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A308796
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Primes p such that A001177(p) = (p-1)/3.
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8
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139, 151, 331, 619, 811, 1231, 1279, 1291, 1471, 1579, 1699, 1999, 2239, 2251, 2371, 2659, 3271, 3331, 3391, 3499, 3631, 3919, 4051, 4159, 4231, 4759, 5059, 5839, 6079, 6619, 6691, 6991, 7219, 7639, 8059, 8599, 8731, 8971, 9151, 9319, 9679, 9739, 10099, 10459, 10771
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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(-(3+sqrt(5))/2,p) = (p-1)/3, 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 = 3. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
N | Number | Decomposing primes*
3 | 5 | 78
4 | 42 | 609
5 | 312 | 4777
6 | 2490 | 39210
7 | 20958 | 332136
8 | 181493 | 2880484
* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (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, Return[k]]];
Reap[For[p = 2, p < 10000, p = NextPrime[p], If[Mod[p, 3] == 1, If[pn[p] == (p - 1)/3, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)
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PROG
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(PARI) Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); if(isprime(p)&&kronecker(k^2+4, p)==1, my(v=divisors(p-1)); for(d=1, #v, if((Mod(M, p)^v[d])[2, 1]==0, return(v[d]))))
forprime(p=2, 11000, if(Entry_for_decomposing_prime(p)==(p-1)/3, print1(p, ", ")))
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CROSSREFS
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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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