 2019

Keywords

• nonn

Comments

Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/6, 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 = 6.

Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 12) for primes p in this sequence.

Number of terms below 10^N:

N | Number | Decomposing primes*

3 | 2 | 78

4 | 11 | 609

5 | 112 | 4777

6 | 898 | 39210

7 | 7777 | 332136

8 | 68115 | 2880484

* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Crossrefs

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), this sequence (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

Programs

Mathematica

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 < 32000, p = NextPrime[p], If[Mod[p, 6] == 1, If[pn[p] == (p - 1)/6, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

PARI

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, 32000, if(Pisano_for_decomposing_prime(p)==(p-1)/6, print1(p, ", ")))