cp's OEIS Frontend

A308789 Primes p such that A001175(p) = (p-1)/4.

%I A308789 #22 Jan 05 2025 19:51:41
%S A308789 769,809,1049,1289,1721,2729,3049,3929,4289,4649,5009,5441,5689,6361,
%T A308789 6961,7321,7841,8209,8329,8369,8681,9689,9769,11161,11489,11969,12049,
%U A308789 12281,12601,12721,13649,13721,14969,15241,15569,16649,17489,18329,19961,21169,21881
%N A308789 Primes p such that A001175(p) = (p-1)/4.
%C A308789 Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/4, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C A308789 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 A308789 For an odd prime p:
%C A308789 (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, ...;
%C A308789 (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, ...
%C A308789 Here k = 1, and this sequence gives primes such that (a) holds and s = 4.
%C A308789 Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 8) for primes p in this sequence.
%C A308789 Number of terms below 10^N:
%C A308789   N | Number | Decomposing primes*
%C A308789   3 |      2 |            78
%C A308789   4 |     23 |           609
%C A308789   5 |    165 |          4777
%C A308789   6 |   1290 |         39210
%C A308789   7 |  10958 |        332136
%C A308789   8 |  95746 |       2880484
%C A308789   * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
%H A308789 Bob Bastasz, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/58-5/bastasz.pdf">Lyndon words of a second-order recurrence</a>, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
%t A308789 pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
%t A308789 Reap[For[p=2, p <= 21881, p = NextPrime[p], If[pn[p] == (p-1)/4, Print[p]; Sow[p]]]][[2, 1]] (* _Jean-François Alcover_, Jul 01 2019 *)
%o A308789 (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]))))
%o A308789 forprime(p=2, 22000, if(Pisano_for_decomposing_prime(p)==(p-1)/4, print1(p, ", ")))
%Y A308789 Similar sequences that give primes such that (a) holds: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), this sequence (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).
%K A308789 nonn
%O A308789 1,1
%A A308789 _Jianing Song_, Jun 25 2019