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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.

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.

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.

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.

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 = 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).

Number of terms below 10^N:

N | 1 mod 4 | 3 mod 4 | Total | Inert primes*

3 | 4 | 6 | 10 | 88

4 | 41 | 43 | 84 | 618

5 | 330 | 353 | 683 | 4813

6 | 2745 | 2736 | 5481 | 39286

7 | 23219 | 23250 | 46469 | 332441

8 | 201805 | 201547 | 403352 | 2880969

* Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

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.

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.

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.

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.

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 = 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).

Number of terms below 10^N:

N | 1 mod 4 | 3 mod 4 | Total | Inert primes*

3 | 3 | 0 | 3 | 88

4 | 6 | 4 | 10 | 618

5 | 36 | 28 | 64 | 4813

6 | 313 | 300 | 613 | 39286

7 | 2563 | 2597 | 5160 | 332441

8 | 22377 | 22350 | 44727 | 2880969

* Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).