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%I A000032 M0155
%S A000032 2,1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,
%T A000032 9349,15127,24476,39603,64079,103682,167761,271443,439204,710647,
%U A000032 1149851,1860498,3010349,4870847,7881196,12752043,20633239,33385282
%N A000032 Lucas numbers (beginning at 2): L(n) = L(n-1) + L(n-2). (Cf. A000204.)
%C A000032 This is also the Horadam sequence (2,1,1,1). - Ross La Haye (rlahaye(AT)new.rr.com), 
               Aug 18 2003
%C A000032 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.
%C A000032 a(n)=sum(P(3;n-1-k,k),k=0..ceiling((n-1)/2)), 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 (pbarry(AT)wit.ie), 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).
%C A000032 Suppose psi=ln(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 (Hieronymus.Fischer(AT)gmx.de), Apr 18 2007
%C A000032 Comments from John Blythe Dobson (j.dobson(AT)uwinnipeg.ca), 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.
%C A000032 The first six multiplication formulae are:
%C A000032 L(2n) = (L(n))^2 - 2*(-1)^n
%C A000032 L(3n) = (L(n))^3 - 3*((-1)^n)*L(n)
%C A000032 L(4n) = (L(n))^4 - 4*((-1)^n)*(L(n))^2 + 2
%C A000032 L(5n) = (L(n))^5 - 5*((-1)^n)*(L(n))^3 + 5*L(n)
%C A000032 L(6n) = (L(n))^6 - 6*((-1)^n)*(L(n))^4 + 9*(L(n))^2 - 2*(-1)^n
%C A000032 Generally, L(n) | L(mn) iff m is odd. (End)
%C A000032 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)
%C A000032 The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci 
               numbers (A00045) 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). - John Blythe Dobson (j.dobson(AT)uwinnipeg.ca), 
               Nov 15 2007. 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 nu 
               mbers repeat in a cycle of length 60.
%C A000032 Further comments from John Blythe Dobson (j.dobson(AT)uwinnipeg.ca), 
               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:
%C A000032 L(n) - L(n-3) = 2*L(n-2)
%C A000032 L(n) - L(n-4) = 5*F(n-2)
%C A000032 L(n) - L(n-6) = 4*L(n-3)
%C A000032 L(n) - L(n-12) = 40*F(n-6)
%C A000032 L(n) - L(n-60) = 4160200*F(n-30).
%C A000032 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).
%C A000032 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)
%C A000032 Sum_{n>0} a(n)/(n*2^n) = 2*log(2) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), 
               Oct 11 2009]
%D A000032 P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 
               1968, vol. 2, p. 69.
%D A000032 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of 
               combinatorial proof, M.A.A. 2003, id. 32,50.
%D A000032 L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.
%D A000032 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 
               3rd ed., Oxford Univ. Press, 1954, p. 148.
%D A000032 V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 
               1969.
%D A000032 Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley 
               and Sons, 2001.
%D A000032 A. McLeod and W. O. J. Moser, Counting cyclic binary strings, Math. Mag., 
               80 (No. 1, 2007), 29-37.
%D A000032 Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas 
               n-step Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 
               05.4.4.
%D A000032 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A000032 Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the 
               Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 
               06.1.1.
%D A000032 S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood 
               Ltd., Chichester, 1989.
%H A000032 N. J. A. Sloane, <a href="b000032.txt">The first 500 Lucas numbers: Table 
               of n, L(n) for n = 0..500</a>
%H A000032 G. Everest, Y. Puri and T. Ward, <a href="http://arXiv.org/abs/math.NT/
               0204173">Integer sequences counting periodic points</a>
%H A000032 R. Javonovic, <a href="http://milan.milanovic.org/math/english/function1/
               function1.html">Lucas Function Calculator</a>
%H A000032 B. Kelly, <a href="http://home.att.net/~blair.kelly/mathematics/fibonacci/
               lucas.txt">Factorizations of Lucas numbers</a>
%H A000032 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A000032 R. Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/
               lucasNbs.html">The Lucas numbers</a>
%H A000032 R. Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/
               lucas200.html">The First 200 Lucas numbers and their factors</a>
%H A000032 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/
               matha1/matha113.htm">Factorizations of many number sequences</a>
%H A000032 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/
               matha1/matha114.htm">Factorizations of many number sequences</a>
%H A000032 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/
               matha1/matha115.htm">Factorizations of many number sequences</a>
%H A000032 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/
               matha1/matha116.htm">Factorizations of many number sequences</a>
%H A000032 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/
               matha1/matha117.htm">Factorizations of many number sequences</a>
%H A000032 B. Rittaud, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL10/Rittaud2/
               rittaud11.pdf">On the Average Growth of Random Fibonacci Sequences</
               a>, Journal of Integer Sequences, 10 (2007), Article 07.2.4.
%H A000032 Zhi-Hong Sun, <a href="http://202.195.112.2/xsjl/szh/ConFn.pdf">Congruences 
               for Fibonacci Numbers</a> [PDF] (Lecture notes, 2003)
%H A000032 Dan Sewell Ward, <a href="http://www.halexandria.org/dward094.htm">Modified 
               Fibonacci Sequence</a>.
%H A000032 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               LucasNumber.html">Link to a section of The World of Mathematics.</
               a>
%H A000032 <a href="Sindx_Ch.html#Cheby">Index entries for sequences related to 
               Chebyshev polynomials.</a>
%H A000032 <a href="Sindx_Rea.html#recur1">Index entries for recurrences a(n) = 
               k*a(n - 1) +/- a(n - 2)</a>
%H A000032 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
               a>
%H A000032 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A000032 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 (gerald.mcgarvey(AT)comcast.net), 
               Dec 21 2007
%F A000032 G.f.: (2-x)/(1-x-x^2). L(n)=((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n.
%F A000032 L(n) = L(n-1) + L(n-2) = (-1)^n L(-n).
%F A000032 E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley (smiley(AT)math.uaa.alaska.edu), 
               Nov 30 2001
%F A000032 L(n) = Fibonacci(2n)/Fibonacci(n) [ Jeff Burch (gburch(AT)erols.com) 
               ]
%F A000032 L(n) = Fib(n) + 2*Fib(n-1) = Fib(n + 1) + Fib(n-1) - Henry Bottomley 
               (se16(AT)btinternet.com), Apr 12 2000
%F A000032 a(n)=sqrt(F(n-1)^2+4*F(n)*F(n-2)) - Benoit Cloitre (benoit7848c(AT)orange.fr), 
               Jan 06 2003
%F A000032 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 (pbarry(AT)wit.ie), Nov 15 2003
%F A000032 L(n)=2*Fib(n+1)-Fib(n) - Paul Barry (pbarry(AT)wit.ie), Mar 22 2004
%F A000032 a(n)=floor((phi)^n+(-phi)^(-n)) - Paul Barry (pbarry(AT)wit.ie), Mar 
               12 2005
%F A000032 Comments from Miklos Kristof (kristmikl(AT)freemail.hu), Mar 19 2007: 
               (Start)
%F A000032 Let F(n)=A000045=Fibonacci numbers, L(n)=a(n)=Lucas numbers:
%F A000032 L(n+m)+(-1)^m*L(n-m)=L(n)*L(m)
%F A000032 L(n+m)-(-1)^m*L(n-m)=8*F(n)*F(m)
%F A000032 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)
%F A000032 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)
%F A000032 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)
%F A000032 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)
%F A000032 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 (Hieronymus.Fischer(AT)gmx.de), May 02 
               2007
%F A000032 Starting (1, 3, 4, 7, 11,...) = row sums of triangle A131774. - Gary 
               W. Adamson (qntmpkt(AT)yahoo.com), Jul 14 2007
%F A000032 a(n)=2*fibonacci(n-1)+fibonacci(n), n>=0 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Oct 05 2007
%F A000032 a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n - Gary W. Adamson (qntmpkt(AT)yahoo.com), 
               Mar 02 2008
%F A000032 Comments from Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), 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 suffices phi-phi^(-1)=1 (see 
               general formula described in A001622).
%F A000032 a(n)=nint(1/(min(fract(phi^n), 1-fract(phi^n))), for n>1, where fract(x)=x-floor(x). 
               (End)
%p A000032 with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
%p A000032 a:=n->2*fibonacci(n-1)+fibonacci(n): seq(a(n), n=0..36); - Zerinvary 
               Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007
%t A000032 a[0] := 2; a[n] := Nest[{Last[ # ], First[ # ] + Last[ # ]} &, {2, 1}, 
               n] // Last
%t A000032 Array[2 Fibonacci[ #+1] - Fibonacci[ # ] &, 50, 0] - Joseph Biberstine 
               (jrbibers(AT)indiana.edu), Dec 26 2006
%t A000032 Table[LucasL[n, 1], {n, 0, 36}] [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Jul 09 2009]
%t A000032 a=1;lst={2,a};s=5;Do[a=s-(a+1);AppendTo[lst,a];s+=a,{n,5!}];lst [From 
               Vladimir Orlovsky (4vladimir(AT)gmail.com), Oct 27 2009]
%o A000032 (MAGMA) [ Lucas(n) : n in [0..120]];
%o A000032 (PARI) a(n)=if(n<0,(-1)^n*a(-n),if(n<2,2-n,a(n-1)+a(n-2)))
%o A000032 (PARI) a(n)=if(n<0,(-1)^n*a(-n),polsym(x^2-x-1,n)[n+1])
%o A000032 (PARI) a(n)=real((2+quadgen(5))*quadgen(5)^n)
%o A000032 sage: [lucas_number2(n,1,-1) for n in range(37)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Jun 25 2008
%Y A000032 Cf. A000204. A000045(n)=(2L(n+1)-L(n))/5.
%Y A000032 First row of array A103324.
%Y A000032 a(n) = A101220(2, 0, n), for n > 0.
%Y A000032 a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k-1), for k > 0.
%Y A000032 Cf. A131774.
%Y A000032 Cf. A001622, A006497, A080039.
%Y A000032 Sequence in context: A058658 A070827 A160191 this_sequence A061084 A055391 
               A134876
%Y A000032 Adjacent sequences: A000029 A000030 A000031 this_sequence A000033 A000034 
               A000035
%K A000032 nonn,nice,easy,core
%O A000032 0,1
%A A000032 N. J. A. Sloane (njas(AT)research.att.com).

    
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