|
|
A308795
|
|
Primes p such that A001177(p) = (p-1)/2.
|
|
8
|
|
|
29, 41, 101, 181, 229, 241, 349, 409, 449, 509, 569, 601, 641, 929, 941, 1021, 1061, 1109, 1129, 1201, 1229, 1321, 1481, 1489, 1549, 1609, 1621, 1669, 1709, 1741, 1789, 1801, 1861, 1889, 2029, 2069, 2129, 2609, 2621, 2861, 3209, 3301, 3361, 3389, 3449, 3461, 3581
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/2, 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 entry point 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), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 2. For even s, all terms are congruent to 1 modulo 4.
Number of terms below 10^N:
N | Number | Decomposing primes*
3 | 15 | 78
4 | 115 | 609
5 | 839 | 4777
6 | 6913 | 39210
7 | 58891 | 332136
8 | 510784 | 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, Return[k]]];
Reap[For[p = 2, p < 3600, p = NextPrime[p], If[pn[p] == (p - 1)/2, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)
|
|
PROG
|
(PARI) Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); if(isprime(p)&&kronecker(k^2+4, p)==1, my(v=divisors(p-1)); for(d=1, #v, if((Mod(M, p)^v[d])[2, 1]==0, return(v[d]))))
forprime(p=2, 3600, if(Entry_for_decomposing_prime(p)==(p-1)/2, print1(p, ", ")))
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|