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%I #22 Jul 02 2019 08:44:27
%S 29,89,101,181,229,349,401,509,761,941,1021,1061,1109,1229,1249,1361,
%T 1409,1549,1621,1669,1709,1741,1789,1861,2029,2069,2089,2441,2621,
%U 2801,2861,3089,3169,3301,3389,3461,3581,3821,3881,3989,4001,4049,4201,4229,4549,4729
%N Primes p such that A001175(p) = (p-1)/2.
%C Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/2, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
%C 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), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
%C Here k = 1, and this sequence gives primes such that (a) holds and s = 2.
%C Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 4) for primes p in this sequence.
%C The number of terms below 10^N:
%C N | Number | Decomposing primes*
%C 3 | 10 | 78
%C 4 | 89 | 609
%C 5 | 630 | 4777
%C 6 | 5207 | 39210
%C 7 | 44296 | 332136
%C 8 | 382966 | 2880484
%C * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
%t pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k+1], n] == 1, Return[k]]];
%t Reap[For[p = 2, p <= 4729, p = NextPrime[p], If[pn[p] == (p-1)/2, Print[p]; Sow[p]]]][[2, 1]] (* _Jean-François Alcover_, Jul 01 2019 *)
%o (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 forprime(p=2, 4800, if(Pisano_for_decomposing_prime(p)==(p-1)/2, print1(p, ", ")))
%Y Similar sequences that give primes such that (a) holds: A003147/{5} (s=1), this sequence (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).
%K nonn
%O 1,1
%A _Jianing Song_, Jun 25 2019