|
| |
|
|
A308788
|
|
Primes p such that A001175(p) = (p-1)/3.
|
|
8
|
|
|
|
139, 151, 331, 619, 661, 811, 829, 1069, 1231, 1279, 1291, 1381, 1471, 1579, 1699, 1999, 2239, 2251, 2281, 2371, 2659, 2689, 2749, 3271, 3331, 3391, 3499, 3631, 3919, 4051, 4159, 4231, 4261, 4759, 4909, 5059, 5581, 5701, 5821, 5839, 6079, 6229, 6469, 6619, 6691
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/3, 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 = 3.
The number of terms below 10^N:
N | Number | Decomposing primes*
3 | 7 | 78
4 | 64 | 609
5 | 455 | 4777
6 | 3688 | 39210
7 | 31412 | 332136
8 | 272318 | 2880484
* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
|
|
|
LINKS
|
|
|
|
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 <= 6691, p = NextPrime[p], If[pn[p] == (p-1)/3, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 01 2019 *)
|
|
|
PROG
|
(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, 7000, if(Pisano_for_decomposing_prime(p)==(p-1)/3, print1(p, ", ")))
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
