A001177 -id:A001177 - cp's OEIS frontend

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

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, ", ")))

A308797 Primes p such that A001177(p) = (p-1)/4.

Original entry on oeis.org

61, 109, 149, 269, 389, 401, 701, 809, 821, 1049, 1181, 1249, 1289, 1301, 1361, 1409, 1429, 1721, 1901, 1949, 2141, 2309, 2341, 2381, 2549, 2729, 2741, 2801, 2909, 3049, 3061, 3089, 3109, 3169, 3181, 3221, 3229, 3541, 3701, 3709, 3929, 4001, 4049, 4349, 4421, 4649

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+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 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 = 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      9 |            78
  4 |     81 |           609
  5 |    651 |          4777
  6 |   5268 |         39210
  7 |  44188 |        332136
  8 | 383224 |       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), this sequence (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 < 6000, p = NextPrime[p], If[Mod[p, 4] == 1, If[pn[p] == (p - 1)/4, 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, 5000, if(Entry_for_decomposing_prime(p)==(p-1)/4, 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, ", ")))

A233283 Record values in partial LCM-products (A233287) of Fibonacci entry points (A001177).

Original entry on oeis.org

1, 3, 12, 60, 120, 840, 2520, 12600, 239400, 2633400, 5266800, 15800400, 458211600, 7789597200, 288215096400, 3746796253200, 26227573772400, 131137868862000, 1049102950896000, 24129367870608000, 1906220061778032000, 78155022532899312000, 6955797005428038768000

Offset: 1

Antti Karttunen, Dec 13 2013

nonn

Conjecture: sequence gives the positions of records in A233284.

a(n+1)/a(n) = 3,4,5,2,7,3,5,19,11,2,3,29,17,37,13,7,5,8,23,79... - Ralf Stephan, Dec 17 2013

Antti Karttunen, Table of n, a(n) for n = 1..101

Cf. A233282, A233284, A233285, A233287.

(define (A233283 n) (A233287 (A233282 n)))

a(n) = A233287(A233282(n)).

A120256 a(n) = number of terms in the n-th row of A120255(n) = number of terms in A001177 equal to n.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 2, 2, 2, 1, 10, 1, 2, 5, 4, 1, 10, 3, 11, 5, 2, 1, 55, 4, 2, 12, 11, 1, 52, 3, 8, 5, 2, 5, 133, 7, 4, 5, 46, 3, 52, 1, 27, 22, 6, 1, 260, 6, 40, 5, 11, 3, 100, 13, 78, 27, 6, 3, 874, 3, 4, 22, 48, 5, 52, 7, 27, 29, 116, 3, 1319, 3, 8, 36, 23, 13, 116, 3, 444, 112, 4, 1, 1834

Offset: 1

Leroy Quet, Jun 13 2006

nonn

			Fibonacci(9) = 34; and the divisors of 34 are 1, 2, 17 and 34. Of these divisors, 1 and 2 divide earlier Fibonacci numbers, 17 and 34 do not. So a(9) = 2.

T. D. Noe, Table of n, a(n) for n=1..300

Cf. A120255, A001177.

f[t_] := Append[t, Select[Divisors[Fibonacci[Length[t] + 1]], FreeQ[Flatten[t], # ] &]]; Length /@ Nest[f, {}, 85] (* Ray Chandler, Jun 14 2006 *)

Extended by Ray Chandler, Jun 14 2006

cp's OEIS Frontend

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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A308797 Primes p such that A001177(p) = (p-1)/4.

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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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A233283 Record values in partial LCM-products (A233287) of Fibonacci entry points (A001177).

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