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 A308785 Primes p such that A001175(p) = 2*(p+1)/7. 2
 307, 797, 1483, 3023, 4157, 4283, 6397, 6733, 7027, 7433, 7867, 9337, 9743, 9883, 10177, 10303, 10597, 11423, 12823, 14293, 18493, 19963, 20593, 20873, 24247, 24793, 25703, 28433, 29917, 30113, 31387, 31723, 31793, 32353, 33347, 34537, 34747, 37057, 38653, 38723 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. 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. 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. 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. 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). For an odd prime p: (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, ...; (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, ... 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). 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). Number of terms below 10^N: N | 1 mod 4 | 3 mod 4 | Total | Inert primes* 3 | 1 | 1 | 2 | 88 4 | 6 | 8 | 14 | 618 5 | 48 | 42 | 90 | 4813 6 | 371 | 350 | 721 | 39286 7 | 3098 | 3086 | 6184 | 332441 8 | 27035 | 26989 | 54024 | 2880969 * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5). LINKS Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29. Wikipedia, Pisano period MATHEMATICA 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 *) PROG (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])))) forprime(p=2, 40000, if(Pisano_for_inert_prime(p)==2*(p+1)/7, print1(p, ", "))) CROSSREFS Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), this sequence (r=7), A308786 (r=9). Sequence in context: A309101 A098042 A125252 * A142572 A204477 A176091 Adjacent sequences: A308782 A308783 A308784 * A308786 A308787 A308788 KEYWORD nonn AUTHOR Jianing Song, Jun 25 2019 STATUS approved

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Last modified December 6 18:58 EST 2022. Contains 358644 sequences. (Running on oeis4.)