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%I #29 Mar 31 2021 19:10:48
%S 307,797,1483,3023,4157,4283,6397,6733,7027,7433,7867,9337,9743,9883,
%T 10177,10303,10597,11423,12823,14293,18493,19963,20593,20873,24247,
%U 24793,25703,28433,29917,30113,31387,31723,31793,32353,33347,34537,34747,37057,38653,38723
%N Primes p such that A001175(p) = 2*(p+1)/7.
%C Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/7, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/7, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
%C Also, primes p such that A001177(p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then A001177(p) = (p+1)/14, otherwise (p+1)/7.
%C Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/14, otherwise (p+1)/7.
%C 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 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 Pisano period 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) (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 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 Here k = 1, and this sequence gives primes such that (b) holds and r = 7. 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 Number of terms below 10^N:
%C N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
%C 3 | 1 | 1 | 2 | 88
%C 4 | 6 | 8 | 14 | 618
%C 5 | 48 | 42 | 90 | 4813
%C 6 | 371 | 350 | 721 | 39286
%C 7 | 3098 | 3086 | 6184 | 332441
%C 8 | 27035 | 26989 | 54024 | 2880969
%C * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).
%H Bob Bastasz, <a href="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 Wikipedia, <a href="https://en.wikipedia.org/wiki/Pisano_period">Pisano period</a>
%t Select[Prime@ Range[1000], Function[n, Mod[Last@ NestWhile[{Mod[#2, n], Mod[#1 + #2, n], #3 + 1} & @@ # &, {1, 1, 1}, #[[1 ;; 2]] != {0, 1} &], n] == Mod[2 (n + 1)/7, n] ]] (* _Michael De Vlieger_, Mar 31 2021, after _Leo C. Stein_ at A001175 *)
%o (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 forprime(p=2, 40000, if(Pisano_for_inert_prime(p)==2*(p+1)/7, print1(p, ", ")))
%Y Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), this sequence (r=7), A308786 (r=9).
%K nonn
%O 1,1
%A _Jianing Song_, Jun 25 2019
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