A308797 -id:A308797 - cp's OEIS frontend

A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

11, 19, 31, 59, 71, 79, 131, 179, 191, 239, 251, 271, 311, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 631, 659, 719, 739, 751, 839, 971, 1019, 1039, 1051, 1091, 1171, 1259, 1319, 1399, 1439, 1451, 1459, 1499, 1531, 1559, 1571, 1619, 1759, 1811, 1831

Offset: 1

Peter K. Pearson (ppearson+att(AT)spamcop.net), May 06 2005

nonn

This is a sister sequence to A000057, because this sequence, since {k : A001177(k) = k-1}, might be called a subdiagonal sequence of A001177, and {k : A001177(k) = k+1}, which might be called a superdiagonal sequence of A001177. Sequences A000057 and A106535 are disjoint. Is this sequence the set of all divisors of some family of sequences, like A000057 is? - Art DuPre, Jul 11 2012
Are all members of this sequence prime? Using A069106, any composite members must exceed 89151931. - Robert Israel, Oct 13 2015
From Jianing Song, Jul 02 2019: (Start)
Yes, all terms are primes. See a brief proof below.
Also, if p == 1 (mod 4) then b(p) divides (p-Legendre(p,5))/2. So terms in this sequence are congruent to 11 or 19 modulo 20.
Primes p such that ord(-(3+sqrt(5))/2,p) = p-1, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer. (End)
Comments from Amiram Eldar, Jan 30 2022 (Start)
Sequence A003147, "Primes p with a Fibonacci primitive root", is defined in the paper: Daniel Shanks, Fibonacci primitive roots, Fibonacci Quarterly, Vol.  10, No. 2 (1972), pp. 163-168, and 181.
A second paper on this subject Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160,
deals with terms p == 3 (mod 4) of A003147, i.e., the intersection of A003147 and A002145 (or A004767).
It states that if g is a Fibonacci primitive root of a prime p such that p == 3 (mod 4) then g-1 and g-2 are also primitive roots of p.
The first 2000 terms of (A003147 intersect A002145) agree with the present sequence, although the definitions are quite different. Are these two sequences the same? (End)

Robert Israel, Table of n, a(n) for n = 1..2000
Alfred Brousseau, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972.
Jianing Song, Proof that all terms are prime

Cf. A000057, A001175, A069106.

Similar sequences that give primes p such that A001177(p) = (p-1)/s: this sequence (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7),A308801 (s=8), A308802 (s=9).

Filtered([2..2000], n -> Fibonacci(n-1) mod n = 0 and Filtered( [2..n-2], x -> Fibonacci(x) mod n = 0 ) = [] );

A106535 := proc(n)
        option remember;
        if n = 1 then
                11;
        else
                for a from procname(n-1)+1 do
                        if A001177(a) = a-1 then
                                return a;
                        end if;
                end do:
        end if;
end proc: # R. J. Mathar, Jul 09 2012
# Alternative:
fmod:= proc(a,b) local A;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1,1>|<1,0>>,integer[8]);
  MatrixPower(b,A,a)[1,2];
end proc:
filter:= proc(n)
  local cands;
  if fmod(n-1,n) <> 0 then return false fi;
  cands:= map(t -> (n-1)/t, numtheory:-factorset(n-1));
  andmap(c -> (fmod(c,n) > 0), cands);
end proc:
select(filter, [$2..10^4]); # Robert Israel, Oct 13 2015

f[n_] := Block[{x = 2}, While[Mod[Fibonacci@ x, n] != 0, x++]; x];Select[Range@ 1860, f@ # == # - 1 &] (* Michael De Vlieger, Oct 13 2015 *)

isok(n) = {x = 2; while(fibonacci(x) % n, x++); x == n-1;} \\ Michel Marcus, Oct 20 2015

{n: A001177(n) = n-1}. - R. J. Mathar, Jul 09 2012

Corrected by T. D. Noe, Oct 25 2006

A308800 Primes p such that A001177(p) = (p-1)/7.

Original entry on oeis.org

2731, 3739, 4831, 6091, 11159, 13679, 14771, 16871, 19559, 20399, 24179, 26111, 29191, 31039, 33811, 34511, 34679, 35911, 40111, 41651, 49211, 55259, 56099, 60859, 62819, 69539, 71191, 71359, 71471, 73291, 74831, 85751, 87991, 96979, 97231, 97931, 104959, 108179

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/7, 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 = 7. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      0 |            78
  4 |      4 |           609
  5 |     36 |          4777
  6 |    347 |         39210
  7 |   2801 |        332136
  8 |  24291 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Amiram Eldar, Table of n, a(n) for n = 1..10000

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), this sequence (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 50000, p = NextPrime[p], If[Mod[p, 7] == 1, If[pn[p] == (p - 1)/7, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

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, 109000, if(Entry_for_decomposing_prime(p)==(p-1)/7, print1(p, ", ")))

A308795 Primes p such that A001177(p) = (p-1)/2.

Original entry on oeis.org

29, 41, 101, 181, 229, 241, 349, 409, 449, 509, 569, 601, 641, 929, 941, 1021, 1061, 1109, 1129, 1201, 1229, 1321, 1481, 1489, 1549, 1609, 1621, 1669, 1709, 1741, 1789, 1801, 1861, 1889, 2029, 2069, 2129, 2609, 2621, 2861, 3209, 3301, 3361, 3389, 3449, 3461, 3581

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/2, 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 = 2. For even s, all terms are congruent to 1 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |     15 |            78
  4 |    115 |           609
  5 |    839 |          4777
  6 |   6913 |         39210
  7 |  58891 |        332136
  8 | 510784 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A106535 (s=1), this sequence (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 3600, p = NextPrime[p], If[pn[p] == (p - 1)/2, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

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, 3600, if(Entry_for_decomposing_prime(p)==(p-1)/2, print1(p, ", ")))

A308796 Primes p such that A001177(p) = (p-1)/3.

Original entry on oeis.org

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

Offset: 1

Jianing Song, Jun 25 2019

nonn

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).

Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), this sequence (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

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 *)

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, ", ")))

A308798 Primes p such that A001177(p) = (p-1)/5.

Original entry on oeis.org

211, 691, 991, 1031, 1151, 1511, 1871, 1951, 2591, 3251, 3851, 4391, 4651, 4691, 4751, 4871, 5531, 5591, 6011, 6211, 6271, 6491, 7211, 7451, 8011, 8171, 8831, 9011, 9091, 9371, 9431, 9931, 10391, 10531, 10691, 10891, 11071, 12011, 12071, 12911, 14051, 14251, 14591

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/5, 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 = 5. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      3 |            78
  4 |     32 |           609
  5 |    192 |          4777
  6 |   1521 |         39210
  7 |  12542 |        332136
  8 | 109034 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), this sequence (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 15000, p = NextPrime[p], If[Mod[p, 5] == 1, If[pn[p] == (p - 1)/5, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

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, 15000, if(Entry_for_decomposing_prime(p)==(p-1)/5, print1(p, ", ")))

A308799 Primes p such that A001177(p) = (p-1)/6.

Original entry on oeis.org

541, 709, 2281, 2389, 2689, 4861, 5869, 7069, 8089, 8761, 8821, 8929, 9049, 9601, 10009, 10321, 10789, 12421, 12781, 13309, 13681, 14341, 14869, 14929, 16981, 19309, 19429, 19501, 19609, 20389, 21841, 22741, 23629, 24181, 24481, 25189, 26821, 27109, 27361, 27961

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/6, 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 = 6. For even s, all terms are congruent to 1 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      2 |            78
  4 |     14 |           609
  5 |    147 |          4777
  6 |   1216 |         39210
  7 |  10477 |        332136
  8 |  90720 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), this sequence (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 28000, p = NextPrime[p], If[Mod[p, 6] == 1, If[pn[p] == (p - 1)/6, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

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, 28000, if(Entry_for_decomposing_prime(p)==(p-1)/6, print1(p, ", ")))

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

Original entry on oeis.org

89, 761, 769, 1009, 2089, 2441, 3881, 4201, 4289, 4729, 5209, 5441, 5849, 6521, 6761, 7369, 7841, 8009, 8081, 9929, 10601, 11489, 11689, 11801, 11969, 12401, 12409, 12569, 12889, 14009, 14249, 15889, 17449, 17609, 17881, 17929, 18121, 18169, 20201, 20249, 21929

Offset: 1

Jianing Song, Jun 25 2019

nonn

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.
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 = 8.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      3 |            78
  4 |     20 |           609
  5 |    154 |          4777
  6 |   1278 |         39210
  7 |  11063 |        332136
  8 |  95613 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

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).

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

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, 22000, if(Entry_for_decomposing_prime(p)==(p-1)/8, print1(p, ", ")))

A308802 Primes p such that A001177(p) = (p-1)/9.

Original entry on oeis.org

199, 919, 6679, 12979, 17011, 17659, 20431, 23059, 23599, 24391, 24859, 39079, 39439, 43399, 48619, 53479, 54091, 62011, 62191, 67411, 69499, 72019, 72091, 77419, 81019, 82279, 91099, 91459, 92179, 97579, 98731, 102259, 103231, 105211, 108271, 111439, 114679, 125119

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/9, 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 = 9. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      2 |            78
  4 |      3 |           609
  5 |     31 |          4777
  6 |    274 |         39210
  7 |   2293 |        332136
  8 |  20097 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

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), A308801 (s=8), this sequence (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 50000, p = NextPrime[p], If[Mod[p, 9] == 1, If[pn[p] == (p - 1)/9, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

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, 126000, if(Entry_for_decomposing_prime(p)==(p-1)/9, print1(p, ", ")))

cp's OEIS Frontend

A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

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A308800 Primes p such that A001177(p) = (p-1)/7.

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A308795 Primes p such that A001177(p) = (p-1)/2.

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A308796 Primes p such that A001177(p) = (p-1)/3.

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A308798 Primes p such that A001177(p) = (p-1)/5.

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A308799 Primes p such that A001177(p) = (p-1)/6.

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

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A308802 Primes p such that A001177(p) = (p-1)/9.

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