cp's OEIS Frontend

A308786 Primes p such that A001175(p) = 2*(p+1)/9.

%I A308786 #27 Jan 05 2025 19:51:41
%S A308786 233,557,953,4013,4733,5147,6983,7307,7883,9377,10133,12923,14867,
%T A308786 15767,17747,19403,20753,22877,23813,26387,26783,27737,29483,32057,
%U A308786 33533,35117,39383,40013,40787,41543,41903,42767,43613,45557,46187,48473,48563,50993,51263,53927
%N A308786 Primes p such that A001175(p) = 2*(p+1)/9.
%C A308786 Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/9, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C A308786 Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/9, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
%C A308786 Also, primes p such that A001177(p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then A001177(p) = (p+1)/18, otherwise (p+1)/9.
%C A308786 Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/18, otherwise (p+1)/9.
%C A308786 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 A308786 For an odd prime p:
%C A308786 (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 A308786 (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 A308786 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 A308786 Here k = 1, and this sequence gives primes such that (b) holds and r = 9. 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 A308786 Number of terms below 10^N:
%C A308786   N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
%C A308786   3 |       3 |       0 |     3 |         88
%C A308786   4 |       6 |       4 |    10 |        618
%C A308786   5 |      36 |      28 |    64 |       4813
%C A308786   6 |     313 |     300 |   613 |      39286
%C A308786   7 |    2563 |    2597 |  5160 |     332441
%C A308786   8 |   22377 |   22350 | 44727 |    2880969
%C A308786   * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).
%H A308786 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 A308786 Wikipedia, <a href="https://en.wikipedia.org/wiki/Pisano_period">Pisano period</a>
%o A308786 (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 A308786 forprime(p=2, 55000, if(Pisano_for_inert_prime(p)==2*(p+1)/9, print1(p, ", ")))
%Y A308786 Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), A308785 (r=7), this sequence (r=9).
%K A308786 nonn
%O A308786 1,1
%A A308786 _Jianing Song_, Jun 25 2019