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%I #15 Jun 04 2021 03:14:48
%S 2731,3739,4831,6091,11159,13679,14771,16871,19559,20399,24179,26111,
%T 29191,31039,33811,34511,34679,35911,40111,41651,49211,55259,56099,
%U 60859,62819,69539,71191,71359,71471,73291,74831,85751,87991,96979,97231,97931,104959,108179
%N Primes p such that A001177(p) = (p-1)/7.
%C Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/7, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C 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).
%C For an odd prime p:
%C (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, ...;
%C (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, ...
%C Here k = 1, and this sequence gives primes such that (a) holds and s = 7. For odd s, all terms are congruent to 3 modulo 4.
%C Number of terms below 10^N:
%C N | Number | Decomposing primes*
%C 3 | 0 | 78
%C 4 | 4 | 609
%C 5 | 36 | 4777
%C 6 | 347 | 39210
%C 7 | 2801 | 332136
%C 8 | 24291 | 2880484
%C * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
%H Amiram Eldar, <a href="/A308800/b308800.txt">Table of n, a(n) for n = 1..10000</a>
%t pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
%t Reap[For[p = 2, p < 50000, p = NextPrime[p], If[Mod[p, 7] == 1, If[pn[p] == (p - 1)/7, Print[p]; Sow[p]]]]][[2, 1]] (* _Jean-François Alcover_, Jul 05 2019 *)
%o (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]))))
%o forprime(p=2, 109000, if(Entry_for_decomposing_prime(p)==(p-1)/7, print1(p, ", ")))
%Y Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), this sequence (s=7), A308801 (s=8), A308802 (s=9).
%K nonn
%O 1,1
%A _Jianing Song_, Jun 25 2019
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