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A308801 Primes p such that A001177(p) = (p-1)/8. 8

%I #11 Jul 05 2019 16:02:00

%S 89,761,769,1009,2089,2441,3881,4201,4289,4729,5209,5441,5849,6521,

%T 6761,7369,7841,8009,8081,9929,10601,11489,11689,11801,11969,12401,

%U 12409,12569,12889,14009,14249,15889,17449,17609,17881,17929,18121,18169,20201,20249,21929

%N Primes p such that A001177(p) = (p-1)/8.

%C 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.

%C 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).

%C For an odd prime p:

%C (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, ...;

%C (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, ...

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

%C Number of terms below 10^N:

%C N | Number | Decomposing primes*

%C 3 | 3 | 78

%C 4 | 20 | 609

%C 5 | 154 | 4777

%C 6 | 1278 | 39210

%C 7 | 11063 | 332136

%C 8 | 95613 | 2880484

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

%t pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];

%t Reap[For[p = 2, p < 22000, p = NextPrime[p], If[Mod[p, 8] == 1, If[pn[p] == (p - 1)/8, Print[p]; Sow[p]]]]][[2, 1]] (* _Jean-François Alcover_, Jul 05 2019 *)

%o (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]))))

%o forprime(p=2, 22000, if(Entry_for_decomposing_prime(p)==(p-1)/8, print1(p, ", ")))

%Y Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), this sequence (s=8), A308802 (s=9).

%K nonn

%O 1,1

%A _Jianing Song_, Jun 25 2019

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