A001175 -id:A001175 - cp's OEIS frontend

A007887 a(n) = Fibonacci(n) mod 9.

0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
G. Wulczy, Unity with Fibonacci, Problem H-247 and solution, Fib. Quarter. p. 89_90, Vol 15, 1, Feb. 1977. Mentions this sequence.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

[Fibonacci(n) mod 9: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Mod[Fibonacci[Range[0,80]],9] (* Harvey P. Dale, Mar 23 2012 *)

a(n)=fibonacci(n)%9 \\ Charles R Greathouse IV, Oct 07 2015

Period 24 = A001175(9). Proof: F_{n+24} = F_n + 9*(5152 F_{n+1} + 3184 F_n). - Olivier Wittenberg, following a conjecture by Ralf Stephan, Sep 28 2004

The numbers have period 24 since F_{n+24} = F_n + 9*(5152 F_{n+1} + 3184 F_n). - Olivier Wittenberg, Sep 28 2004

A022099 Fibonacci sequence beginning 1, 9.

Original entry on oeis.org

1, 9, 10, 19, 29, 48, 77, 125, 202, 327, 529, 856, 1385, 2241, 3626, 5867, 9493, 15360, 24853, 40213, 65066, 105279, 170345, 275624, 445969, 721593, 1167562, 1889155, 3056717, 4945872, 8002589, 12948461, 20951050, 33899511, 54850561, 88750072, 143600633, 232350705

Offset: 0

N. J. A. Sloane, Jun 14 1998

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have: b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths:  1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - Klaus Purath, Jun 26 2022

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093644, A101220, A109754.

a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 9}, 36] (* Robert G. Wilson v, Apr 11 2014 *)

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) =  8*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 10*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+3) + Fibonacci(n-4). - Greg Dresden and Mary Beth Pittman, Mar 12 2022

A296240 Pisano quotients: a(n) = (p-1)/k(p) if p == +- 1 mod 5, = (2*p+2)/k(p) if p == +- 2 mod 5, where p = prime(n) and k(p) = Pisano period(p).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 9, 5, 1, 1, 2, 9, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 7, 1, 1, 1, 3, 1, 3, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 5, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 20, 1, 6, 1, 9, 3, 1, 1, 1, 1, 1, 1

Offset: 4

Jonathan Sondow, Dec 09 2017

nonn

Wall (1960) in Theorems 6 and 7 proved that a(n) is an integer for n >= 4. Jarden (1946) proved that the sequence is unbounded. See Elsenhans and Jahnel (2010), pp. 1-2.

A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, arXiv 1006.0824 [math.NT], 2010.
D. Jarden, Two theorems on Fibonacci's sequence, Amer. Math. Monthly, 53 (1946), 425-427.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Wikipedia, Wall-Sun-Sun prime

Cf. A001175, A092330, A060305, A222361, A222413.

With[{p = Prime[n]}, T = Table[a = {1, 0}; a0 = a; k = 0; While[k++; s = Mod[Plus @@ a, p]; a = RotateLeft[a]; a[[2]] = s; a != a0]; k, {n, 1, 130}]; Table[L = KroneckerSymbol[p, 5]; (3 - L)/2 (p - L)/T[[n]], {n, 4, 130}]] (* after T. D. Noe *)

a(n) = (3 - L(p))/2 * (p - L(p)) / k(p), where p = prime(n), L(p) = Legendre(p|5), and k(p) = Pisano period(p) = A001175(p).

a(n) > 1 if and only if prime(n) is in A222413.

A094401 Composite n such that n divides both Fibonacci(n-1) and Fibonacci(n) - 1.

Original entry on oeis.org

2737, 4181, 6721, 13201, 15251, 34561, 51841, 64079, 64681, 67861, 68251, 90061, 96049, 97921, 118441, 146611, 163081, 179697, 186961, 194833, 197209, 219781, 252601, 254321, 257761, 268801, 272611, 283361, 302101, 303101, 327313, 330929

Offset: 1

Eric Rowland, May 01 2004

nonn

Composite n such that Q^(n-1) = I (mod n), where Q is the Fibonacci matrix {{1,1},{1,0}} and I is the identity matrix. The identity is also true for the primes congruent to 1 or 4 (mod 5), which is sequence A045468. The period of Q^k (mod n) is the same as the period of the Fibonacci numbers F(k) (mod n), A001175. Hence the terms in this sequence are the composite n such that A001175(n) divides n-1. [T. D. Noe, Jan 09 2009]

Giovanni Resta, Table of n, a(n) for n = 1..1000 (first 200 terms from T. D. Noe)
Eric Weisstein, MathWorld: Fibonacci Q-Matrix.

Cf. A005845, A069106, A094394, A094400.

Select[Range[2, 50000], ! PrimeQ[ # ] && Mod[Fibonacci[ # - 1], # ] == 0 && Mod[Lucas[ # ] - 1, # ] == 0 &]

More terms from Ryan Propper, Sep 24 2005

A105870 Fibonacci sequence (mod 7).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3

Offset: 0

Shyam Sunder Gupta, May 05 2005

Sequence is periodic with Pisano period 16 = A001175(7).

			a(5) = 5 because Fibonacci(5) = 5.
a(6) = 1 because Fibonacci(6) = 8 and 8 mod 7 = 1.
a(7) = 6 because Fibonacci(7) = 13 and 13 mod 7 = 6.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Brady Haran, Fibonacci Tartan and Bagpipes, Numberphile video (2013). The music by Alan Stewart at 1:53 to 3:20 has pitch based on this sequence.
Wayne Peng, ABC Implies There are Infinitely Many non-Fibonacci-Wieferich Primes - An Application of ABC Conjecture over Number Fields, arXiv:1511.05645 [math.NT], 2015.
Diana Savin and Elif Tan, On Companion sequences associated with Leonardo quaternions: Applications over finite fields, arXiv:2403.01592 [math.CO], 2024. See p. 10.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

a105870 n = a105870_list !! (n-1)
a105870_list = 1 : 1 : zipWith (\u v -> (u + v) `mod` 7)
                               (tail a105870_list) a105870_list
-- Reinhard Zumkeller, Jan 15 2014

[Fibonacci(n) mod 7: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Table[Mod[Fibonacci[n], 7], {n, 0, 100}] (* Alonso del Arte, Jul 29 2013 *)

a(n)=fibonacci(n)%7 \\ Charles R Greathouse IV, Jun 04 2013

a(n)=lift(((Mod([1,1;1,0],7))^n)[1,2]) \\ Charles R Greathouse IV, Jun 04 2013

a(n)=fibonacci(n%16)%7 \\ Charles R Greathouse IV, Jan 06 2016

A105870_list, a, b, = [], 0, 1
for _ in range(10**3):
    A105870_list.append(a)
    a, b = b, (a+b) % 7 # Chai Wah Wu, Nov 26 2015

G.f.: - x*(1 + x + 2*x^2 + 3*x^3 + 5*x^4 + x^5 + 6*x^6 + 6*x^8 + 6*x^9 + 5*x^10 + 4*x^11 + 2*x^12 + 6*x^13 + x^14)/((x - 1)*(1 + x)*(1 + x^2)*(1 + x^4)*(1 + x^8)). - R. J. Mathar, Jul 14 2012

a(1) = a(2) = 1, then a(n) = (a(n - 2) + a(n - 1)) mod 7. - Alonso del Arte, Jul 30 2013

a(0)=0 from Vincenzo Librandi, Feb 04 2014

A161553 Table which contains in row n the fundamental Pisano period of the Fibonacci sequence (mod n).

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8

Offset: 1

Alexander Adamchuk, Jun 13 2009

The length of the n-th row (the length of the period) is A001175(n).

			F(n) mod 1 {0},
F(n) mod 2 {0,1,1},
F(n) mod 3 {0,1,1,2,0,2,2,1},
F(n) mod 4 {0,1,1,2,3,1},
F(n) mod 5 {0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1},
F(n) mod 6 {0,1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1,4,5,3,2,5,1},
F(n) mod 7 {0,1,1,2,3,5,1,6,0,6,6,5,4,2,6,1},
F(n) mod 8 {0,1,1,2,3,5,0,5,5,2,7,1},
F(n) mod 9 {0,1,1,2,3,5,8,4,3,7,1,8,0,8,8,7,6,4,1,5,6,2,8,1},
F(n) mod 10 {0,1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,1,7,8,5,3,8, 1,9,0,9,9,8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6,9,5,4,9,3,2,5,7,2,9,1}.

Alois P. Heinz, Rows n = 1..200, flattened
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithm. 16 (1969) 106-110.
Wayne Peng, ABC Implies There are Infinitely Many non-Fibonacci-Wieferich Primes - An Application of ABC Conjecture over Number Fields, arXiv:1511.05645 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Pisano Period.
Wikipedia, Pisano period.

Cf. A000045, A001175.
Main diagonal gives A002708.
Row sums give A214300.

per[1] = 1; per[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[Fibonacci[k + 1], n] == 1, Return[k]]];
row[n_] := Table[Mod[Fibonacci[k], n], {k, 0, per[n]-1}];
Array[row, 9] // Flatten (* Jean-François Alcover, Oct 30 2018 *)

row(n)={my(L=List([0]), X=Mod([1,1;1,0],n), I=Mod([1,0;0,1],n), M=X); while(M<>I, M*=X; listput(L, lift(M[2,2]))); Vec(L)} \\ Andrew Howroyd, Mar 05 2023

Moved into the keyword:tabf category by R. J. Mathar, Oct 04 2009

A022388 Fibonacci sequence beginning 6, 13.

Original entry on oeis.org

6, 13, 19, 32, 51, 83, 134, 217, 351, 568, 919, 1487, 2406, 3893, 6299, 10192, 16491, 26683, 43174, 69857, 113031, 182888, 295919, 478807, 774726, 1253533, 2028259, 3281792, 5310051, 8591843, 13901894, 22493737, 36395631, 58889368, 95284999, 154174367, 249459366, 403633733, 653093099

Offset: 0

N. J. A. Sloane

The Pisano periods for this sequence are different from those for the Fibonacci numbers (A001175) for modulus 11 and 22. Furthermore, its Pisano periods are exactly the same as those of the Lucas sequence (A000032), given in A106291. - Klaus Purath, Apr 20 2019
a(n) is the alternating sum of 5 consecutive Lucas numbers (A000032). Also the sum of 4*k consecutive terms of A000285 divided by Fibonacci(2*k) (A000045), k = {1, 2, …}. All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Jul 29 2019
From Wajdi Maaloul, Jun 25 2022: (Start)
For n > 0, a(n) is the number of ways to tile the following figure (a T-shaped horizontal strip of length n beginning with a vertical strip of length 6) with squares and dominoes.
  ._
  |_|
  |_|
  ||______     _
  |||_|||...|_|
  |_|
  |_|
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045.

List([0..40], n-> 6*Fibonacci(n+2) + Fibonacci(n)); # G. C. Greubel, Jun 30 2019

[6*Fibonacci(n+2) + Fibonacci(n): n in [0..40]]; // G. C. Greubel, Mar 02 2018

Table[6*Fibonacci[n+2] + Fibonacci[n], {n, 0, 40}] (* or *) LinearRecurrence[{1,1}, {6,13}, 40] (* G. C. Greubel, Mar 02 2018 *)

vector(40, n, n--; 6*fibonacci(n+2) + fibonacci(n)) \\ G. C. Greubel, Mar 02 2018

[6*fibonacci(n+2) + fibonacci(n) for n in (0..40)] # G. C. Greubel, Jun 30 2019

G.f.: (6+7*x)/(1-x-x^2). - Philippe Deléham, Nov 20 2008
a(n) = 6*Fibonacci(n+2) + Fibonacci(n) = 6*Fibonacci(n-1) + 13*Fibonacci(n). - G. C. Greubel, Mar 02 2018
From Klaus Purath, Jul 29 2019: (Start)
L = Lucas (A000032), F = Fibonacci (A000045). All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
a(n+1) - a(n-4) = L(n)*11.
a(n) = L(n-1) + L(n+4).
a(n) = 3*L(n+1) + L(n+2) = L(n) + 4*L(n+1) = L(n+6) - 4*L(n+2).
a(n) = L(n+1) + 5*F(n+2) = L(n+5) - 5*F(n+1).
a(n) = (7*L(n+1) + 5*F(n+1))/2.
a(n) = (13*L(n+1) + L(n+5) - 5*F(n))/4.
a(n) = 7*F(n) + 6*F(n+1) = 7*F(n+2) - F(n+1).
a(n) = 8*F(n+2) - F(n+3) = 17*F(n+4) - 9*F(n+5).
The following six formulas apply for all sequences of the Fibonacci type.
a(n) = L(2*m)*a(n+2*m) - a(n+4*m).
a(n) = (F(m+2)*a(n+2) - a(m+n+2))/F(m).
a(n) = F(n-m-1)*a(m) + F(n-m)*a(m+1).
a(n)^2 + a(n+3)^2 = 2*(a(n+1)^2 + a(n+2)^2).
a(n)^2 + a(n+2)^2 + a(n+1)^2 + a(n+3)^2 = 3*(a(n)*a(n+2) + a(n+1)*a(n+3)).
3*a(n+2)*a(n+1)*a(n) = a(n+2)^3 - a(n+1)^3 - a(n)^3. (End)
E.g.f.: exp(-2*x/(1+sqrt(5)))*(-15-sqrt(5)+(45+19*sqrt(5))*exp(sqrt(5)*x))/(5+3*sqrt(5)). - Stefano Spezia, Aug 16 2019

Terms a(36) onward added by G. C. Greubel, Mar 02 2018

A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

Original entry on oeis.org

8, 144

Offset: 1

Robert C. Lyons, Jan 08 2014

This sequence and A229037 and A235265 are winners in the contest held at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
Carmichael's theorem implies that 8 and 144 are the only terms of this sequence.
First two terms of A061899, A111687, A172150, A212703, and A231851. - Omar E. Pol, Jan 21 2014
Saha and Karthik conjectured (without reference to Carmichael's theorem) that the only positive integers k for which A001175(k^2) = A001175(k) are 6 and 12. (A000045(6) = 8 and A000045(12) = 144.) - L. Edson Jeffery, Feb 13 2014
Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only nontrivial perfect power Fibonacci numbers. - Robert C. Lyons, Dec 23 2015

			The Fibonacci number 8 is in the sequence because 8=2*2*2, and 2 is a Fibonacci number that is not equal to 8. The Fibonacci number 144 is in the sequence because 144=3*3*2*2*2*2, and both 2 and 3 are Fibonacci numbers that are not equal to 144.

Yann Bugeaud, Maurice Mignotte, and Samir Siksek, Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers, Annals of Mathematics, 163 (2006), pp. 969-1018.
Arpan Saha and Karthik C S, A few equivalences of Wall-Sun-Sun prime conjecture, arXiv:1102.1636 [math.NT], 2011.
Wikipedia, Carmichael's theorem.

Cf. A000045, A061899, A065108, A227875.

A022101 Fibonacci sequence beginning 1, 11.

Original entry on oeis.org

1, 11, 12, 23, 35, 58, 93, 151, 244, 395, 639, 1034, 1673, 2707, 4380, 7087, 11467, 18554, 30021, 48575, 78596, 127171, 205767, 332938, 538705, 871643, 1410348, 2281991, 3692339, 5974330, 9666669, 15640999, 25307668, 40948667, 66256335, 107205002, 173461337, 280666339

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(11;n-1-k,k) with n >= 1, a(-1)=10. These are the SW-NE diagonals in P(11;n,k), the (11,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000032, A000045.

a(n) = A109754(10, n+1) = A101220(10, 0, n+1).

a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,11},40] (* Harvey P. Dale, Aug 16 2015 *)

a(n) = 10*fibonacci(n)+fibonacci(n+1) \\ Charles R Greathouse IV, Jun 11 2015

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=11. a(-1)=10.
G.f.: (1+10*x)/(1-x-x^2).
a(n-1) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + 5*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 10*A000045(n) + A000045(n+1). - R. J. Mathar, Apr 07 2011
a(n) = 12*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = A000045(n+4) + A000032(n-4) for n > 0. - Bruno Berselli, Sep 27 2017

A104217 Period of Perrin (0,2,3,2,5,5,..., A001608) sequence mod n.

Original entry on oeis.org

1, 7, 13, 14, 24, 91, 48, 28, 39, 168, 120, 182, 183, 336, 312, 56, 288, 273, 180, 168, 624, 840, 22, 364, 120, 1281, 117, 336, 871, 2184, 993, 112, 1560, 2016, 48, 546, 1368, 1260, 2379, 168, 1723, 4368, 231, 840, 312, 154, 2257, 728, 336, 840, 3744, 2562

Offset: 1

Anthony C Robin, Mar 14 2005

nonn

Analogy to A001175, Pisano periods (or Pisano numbers): period of Fibonacci numbers mod n AND to A046738 for Perrin sequence, where a(n)=a(n-2)+a(n-3)

It appears that the n such that n-1 divides a(n) is the set of primes of the form x^2+23*y^2 (A033217). The discriminant of the characteristic polynomial of the Perrin sequence is -23. - T. D. Noe, Feb 23 2007

T. D. Noe, Table of n, a(n) for n=1..1000

Cf. A001175, A046738 and Perrin sequence A001608.

Table[a={0,2,3}; a=a0=Mod[a, n]; k=0; While[k++; s=a[[2]]+a[[1]]; a=RotateLeft[a]; a[[ -1]]=Mod[s,n]; a!=a0]; k, {n,100}] (* T. D. Noe, Oct 10 2006 *)

from math import lcm
from functools import lru_cache
from sympy import factorint
@lru_cache(maxsize=None)
def A104217(n):
    if n < 4:
        return (1,7,13)[n-1]
    f = factorint(n).items()
    if len(f) > 1:
        return lcm(*(A104217(a**b) for a,b in f))
    else:
        k,x = 1, (0,2,3)
        while x != (3,0,2):
            k += 1
            x = (x[1], x[2], (x[0]+x[1]) % n)
        return k # Chai Wah Wu, Apr 25 2025

More terms from T. D. Noe, Oct 10 2006

cp's OEIS Frontend

A007887 a(n) = Fibonacci(n) mod 9.

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Magma

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A022099 Fibonacci sequence beginning 1, 9.

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Magma

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A296240 Pisano quotients: a(n) = (p-1)/k(p) if p == +- 1 mod 5, = (2*p+2)/k(p) if p == +- 2 mod 5, where p = prime(n) and k(p) = Pisano period(p).

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A094401 Composite n such that n divides both Fibonacci(n-1) and Fibonacci(n) - 1.

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Mathematica

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A105870 Fibonacci sequence (mod 7).

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Haskell

Magma

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A161553 Table which contains in row n the fundamental Pisano period of the Fibonacci sequence (mod n).

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Mathematica

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A022388 Fibonacci sequence beginning 6, 13.

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GAP