0,1

This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003

For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.

a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).

Suppose psi = log(phi). We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007

From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)

The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.

The first six multiplication formulae are:

L(2n) = (L(n))^2 - 2*(-1)^n;

L(3n) = (L(n))^3 - 3*((-1)^n)*L(n);

L(4n) = (L(n))^4 - 4*((-1)^n)*(L(n))^2 + 2;

L(5n) = (L(n))^5 - 5*((-1)^n)*(L(n))^3 + 5*L(n);

L(6n) = (L(n))^6 - 6*((-1)^n)*(L(n))^4 + 9*(L(n))^2 - 2*(-1)^n.

Generally, L(n) | L(mn) if and only if m is odd.

In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities which have close analogues in the Lucas numbers:

For a >= b and odd b, L(a + b) + L(a - b) = 5*F(a)*F(b).

For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).

For a >= b and odd b, L(a + b) - L(a - b) = L(a)*L(b).

For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).

A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:

L(n) - L(n - 3) = 2*L(n - 2);

L(n) - L(n - 4) = 5*F(n - 2);

L(n) - L(n - 6) = 4*L(n - 3);

L(n) - L(n - 12) = 40*F(n - 6);

L(n) - L(n - 60) = 4160200*F(n - 30).

These formulae establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).

The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)

Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009

A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011

For n >= 3, also the number of vertex covers for the cycle graph C_n. - Eric W. Weisstein, Jan 04 2014

The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014

Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0,  regardless of initial values which may be non-integers. The corresponding sequences for other values of m are: for m = 2, 2*A001333;   for m = 3, A006497;   for m = 4, 2*A001077;  for m = 5, A087130;   for m = 6, 2*A005667;  for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014

If x=a(n), y=a(n+1), z=a(n+2), then -x^2 -z*x -3*y*x -y^2 +y*z +z^2=5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015

P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 69.

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 32,50.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 148.

Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

N. J. A. Sloane, The first 500 Lucas numbers: Table of n, L(n) for n = 0..500

A. Akbary, Q. Wang, A generalized Lucas sequence and permutation binomials, Proc. Amer. Math. Soc. 134 (2006) 15-22, sequence a(n) for l=5.

A. Aksenov, The Newman phenomenon and Lucas sequence, arXiv:1108.5352, 2011

Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.

Paula Burkhardt, Alice Zhuo-Yu Chan, Gabriel Currier, Stephan Ramon Garcia, Florian Luca, Hong Suh, Visual properties of generalized Kloosterman sums, arXiv preprint, 2015.

Charles Cratty, Samuel Erickson, Frehiwet Negass, Lara Pudwell, Pattern Avoidance in Double Lists, preprint, 2015.

G. Everest, Y. Puri and T. Ward, Integer sequences counting periodic points, arXiv:math/0204173 [math.NT], 2012.

Sergio Falcon, On the Lucas triangle and its relationship with the k-Lucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, pp. 425-434.

A. M. Hinz, S. Klavžar, U. Milutinović, C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 12. Book's website

V. E. Hoggat, Jr., Marjorie Bicknell, Some congruences of the Fibonacci numbers modulo a prime p, Math. Mag. 47 (4) (1974) 210-214.

R. Javonovic, Lucas Function Calculator

Blair Kelly, Factorizations of Lucas numbers

Tanya Khovanova, Recursive Sequences

R. Knott, The Lucas numbers

R. Knott, The First 200 Lucas numbers and their factors

Dmitry Kruchinin, Superposition's properties of logarithmic generating functions, arXiv:1109.1683, 2011.

Matthew Macauley , Jon McCammond, Henning S. Mortveit, Dynamics groups of asynchronous cellular automata, Journal of Algebraic Combinatorics, Vol 33, No 1 (2011), p 11-35.

T. Mansour, M. Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) Vol. 124, No. 2, May 2014, pp. 127-140.

A. McLeod and W. O. J. Moser, Counting cyclic binary strings, Math. Mag., 80 (No. 1, 2007), 29-37.

R. Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820, 2014

Hisanori Mishima, Factorizations of Lucas Numbers: m=1..100, m=101..200, n=201..300, n=301..400, n=401..478

Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4

Arzu Özkoç, Some algebraic identities on quadra Fibona-Pell integer sequence, Advances in Difference Equations, 2015, 2015:148.

J. L. Ramirez and V. F. Sirvent, Incomplete Tribonacci Numbers and Polynomials, Journal of Integer Sequences, Vol. 17, 2014, #14.4.2.

B. Rittaud, On the Average Growth of Random Fibonacci Sequences, Journal of Integer Sequences, 10 (2007), Article 07.2.4.

M. Z. Spivey and L. L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.

Zhi-Hong Sun, Congruences for Fibonacci Numbers [PDF] (Lecture notes, 2009)

M. Waldschmidt, Lectures on Multiple Zeta Values, IMSC 2011.

Eric Weisstein's World of Mathematics, Lucas Number

Roman Witula, Damian Slota and Edyta Hetmaniok, Bridges between different known integer sequences, Annales Mathematicae et Informaticae, 41 (2013) pp. 255-263.

Index entries for sequences related to Chebyshev polynomials.

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)

Index entries for linear recurrences with constant coefficients, signature (1,1).

Index entries for two-way infinite sequences

G.f.: (2 - x)/(1 - x - x^2).

L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n.

L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).

L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999

E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley (smiley(AT)math.uaa.alaska.edu), Nov 30 2001

L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000

a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]

a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003

L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004

a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005

From Miklos Kristof, Mar 19 2007: (Start)

Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:

L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).

L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).

L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).

L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).

L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).

L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)

Inverse: floor(log_phi(a(n)) + 0.5) = n, for n>1. Also for n >= 0, floor(1/2*log_phi(a(n)*a(n + 1))) = n. Extension valid for all integers n: floor(1/2*sign(a(n)*a(n + 1))*log_phi|a(n)*a(n + 1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007

Conjecture: Let f(n) = phi^n + phi^( - n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k=0..infinity} C(k + 1)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007

Starting (1, 3, 4, 7, 11,...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007

a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008

From Hieronymus Fischer, Jan 02 2009: (Start)

For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).

a(n) = round(1/(min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)

E.g.f.: exp(phi*x) + exp( - x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010

L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010

a(n) = 2*a(n - 2) + a(n - 3), n>2. - Gary Detlefs, Sep 09 2010

L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010

L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010

L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs Nov 29 2010

L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)). - Gary Detlefs Dec 13 2010

L(n) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n + 1)). - Gary Detlefs Dec 26 2010

L(n) = floor(phi^n) + ((-1)^n + 1)/2, where phi = A001662. - Gary Detlefs Jan 20 2011

L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs May 19 2011

For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012

a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012

From Gary Detlefs, Dec 21 2012: (Start)

L(2*k*n) = phi^(n*k) + (2 - phi)^(n*k).

L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.

L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.

where phi = (1 + sqrt(5))/2, F(n) = A000045(n).

(End)

L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H.W. Gould]. - Gary Detlefs, Jan 20 2013

G.f.: G(0), where G(k)= 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013

L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013

Let b(n) = b(n-1) + b(n-2), with b(0) = 1, b(1) = phi. Then, for n>=1, L(n)= floor(b(n-1)) if n is odd, L(n) = ceiling(b(n-1)), if n is even, with convergence. - Richard R. Forberg, Jan 20 2014

L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014

L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014

L(n+2) = 1 + Sum_{k=0..n}L(k). - Tom Edgar, Apr 15 2015

G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x65 + 18*x^6 + 29*x^7 + ...

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);

seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[ # ], First[ # ] + Last[ # ]} &, {2, 1}, n] // Last

Array[2 Fibonacci[ #+1] - Fibonacci[ # ] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)

Table[LucasL[n, 1], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)

a = 1; lst = {2, a}; s = 5; Do[a = s - (a + 1); AppendTo[lst, a]; s += a, {n, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Oct 27 2009 *)

LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)

(MAGMA) [ Lucas(n) : n in [0..120]];

(PARI) {a(n) = if( n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

(PARI) {a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

(PARI) {a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

(PARI) a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

(PARI) polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

(Sage) [lucas_number2(n, 1, -1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

(Haskell)

a000032 n = a000032_list !! n

a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)

-- Reinhard Zumkeller, Aug 20 2011

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.

First row of array A103324.

a(n) = A101220(2, 0, n), for n > 0.

a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.

Cf. A131774, A001622, A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods).

Sequence in context: A058658 A070827 A160191 * A061084 A055391 A177940

Adjacent sequences:  A000029 A000030 A000031 * A000033 A000034 A000035

nonn,nice,easy,core

N. J. A. Sloane, May 24 1994

approved