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Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/6, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.

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 entry point 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), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...

Here k = 1, and this sequence gives primes such that (a) holds and s = 6. For even s, all terms are congruent to 1 modulo 4.

Number of terms below 10^N:

N | Number | Decomposing primes*

3 | 2 | 78

4 | 14 | 609

5 | 147 | 4777

6 | 1216 | 39210

7 | 10477 | 332136

8 | 90720 | 2880484

* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/8, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.

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 entry point 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), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...

Here k = 1, and this sequence gives primes such that (a) holds and s = 8.

Number of terms below 10^N:

N | Number | Decomposing primes*

3 | 3 | 78

4 | 20 | 609

5 | 154 | 4777

6 | 1278 | 39210

7 | 11063 | 332136

8 | 95613 | 2880484

* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/9, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.

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 entry point 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), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...

Here k = 1, and this sequence gives primes such that (a) holds and s = 9. For odd s, all terms are congruent to 3 modulo 4.

Number of terms below 10^N:

N | Number | Decomposing primes*

3 | 2 | 78

4 | 3 | 609

5 | 31 | 4777

6 | 274 | 39210

7 | 2293 | 332136

8 | 20097 | 2880484

* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

The n-th prime p is in this sequence iff A001602(n) = p-(5/p) (that is the maximum possible value of A001602(n)).

Subset of A033949 and A175594 (essentially the same sequence).

Numbers of the form 2^k, with k>=3, appear to be part of the sequence.

The file "List of indexes and steps (k, x, y)" (see Links) for k = 1, 2, 3, 4, ... consecutive Fibonacci numbers gives the minimum index to start to calculate the average ( x ) and the step to add to get all the other averages ( y ).

E.g.: for k = 7 we have 7, 6, 8. This means that we must start from the 6th Fibonacci number to add 7 consecutive Fibonacci numbers and get an average that is an integer. Fibonacci(6) + Fibonacci(7) + ... + Fibonacci(12) = 8 + 13 + 21 + 34 + 55 + 89 + 144 = 364 and 364 / 7 = 52.

Then 6 + 1*8 = 14, 6 + 2*8 = 22, 6 + 3*8 = 30, etc. are the other indexes:

Fibonacci(14) + Fibonacci (15) + ... + Fibonacci(20) = 377 + 610 + 987 + 1597 + 2584 + 4181 + 6765 = 17101 and 17101 / 7 = 2443;

Fibonacci(22) + Fibonacci(23) + ... + Fibonacci(28) = 17711 + 28657 + 46368 + 75025 + 121393 + 196418 + 317811 = 803383 and 803383 / 7 = 114769;

Fibonacci(30) + Fibonacci(31) + ... + Fibonacci(36) = 832040 + 1346269 + 2178309 + 3524578 + 5702887 + 9227465 + 14930352 = 37741900 and 37741900 / 7 = 5391700; etc.

In particular we note that:

x = 0 is A219612; x = 1 is A124456; x = 0 and y = k - 1 is A106535;

y = 1 is A141767; x = k - 1 and y = k + 1 is A000057;

x = y - 1 or y|k is A023172; y = k is A000351;

x = y - k + 1 appears to give only prime numbers: 3,11,19,31,59,71,79,131,179,191,239,251,271,311,359,379,419,431,439,479,491,499,571,599,631,659,719,739,751,839,971, etc.