cp's OEIS Frontend

A308798 Primes p such that A001177(p) = (p-1)/5.

%I A308798 #11 Jul 05 2019 16:01:50
%S A308798 211,691,991,1031,1151,1511,1871,1951,2591,3251,3851,4391,4651,4691,
%T A308798 4751,4871,5531,5591,6011,6211,6271,6491,7211,7451,8011,8171,8831,
%U A308798 9011,9091,9371,9431,9931,10391,10531,10691,10891,11071,12011,12071,12911,14051,14251,14591
%N A308798 Primes p such that A001177(p) = (p-1)/5.
%C A308798 Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/5, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C A308798 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 A308798 For an odd prime p:
%C A308798 (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 A308798 (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 A308798 Here k = 1, and this sequence gives primes such that (a) holds and s = 5. For odd s, all terms are congruent to 3 modulo 4.
%C A308798 Number of terms below 10^N:
%C A308798   N | Number | Decomposing primes*
%C A308798   3 |      3 |            78
%C A308798   4 |     32 |           609
%C A308798   5 |    192 |          4777
%C A308798   6 |   1521 |         39210
%C A308798   7 |  12542 |        332136
%C A308798   8 | 109034 |       2880484
%C A308798   * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
%t A308798 pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
%t A308798 Reap[For[p = 2, p < 15000, p = NextPrime[p], If[Mod[p, 5] == 1, If[pn[p] == (p - 1)/5, Print[p]; Sow[p]]]]][[2, 1]] (* _Jean-François Alcover_, Jul 05 2019 *)
%o A308798 (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 A308798 forprime(p=2, 15000, if(Entry_for_decomposing_prime(p)==(p-1)/5, print1(p, ", ")))
%Y A308798 Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), this sequence (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).
%K A308798 nonn
%O A308798 1,1
%A A308798 _Jianing Song_, Jun 25 2019