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A308784
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Primes p such that A001175(p) = 2*(p+1)/3.
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4
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47, 107, 113, 263, 347, 353, 563, 677, 743, 977, 1097, 1217, 1223, 1277, 1307, 1523, 1553, 1733, 1823, 1877, 1913, 1973, 2027, 2237, 2243, 2267, 2333, 2447, 2663, 2687, 2753, 2777, 3323, 3347, 3407, 3467, 3533, 3557, 3617, 3623, 3767, 3947, 4133, 4493, 4547, 4583
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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)/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).
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LINKS
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MATHEMATICA
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pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
Reap[For[p = 2, p <= 4583, p = NextPrime[p], If[pn[p] == 2(p+1)/3, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 02 2019 *)
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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, 4000, if(Pisano_for_inert_prime(p)==2*(p+1)/3, print1(p, ", ")))
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CROSSREFS
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Similar sequences that give primes such that (b) holds: A071774 (r=1), this sequence (r=3), A308785 (r=7), A308786 (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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