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Fibonacci sequence (mod 5).
10

%I #54 Mar 09 2024 11:05:50

%S 0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2,3,0,3,3,1,4,0,4,4,3,

%T 2,0,2,2,4,1,0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2,3,0,3,3,

%U 1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1

%N Fibonacci sequence (mod 5).

%C This sequence contains the complete set of residues modulo 5. See A079002. - _Michel Marcus_, Jan 31 2020

%D S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989. See p. 88. - _N. J. A. Sloane_, Feb 20 2013

%H Vincenzo Librandi, <a href="/A082116/b082116.txt">Table of n, a(n) for n = 0..1000</a>

%H Brandon Avila and Yongyi Chen, <a href="https://www.fq.math.ca/Papers1/51-2/AvilaChen.pdf">On Moduli For Which the Lucas Numbers Contain a Complete Residue System</a>, Fibonacci Quart. 51 (2013), no. 2, 151-152. See p. 151.

%H S. A. Burr, <a href="https://www.fq.math.ca/Scanned/9-5/burr.pdf">On moduli for which the Fibonacci sequence contains a complete system of residues</a>, The Fibonacci Quarterly, 9.5 (1971), 497-504.

%H Diana Savin and Elif Tan, <a href="https://arxiv.org/abs/2403.01592">On Companion sequences associated with Leonardo quaternions: Applications over finite fields</a>, arXiv:2403.01592 [math.CO], 2024. See p. 11.

%H Minjia Shi and Patrick Solé, <a href="https://arxiv.org/abs/1807.08418">The largest number of weights in cyclic codes</a>, arXiv:1807.08418 [cs.IT], 2018.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">Fibonacci Number</a>

%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,-1,1,1,-1,-1,1,0,-1,0,1).

%F Sequence is periodic with Pisano period 20.

%F a(n) = 2 + ((n mod 20) - ((n - 1) mod 20) - ((n - 3) mod 20) - ((n - 4) mod 20) + 3*((n - 5) mod 20) - 3*((n - 6) mod 20) + 2*((n - 8) mod 20) - 3*((n - 9) mod 20) + 4*((n - 10) mod 20) - 4*((n - 11) mod 20) + ((n - 13) mod 20) + ((n - 14) mod 20) + 2*((n - 15) mod 20) - 2*((n - 16) mod 20) - 2*((n - 18) mod 20) + 3*((n - 19) mod 20))/20. - _Hieronymus Fischer_, Jun 30 2007

%F G.f.: (x + x^2 + 2x^3 + 3x^4 + 3x^6 + 3x^7 + x^8 + 4x^9 + 4x^11 + 4x^12 + 3x^13 + 2x^14 + 2x^16 + 2x^17 + 4x^18 + x^19)/(1 - x^20), not reduced. - _Hieronymus Fischer_, Jun 30 2007

%F a(n) = A010073(n) mod 5. - _Hieronymus Fischer_, Jun 30 2007

%F G.f.: -x*(1 + x + x^2 + 2*x^3 + 3*x^6 - x^7 - 2*x^8 - x^4 + x^9 + 4*x^10 + x^11) / ( (x - 1) * (x^4 + x^3 + x^2 + x + 1) * (x^8 - x^6 + x^4 - x^2 + 1) ). - _R. J. Mathar_, Jul 14 2012

%t Table[Mod[Fibonacci[n], 5], {n, 0, 125}] (* _Alonso del Arte_, Jul 29 2013 *)

%o (Magma) [Fibonacci(n) mod 5: n in [0..100]]; // _Vincenzo Librandi_, Feb 04 2014

%o (PARI) a(n)=fibonacci(n)%5 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A000045, A011655, A082115, A079343, A082116, A082117, A079344, A079002.

%K nonn,easy

%O 0,4

%A _Eric W. Weisstein_, Apr 03 2003

%E Added a(0)=0 from _Vincenzo Librandi_, Feb 04 2014