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A308786
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Primes p such that A001175(p) = 2*(p+1)/9.
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3
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233, 557, 953, 4013, 4733, 5147, 6983, 7307, 7883, 9377, 10133, 12923, 14867, 15767, 17747, 19403, 20753, 22877, 23813, 26387, 26783, 27737, 29483, 32057, 33533, 35117, 39383, 40013, 40787, 41543, 41903, 42767, 43613, 45557, 46187, 48473, 48563, 50993, 51263, 53927
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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PROG
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(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, 55000, if(Pisano_for_inert_prime(p)==2*(p+1)/9, print1(p, ", ")))
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CROSSREFS
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Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), A308785 (r=7), this sequence (r=9).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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