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A308784 Primes p such that A001175(p) = 2*(p+1)/3.

%I A308784 #31 Jan 05 2025 19:51:41
%S A308784 47,107,113,263,347,353,563,677,743,977,1097,1217,1223,1277,1307,1523,
%T A308784 1553,1733,1823,1877,1913,1973,2027,2237,2243,2267,2333,2447,2663,
%U A308784 2687,2753,2777,3323,3347,3407,3467,3533,3557,3617,3623,3767,3947,4133,4493,4547,4583
%N A308784 Primes p such that A001175(p) = 2*(p+1)/3.
%C A308784 Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/3, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C A308784 Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/3, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
%C A308784 Also, primes p such that A001177(p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then A001177(p) = (p+1)/6, otherwise (p+1)/3.
%C A308784 Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/6, otherwise (p+1)/3.
%C A308784 In general, 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 A308784 For an odd prime p:
%C A308784 (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 A308784 (b) if p is inert in K, then u^(p+1) == -1 (mod p) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
%C A308784 If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
%C A308784 Here k = 1, and this sequence gives primes such that (b) holds and r = 3. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
%C A308784 Number of terms below 10^N:
%C A308784   N | 1 mod 4 | 3 mod 4 |  Total | Inert primes*
%C A308784   3 |       4 |       6 |     10 |         88
%C A308784   4 |      41 |      43 |     84 |        618
%C A308784   5 |     330 |     353 |    683 |       4813
%C A308784   6 |    2745 |    2736 |   5481 |      39286
%C A308784   7 |   23219 |   23250 |  46469 |     332441
%C A308784   8 |  201805 |  201547 | 403352 |    2880969
%C A308784   * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).
%H A308784 Michael De Vlieger, <a href="/A308784/b308784.txt">Table of n, a(n) for n = 1..714</a>
%H A308784 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.
%H A308784 Wikipedia, <a href="https://en.wikipedia.org/wiki/Pisano_period">Pisano period</a>
%t A308784 pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
%t A308784 Reap[For[p = 2, p <= 4583, p = NextPrime[p], If[pn[p] == 2(p+1)/3, Print[p]; Sow[p]]]][[2, 1]] (* _Jean-François Alcover_, Jul 02 2019 *)
%o A308784 (PARI) Pisano_for_inert_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(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
%o A308784 forprime(p=2, 4000, if(Pisano_for_inert_prime(p)==2*(p+1)/3, print1(p, ", ")))
%Y A308784 Similar sequences that give primes such that (b) holds: A071774 (r=1), this sequence (r=3), A308785 (r=7), A308786 (r=9).
%K A308784 nonn
%O A308784 1,1
%A A308784 _Jianing Song_, Jun 25 2019