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A308789
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Primes p such that A001175(p) = (p-1)/4.
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9
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769, 809, 1049, 1289, 1721, 2729, 3049, 3929, 4289, 4649, 5009, 5441, 5689, 6361, 6961, 7321, 7841, 8209, 8329, 8369, 8681, 9689, 9769, 11161, 11489, 11969, 12049, 12281, 12601, 12721, 13649, 13721, 14969, 15241, 15569, 16649, 17489, 18329, 19961, 21169, 21881
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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) = (p-1)/4, 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 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), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 4.
Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 8) for primes p in this sequence.
Number of terms below 10^N:
N | Number | Decomposing primes*
3 | 2 | 78
4 | 23 | 609
5 | 165 | 4777
6 | 1290 | 39210
7 | 10958 | 332136
8 | 95746 | 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 && Mod[ Fibonacci[k + 1], n] == 1, Return[k]]];
Reap[For[p=2, p <= 21881, p = NextPrime[p], If[pn[p] == (p-1)/4, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 01 2019 *)
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PROG
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(PARI) Pisano_for_decomposing_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(p-1)); for(d=1, #v, if(Mod(M, p)^v[d]==Id, return(v[d]))))
forprime(p=2, 22000, if(Pisano_for_decomposing_prime(p)==(p-1)/4, 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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