A071774 -id:A071774 - cp's OEIS frontend

A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

3, 8, 20, 16, 10, 28, 36, 18, 48, 14, 30, 76, 40, 88, 32, 108, 58, 60, 136, 70, 148, 78, 168, 44, 196, 50, 208, 72, 108, 76, 256, 130, 276, 46, 148, 50, 316, 328, 336, 348, 178, 90, 190, 388, 396, 22, 42, 448, 456, 114, 52, 238, 240, 250, 516, 176, 268, 270, 556

Offset: 1

Louis Mello (mellols(AT)aol.com), Mar 26 2001

nonn

Assuming Wall's conjecture (which is still open) allows one to calculate A001175(m) when m is a prime power since for any k >= 1: A001175(prime(n)^k) = a(n)*prime(n)^(k-1). For example: A001175(2^k) = 3*2^(k-1) = A007283(k-1).

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014), # 14.8.5.
A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, (2004).
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Wikipedia, Pisano period.

Cf. A001175, A000961, A071774, A003147, A296240.

a:= proc(n) option remember; local F, k, p;
      F:=[1,1]; p:=ithprime(n);
      for k while F<>[0,1] do
        F:=[F[2], irem(F[1]+F[2],p)]
      od: k
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 16 2015

Table[p=Prime[n]; a={1,0}; a0=a; k=0; While[k++; s=Mod[Plus@@a,p];a=RotateLeft[a]; a[[2]]=s; a!=a0]; k, {n,100}] (* T. D. Noe, Jun 12 2006 *)

for(n=1,100,s=1; while(sum(i=n,n+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))+sum(i=n+1,n+1+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))>0,s++); print1(s,","))

from itertools import count
from sympy import prime
def A060305(n):
    x, p = (1,1), prime(n)
    for k in count(1):
        if x == (0,1):
            return k
        x = (x[1], (x[0]+x[1]) % p) # Chai Wah Wu, May 31 2022

a(n) = A001175(prime(n)). - Jonathan Sondow, Dec 09 2017

a(n) = (3 - L(p))/2 * (p - L(p)) / A296240(n) for n >= 4, where p = prime(n) and L(p) = Legendre(p|5); so a(n) <= p-1 if p == +- 1 mod 5, and a(n) <= 2*p+2 if p == +- 2 mod 5. See Wall's Theorems 6 and 7. - Jonathan Sondow, Dec 10 2017

Corrected by Benoit Cloitre, Jun 04 2002

Name clarified by Jonathan Sondow, Dec 09 2017

A308784 Primes p such that A001175(p) = 2*(p+1)/3.

Original entry on oeis.org

47, 107, 113, 263, 347, 353, 563, 677, 743, 977, 1097, 1217, 1223, 1277, 1307, 1523, 1553, 1733, 1823, 1877, 1913, 1973, 2027, 2237, 2243, 2267, 2333, 2447, 2663, 2687, 2753, 2777, 3323, 3347, 3407, 3467, 3533, 3557, 3617, 3623, 3767, 3947, 4133, 4493, 4547, 4583

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/3, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/3, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then A001177(p) = (p+1)/6, otherwise (p+1)/3.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/3 or (p+1)/6. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/6, otherwise (p+1)/3.
In general, 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) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 3. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 |  Total | Inert primes*
  3 |       4 |       6 |     10 |         88
  4 |      41 |      43 |     84 |        618
  5 |     330 |     353 |    683 |       4813
  6 |    2745 |    2736 |   5481 |      39286
  7 |   23219 |   23250 |  46469 |     332441
  8 |  201805 |  201547 | 403352 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

Michael De Vlieger, Table of n, a(n) for n = 1..714
Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.
Wikipedia, Pisano period

Similar sequences that give primes such that (b) holds: A071774 (r=1), this sequence (r=3), A308785 (r=7), A308786 (r=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 <= 4583, p = NextPrime[p], If[pn[p] == 2(p+1)/3, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 02 2019 *)

Pisano_for_inert_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(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 4000, if(Pisano_for_inert_prime(p)==2*(p+1)/3, print1(p, ", ")))

A071776 Related to Pisano periods: n such that the period of Fibonacci numbers mod n equals 3*(n+2).

Original entry on oeis.org

6, 14, 26, 74, 86, 134, 146, 194, 206, 254, 314, 326, 386, 446, 554, 566, 626, 674, 734, 746, 794, 866, 914, 926, 974, 1046, 1094, 1154, 1214, 1226, 1286, 1346, 1454, 1466, 1514, 1574, 1646, 1706, 1754, 1766, 1814, 1874, 1994, 2066, 2126, 2186, 2234, 2246

Offset: 1

Benoit Cloitre, Jun 04 2002

a(n)/2 are primes with final digit 3 or 7 among primes in a related sequence: "m such that the period of Fibonacci numbers mod m equals 2*(m+1)".

Cf. A071774, A001175.

for(n=2,4000,t=3*(n+2);good=1;if(fibonacci(t)%n==0, for(s=0,t,if(fibonacci(t+s)%n!=fibonacci(s)%n,good=0;break); if(s>1&&s

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Dec 21 2004

A308785 Primes p such that A001175(p) = 2*(p+1)/7.

Original entry on oeis.org

307, 797, 1483, 3023, 4157, 4283, 6397, 6733, 7027, 7433, 7867, 9337, 9743, 9883, 10177, 10303, 10597, 11423, 12823, 14293, 18493, 19963, 20593, 20873, 24247, 24793, 25703, 28433, 29917, 30113, 31387, 31723, 31793, 32353, 33347, 34537, 34747, 37057, 38653, 38723

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/7, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/7, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then A001177(p) = (p+1)/14, otherwise (p+1)/7.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/7 or (p+1)/14. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/14, otherwise (p+1)/7.
In general, 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) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 7. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
  3 |       1 |       1 |     2 |         88
  4 |       6 |       8 |    14 |        618
  5 |      48 |      42 |    90 |       4813
  6 |     371 |     350 |   721 |      39286
  7 |    3098 |    3086 |  6184 |     332441
  8 |   27035 |   26989 | 54024 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

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

Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), this sequence (r=7), A308786 (r=9).

Select[Prime@ Range[1000], Function[n, Mod[Last@ NestWhile[{Mod[#2, n], Mod[#1 + #2, n], #3 + 1} & @@ # &, {1, 1, 1}, #[[1 ;; 2]] != {0, 1} &], n] == Mod[2 (n + 1)/7, n] ]] (* Michael De Vlieger, Mar 31 2021, after Leo C. Stein at A001175 *)

Pisano_for_inert_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(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 40000, if(Pisano_for_inert_prime(p)==2*(p+1)/7, print1(p, ", ")))

A308786 Primes p such that A001175(p) = 2*(p+1)/9.

Original entry on oeis.org

233, 557, 953, 4013, 4733, 5147, 6983, 7307, 7883, 9377, 10133, 12923, 14867, 15767, 17747, 19403, 20753, 22877, 23813, 26387, 26783, 27737, 29483, 32057, 33533, 35117, 39383, 40013, 40787, 41543, 41903, 42767, 43613, 45557, 46187, 48473, 48563, 50993, 51263, 53927

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/9, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Also, primes p such that the least integer k > 0 such that M^k == I (mod p) is 2*(p+1)/9, where M = [{1, 1}, {1, 0}] and I is the identity matrix.
Also, primes p such that A001177(p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then A001177(p) = (p+1)/18, otherwise (p+1)/9.
Also, primes p such that ord(-(3+sqrt(5))/2,p) = (p+1)/9 or (p+1)/18. If p == 1 (mod 4), then ord(-(3+sqrt(5))/2,p) = (p+1)/18, otherwise (p+1)/9.
In general, 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) (see the Wikipedia link below), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
If (b) holds, then the entry point of {T(n)} modulo p is (p+1)/r if p == 3 (mod 4) and (p+1)/(2r) if p == 1 (mod 4). Proof: let d = ord(u,p) = 2*(p+1)/r, d' = ord(-u^2,p), then (-u^2)^d' == (u^(-p-1)*u^2)^d == u^(d'*(-p+1)) (mod p), so d divides d'*(p-1), d' = d/gcd(d, p-1). It is easy to see that gcd(d, p-1) = 4 if p == 1 (mod 4) and 2 if p == 3 (mod 4).
Here k = 1, and this sequence gives primes such that (b) holds and r = 9. For k = 1, r cannot be a multiple of 5 because if 5 divides p+1 then p decomposes in K = Q[sqrt(5)], which contradicts with (b).
Number of terms below 10^N:
  N | 1 mod 4 | 3 mod 4 | Total | Inert primes*
  3 |       3 |       0 |     3 |         88
  4 |       6 |       4 |    10 |        618
  5 |      36 |      28 |    64 |       4813
  6 |     313 |     300 |   613 |      39286
  7 |    2563 |    2597 |  5160 |     332441
  8 |   22377 |   22350 | 44727 |    2880969
  * Here "Inert primes" means primes p > 2 such that Legendre(5,p) = -1, i.e., p == 2, 3 (mod 5).

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

Similar sequences that give primes such that (b) holds: A071774 (r=1), A308784 (r=3), A308785 (r=7), this sequence (r=9).

Pisano_for_inert_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(2*(p+1))); for(d=1, #v, if(Mod(M,p)^v[d]==Id, return(v[d]))))
forprime(p=2, 55000, if(Pisano_for_inert_prime(p)==2*(p+1)/9, print1(p, ", ")))

A216067 Prime numbers p such that p is odd and is congruent to 2 (mod 5) or 3 (mod 5), but the period of the irreducible polynomial x^2-x-1 in GF(p^2) is not 2*(p+1).

Original entry on oeis.org

47, 107, 113, 233, 263, 307, 347, 353, 557, 563, 677, 743, 797, 953, 967, 977, 1087, 1097, 1103, 1217, 1223, 1277, 1307, 1427, 1483, 1523, 1553, 1597, 1733, 1823, 1877, 1913, 1973, 2027, 2207, 2237, 2243, 2267, 2333, 2417, 2447, 2663, 2687, 2753, 2777

Offset: 1

V. Raman, Sep 01 2012

nonn

			47 is in the sequence because the period of the Fibonacci / Lucas numbers (mod 47) = 32, is not 2*(47+1) = 96.

V. Raman, Table of n, a(n) for n = 1..10000

Cf. A001175, A060305, A071776.

Cf. A071774.

forprime(p=3,3000,if(p%5==2||p%5==3,a=1;b=0;c=1;while(a!=0||b!=1,c++;d=a;a=b;a=(a+d)%p;b=d%p);if(c!=(2*(p+1)),print1(p",")))) \\ V. Raman, Nov 22 2012

Definition corrected by V. Raman, Nov 22 2012

A229466 Numbers k such that the period of Fibonacci numbers mod k is 3*(k+10).

Original entry on oeis.org

10, 30, 70, 130, 370, 430, 670, 730, 970, 1030, 1270, 1570, 1630, 1930, 2230, 2770, 2830, 3130, 3370, 3670, 3730, 3970, 4330, 4570, 4630, 4870, 5230, 5470, 5770, 6070, 6130, 6430, 6730, 7270, 7330, 7570, 7870, 8230, 8530, 8770, 8830, 9070, 9370, 9970

Offset: 1

Matthew Goers, Sep 24 2013

nonn

Related to Pisano periods. Other than the initial term 10, these are a subset of the terms of A071774 multiplied by 10, where A071774 are numbers m such that Fibonacci numbers mod m = 2*(m+1). All A071774 terms multiplied by 10 have Pisano periods 3*(n+10) or (n+10). This sequence is the 3*(n+10) subset. A229467 is the n+10 subset.

			The Pisano period of the Fibonacci numbers mod 30 = 120, which is 3*(30+10).
The Pisano period of the Fibonacci numbers mod 1570 = 4740, which is 3*(1570+10).

Matthew Goers, Table of n, a(n) for n = 1..74

Cf. A000045, A001175, A071774, A229467.

t = {}; Do[a = {1, 0}; a0 = a; k = 0; While[k++; s = Mod[Plus @@ a, n]; a = RotateLeft[a]; a[[2]] = s; k <= 3*(n + 10) && a != a0]; If[k == 3*(n + 10), AppendTo[t, n]], {n, 2, 10000}]; t (* T. D. Noe, Oct 02 2013 *)

A229467 Related to Pisano periods: numbers n such that there are n+10 distinct Fibonacci numbers mod n.

Original entry on oeis.org

170, 230, 530, 830, 1370, 1670, 1730, 1970, 2270, 2570, 2930, 3170, 3830, 4430, 4670, 5030, 5870, 5930, 6170, 6470, 6530, 6830, 7430, 7730, 8270, 8570, 8630, 8870, 9470, 9770, 9830, 10130, 11630, 11870, 11930, 12170, 12830, 13070, 13670, 13730, 14330

Offset: 1

Matthew Goers, Sep 24 2013

nonn

These are a subset of the terms of A071774 multiplied by 10, where A071774 are numbers m such that Fibonacci numbers mod m = 2*(m+1). All A071774 terms multiplied by 10 have Pisano periods 3*(n+10) or (n+10). This sequence is the (n+10) subset.

			The Pisano period of the Fibonacci numbers mod 170 = 180, which is 170+10.
The Pisano period of the Fibonacci numbers mod 1670 = 1680, which is 1670+10.

Matthew Goers, Table of n, a(n) for n = 1..60

Cf. A000045, A001175, A071774, A229466.

Added 3 terms - Matthew Goers, Oct 14 2013

A366951 a(n) = 2(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.

Original entry on oeis.org

2, 1, 0, 1, 2, 1, 1, 2, 1, 4, 2, 1, 2, 1, 3, 1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 4, 1, 3, 2, 3, 1, 2, 1, 6, 2, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 18, 10, 1, 1, 4, 9, 2, 2, 2, 1, 3, 2, 2, 1, 10, 1, 1, 7, 2, 1, 1, 6, 1, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 4, 2, 2, 10, 2, 1, 2, 1

Offset: 1

A.H.M. Smeets, Oct 29 2023

nonn

Cf. A000040, A003147, A060305, A047650, A047652, A071774, A124096, A124253, A296240, A308784 - A308794.

a(n) == 0 (mod 2) for prime(n) == +/- 1 (mod 5) and n > 2.
a(n) == 1 (mod 2) for Prime(n) == +/- 2 (mod 5) and n > 2.
a(n) = 1 iff prime(n) in A071774.
a(n) = 2 iff prime(n) in ({2} union A003147)/{5}.
a(n) = 3 iff prime(n) in A308784.
a(n) = 4 iff prime(n) in A308787.
a(n) = 6 iff prime(n) in A308788.
a(n) = 7 iff prime(n) in A308785.
a(n) = 8 iff prime(n) in A308789.
a(n) = 9 iff prime(n) in A308786.
a(n) = 10 iff prime(n) in A308790.
a(n) = 12 iff prime(n) in A308791.
a(n) = 14 iff prime(n) in A308792.
a(n) = 16 iff prime(n) in A308793.
a(n) = 18 iff prime(n) in A308794.
a(n) = A296240(n) iff prime(n) == +/- 2 (mod 5) and n > 3.
a(n) = 2*A296240(n) iff prime(n) == +/- 1 (mod 5) and n > 3.
a(n) in {2^k: k > 1} iff prime(n) in {A047650}.
a(n) == 3 (mod 6) iff prime(n) in {A124096}.
a(n) == 6 (mod 12) iff prime(n) in {A046652}.
a(n) == 0 (mod 14) iff prime(n) in {A125252}.

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A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

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A308784 Primes p such that A001175(p) = 2*(p+1)/3.

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A071776 Related to Pisano periods: n such that the period of Fibonacci numbers mod n equals 3*(n+2).

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A308785 Primes p such that A001175(p) = 2*(p+1)/7.

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A308786 Primes p such that A001175(p) = 2*(p+1)/9.

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A216067 Prime numbers p such that p is odd and is congruent to 2 (mod 5) or 3 (mod 5), but the period of the irreducible polynomial x^2-x-1 in GF(p^2) is not 2*(p+1).

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A229466 Numbers k such that the period of Fibonacci numbers mod k is 3*(k+10).

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A229467 Related to Pisano periods: numbers n such that there are n+10 distinct Fibonacci numbers mod n.

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A366951 a(n) = 2(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.