A001175 -id:A001175 - cp's OEIS frontend

A308787 Primes p such that A001175(p) = (p-1)/2.

29, 89, 101, 181, 229, 349, 401, 509, 761, 941, 1021, 1061, 1109, 1229, 1249, 1361, 1409, 1549, 1621, 1669, 1709, 1741, 1789, 1861, 2029, 2069, 2089, 2441, 2621, 2801, 2861, 3089, 3169, 3301, 3389, 3461, 3581, 3821, 3881, 3989, 4001, 4049, 4201, 4229, 4549, 4729

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 2.
Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 4) for primes p in this sequence.
The number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |     10 |            78
  4 |     89 |           609
  5 |    630 |          4777
  6 |   5207 |         39210
  7 |  44296 |        332136
  8 | 382966 |       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: A003147/{5} (s=1), this sequence (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

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 <= 4729, p = NextPrime[p], If[pn[p] == (p-1)/2, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 01 2019 *)

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, 4800, if(Pisano_for_decomposing_prime(p)==(p-1)/2, print1(p, ", ")))

A308789 Primes p such that A001175(p) = (p-1)/4.

Original entry on oeis.org

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

Offset: 1

Jianing Song, Jun 25 2019

nonn

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).

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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), this sequence (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

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 *)

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, ", ")))

A308792 Primes p such that A001175(p) = (p-1)/7.

Original entry on oeis.org

2269, 2731, 2969, 3739, 4831, 6091, 6329, 11159, 11789, 13049, 13679, 14281, 14449, 14771, 16871, 19559, 20399, 24179, 26111, 29191, 31039, 33181, 33811, 34511, 34679, 35911, 40111, 41651, 42701, 43961, 49211, 54881, 55259, 55721, 56099, 58129, 60859, 62819, 66809

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 7.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      0 |            78
  4 |      7 |           609
  5 |     55 |          4777
  6 |    507 |         39210
  7 |   4144 |        332136
  8 |  36319 |       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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), this sequence (s=7), A308793 (s=8), A308794 (s=9).

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 < 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 *)

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, 67000, if(Pisano_for_decomposing_prime(p)==(p-1)/7, print1(p, ", ")))

A308793 Primes p such that A001175(p) = (p-1)/8.

Original entry on oeis.org

1009, 3329, 8081, 12401, 15889, 19681, 25601, 25841, 26641, 32321, 33329, 33521, 34369, 36929, 41681, 42929, 47809, 53569, 55249, 64849, 70289, 74209, 76081, 85361, 86209, 87649, 88129, 88801, 90001, 93089, 93329, 97649, 98689, 99089, 100049, 101489, 107441, 117841

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 8.
Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 16) for primes p in this sequence.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      0 |            78
  4 |      3 |           609
  5 |     34 |          4777
  6 |    315 |         39210
  7 |   2751 |        332136
  8 |  23878 |       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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), this sequence (s=8), A308794 (s=9).

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 < 50000, 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 *)

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, 118000, if(Pisano_for_decomposing_prime(p)==(p-1)/8, print1(p, ", ")))

A308788 Primes p such that A001175(p) = (p-1)/3.

Original entry on oeis.org

139, 151, 331, 619, 661, 811, 829, 1069, 1231, 1279, 1291, 1381, 1471, 1579, 1699, 1999, 2239, 2251, 2281, 2371, 2659, 2689, 2749, 3271, 3331, 3391, 3499, 3631, 3919, 4051, 4159, 4231, 4261, 4759, 4909, 5059, 5581, 5701, 5821, 5839, 6079, 6229, 6469, 6619, 6691

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 3.
The number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      7 |            78
  4 |     64 |           609
  5 |    455 |          4777
  6 |   3688 |         39210
  7 |  31412 |        332136
  8 | 272318 |       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: A003147/{5} (s=1), A308787 (s=2), this sequence (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

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 <= 6691, p = NextPrime[p], If[pn[p] == (p-1)/3, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 01 2019 *)

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, 7000, if(Pisano_for_decomposing_prime(p)==(p-1)/3, print1(p, ", ")))

A308790 Primes p such that A001175(p) = (p-1)/5.

Original entry on oeis.org

211, 281, 421, 691, 881, 991, 1031, 1151, 1511, 1871, 1951, 2591, 3251, 3851, 4391, 4651, 4691, 4751, 4871, 5381, 5531, 5591, 5801, 6011, 6101, 6211, 6271, 6491, 7211, 7451, 8011, 8171, 8831, 8861, 9011, 9091, 9241, 9371, 9431, 9931, 10061, 10391, 10531, 10691

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 5.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      6 |            78
  4 |     40 |           609
  5 |    280 |          4777
  6 |   2289 |         39210
  7 |  18903 |        332136
  8 | 163395 |       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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), this sequence (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

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 < 11000, 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 *)

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, 11000, if(Pisano_for_decomposing_prime(p)==(p-1)/5, print1(p, ", ")))

A308791 Primes p such that A001175(p) = (p-1)/6.

Original entry on oeis.org

541, 709, 2389, 3121, 3529, 4561, 4861, 5869, 7069, 8821, 9001, 10789, 12421, 12781, 13309, 14341, 14869, 16981, 18289, 19249, 19309, 19429, 19501, 20389, 20809, 20929, 21649, 22741, 23629, 24181, 25189, 26821, 27109, 27409, 28669, 30181, 30469, 30781, 30949, 31189

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 6.
Note that the Pisano period of {T(n)} modulo p must be even, so we have p == 1 (mod 12) for primes p in this sequence.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      2 |            78
  4 |     11 |           609
  5 |    112 |          4777
  6 |    898 |         39210
  7 |   7777 |        332136
  8 |  68115 |       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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), this sequence (s=6), A308792 (s=7), A308793 (s=8), A308794 (s=9).

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 < 32000, 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 *)

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, 32000, if(Pisano_for_decomposing_prime(p)==(p-1)/6, print1(p, ", ")))

A308794 Primes p such that A001175(p) = (p-1)/9.

Original entry on oeis.org

199, 919, 6679, 7489, 12979, 16921, 17011, 17659, 20089, 20431, 23059, 23599, 24391, 24859, 25309, 28081, 29629, 33301, 36901, 39079, 39439, 41761, 42589, 43399, 43669, 45361, 46261, 48619, 51481, 53479, 54091, 62011, 62191, 67411, 69499, 72019, 72091, 77419, 78301

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+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 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 = 9.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      2 |            78
  4 |      4 |           609
  5 |     49 |          4777
  6 |    405 |         39210
  7 |   3489 |        332136
  8 |  30132 |       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: A003147/{5} (s=1), A308787 (s=2), A308788 (s=3), A308789 (s=4), A308790 (s=5), A308791 (s=6), A308792 (s=7), A308793 (s=8), this sequence (s=9).

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 < 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 *)

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, 80000, if(Pisano_for_decomposing_prime(p)==(p-1)/9, print1(p, ", ")))

A235702 Fixed points of A001175 (Pisano periods).

Original entry on oeis.org

1, 24, 120, 600, 3000, 15000, 75000, 375000, 1875000, 9375000, 46875000, 234375000, 1171875000, 5859375000, 29296875000, 146484375000, 732421875000, 3662109375000, 18310546875000, 91552734375000, 457763671875000, 2288818359375000, 11444091796875000

Offset: 1

Reinhard Zumkeller, Jan 15 2014

Michael De Vlieger, Table of n, a(n) for n = 1..1430
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 26.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
Wikipedia, Pisano period
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A001175, A001178, A008606, A000351.

a235702 n = if n == 1 then 1 else 24 * 5 ^ (n - 2)
a235702_list = 1 : iterate (* 5) 24

LinearRecurrence[{5},{1,24},30] (* or *) Join[{1},NestList[5#&,24,30]] (* Harvey P. Dale, May 07 2017 *)

Vec(-x*(19*x+1)/(5*x-1) + O(x^100)) \\ Colin Barker, Jan 16 2014

A001175(a(n)) = a(n); A001178(a(n)) = 0.
From Colin Barker, Jan 16 2014: (Start)
a(n) = 24*5^(n-2) for n > 1.
a(n) = 5*a(n-1) for n > 2.
G.f.: -x*(19*x+1) / (5*x-1). (End)
E.g.f.: (24*(exp(5*x) - 1) - 95*x)/25. - Stefano Spezia, Nov 09 2024

A222413 All primes p > 5 such that A001175(p) is smaller than the maximal value permitted by Wall's Theorems 6 and 7.

Original entry on oeis.org

29, 47, 89, 101, 107, 113, 139, 151, 181, 199, 211, 229, 233, 263, 281, 307, 331, 347, 349, 353, 401, 421, 461, 509, 521, 541, 557, 563, 619, 661, 677, 691, 709, 743, 761, 769, 797, 809, 811, 829, 859, 881, 911, 919, 941, 953, 967, 977, 991, 1009, 1021, 1031, 1049, 1061, 1069, 1087, 1097, 1103, 1109, 1151, 1217, 1223, 1229, 1231, 1249, 1277

Offset: 1

N. J. A. Sloane, Feb 28 2013

nonn

Included because A001175 is still a mystery (as are many sequences of the same type).
A222414 gives the corresponding values of A001175(a(n)).
The maximal value for a prime p > 5 is p-1 if p == 1 or 9 (mod 10) and 2*(p+1) if p == 3 or 7 (mod 10). See Wall's Theorems 6 and 7. These values are given in A253806. - Wolfdieter Lang, Jan 16 2015
Prime(n) is a member if and only if A296240(n) > 1. - Jonathan Sondow, Dec 10 2017

			From _Wolfdieter Lang_, Jan 16 2015: (Start)
a(1) = 29 because A001175(29) = 14 but the maximal value is 29 - 1 = 28.
a(2) = 47 because A001175(47) = 32 but the maximal value is 2*(47 + 1) = 96.
All other primes p > 5 have A001175(p) = maximal value for p.
E.g., p = 11 has  A001175(11) = 11-1 = 10 and  p = 7 has A001175(7) = 2*(7 + 1) = 16. (End)

D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.

Cf. A001175, A060305, A222414, A296240.

Cf. A001176, A001177. - Wolfdieter Lang, Jan 16 2015

Name corrected by Wolfdieter Lang, Jan 16 2015

cp's OEIS Frontend

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A308793 Primes p such that A001175(p) = (p-1)/8.

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A308788 Primes p such that A001175(p) = (p-1)/3.

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A308790 Primes p such that A001175(p) = (p-1)/5.

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A308791 Primes p such that A001175(p) = (p-1)/6.

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A308794 Primes p such that A001175(p) = (p-1)/9.

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A235702 Fixed points of A001175 (Pisano periods).

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