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%I M0155
%S 2,1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,
%T 9349,15127,24476,39603,64079,103682,167761,271443,439204,710647,
%U 1149851,1860498,3010349,4870847,7881196,12752043,20633239,33385282
%N Lucas numbers (beginning at 2): L(n) = L(n-1) + L(n-2). (Cf. A000204.)
%C This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye (rlahaye(AT)new.rr.com), Aug 18 2003
%C 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 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_, 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 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
%C 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 The first six multiplication formulae are:
%C L(2n) = (L(n))^2 - 2*(-1)^n
%C L(3n) = (L(n))^3 - 3*((-1)^n)*L(n)
%C L(4n) = (L(n))^4 - 4*((-1)^n)*(L(n))^2 + 2
%C L(5n) = (L(n))^5 - 5*((-1)^n)*(L(n))^3 + 5*L(n)
%C L(6n) = (L(n))^6 - 6*((-1)^n)*(L(n))^4 + 9*(L(n))^2 - 2*(-1)^n
%C Generally, L(n) | L(mn) if and only if m is odd. (End)
%C 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 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). - 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 numbers repeat in a cycle of length 60 (see A106291(100)).
%C 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 L(n) - L(n - 3) = 2*L(n - 2)
%C L(n) - L(n - 4) = 5*F(n - 2)
%C L(n) - L(n - 6) = 4*L(n - 3)
%C L(n) - L(n - 12) = 40*F(n - 6)
%C L(n) - L(n - 60) = 4160200*F(n - 30).
%C 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 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 Sum_{n > 0} a(n)/(n*2^n) = 2*log(2) [From _Jaume Oliver Lafont_, Oct 11 2009]
%C A010888(a(n)) = A030133(n). [_Reinhard Zumkeller_, Aug 20 2011]
%D 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.
%D P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 69.
%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 32,50.
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.
%D S. Falcon, On the Lucas triangle and its relationship with the k-Lucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425-434. Available online at http://scik.org. - From _N. J. A. Sloane_, Sep 20 2012
%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 148.
%D V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
%D V. E. Hoggat, Jr., Marjorie Bicknell, Some congruences of the Fibonacci numbers modulo a prime p, Math. Mag. 47 (4) (1974) 210-214.
%D Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
%D A. McLeod and W. O. J. Moser, Counting cyclic binary strings, Math. Mag., 80 (No. 1, 2007), 29-37.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D 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 S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
%D M. Waldschmidt, Lectures on Multiple Zeta Values (IMSC 2011), http://www.math.jussieu.fr/~miw/articles/pdf/MZV2011IMSc.pdf
%H N. J. A. Sloane, <a href="/A000032/b000032.txt">The first 500 Lucas numbers: Table of n, L(n) for n = 0..500</a>
%H A. Akbary, Q. Wang, <a href="http://dx.doi.org/10.1090/S0002-9939-05-08220-1">A generalized Lucas sequence and permutation binomials</a>, Proc. Amer. Math. Soc. 134 (2006) 15-22, sequence a(n) for l=5.
%H A. Aksenov, <a href="http://arxiv.org/abs/1108.5352">The Newman phenomenon and Lucas sequence</a>, arXiv:1108.5352, 2011
%H G. Everest, Y. Puri and T. Ward, <a href="http://arXiv.org/abs/math.NT/0204173">Integer sequences counting periodic points</a>
%H R. Javonovic, <a href="http://milan.milanovic.org/math/english/function1/function1.html">Lucas Function Calculator</a>
%H Blair Kelly, <a href="http://mersennus.net/fibonacci//lucas.txt">Factorizations of Lucas numbers</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H R. Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/lucasNbs.html">The Lucas numbers</a>
%H 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 Dmitry Kruchinin, <a href="http://arxiv.org/abs/1109.1683">Superposition's properties of logarithmic generating functions</a>, arXiv:1109.1683, 2011.
%H A. M. Hinz, S. Klavžar, U. Milutinović, C. Petr, <a href="http://dx.doi.org/10.1007/978-3-0348-0237-6">The Tower of Hanoi - Myths and Maths</a>, Birkhäuser 2013. See page 12. <a href="http://tohbook.info">Book's website</a>
%H Hisanori Mishima, Factorizations of Lucas Numbers: <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha113.htm">m=1..100</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha114.htm">m=101..200</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha115.htm">n=201..300</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha116.htm">n=301..400</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha117.htm">n=401..478</a>
%H Tony D. Noe and Jonathan Vos Post, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Noe/noe5.html">Primes in Fibonacci n-step and Lucas n-step Sequences,</a> J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4
%H 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 Zhi-Hong Sun, <a href="http://www.hytc.edu.cn/xsjl/szh/ConFn.pdf">Congruences for Fibonacci Numbers</a> [PDF] (Lecture notes, 2009)
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LucasNumber.html">Lucas Number</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rea#recLCC">Index entries for sequences related to linear recurrences with constant coefficients</a>, signature (1,1).
%F G.f.: (2 - x)/(1 - x - x^2).
%F L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n.
%F L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
%F E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley (smiley(AT)math.uaa.alaska.edu), Nov 30 2001
%F L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - _Jeff Burch_
%F L(n) = Fib(n) + 2*Fib(n - 1) = Fib(n + 1) + Fib(n - 1) - _Henry Bottomley_, Apr 12 2000
%F 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]
%F 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
%F L(n) = 2*Fib(n + 1) - Fib(n) - _Paul Barry_, Mar 22 2004
%F a(n) = floor((phi)^n + ( - phi)^( - n)) - _Paul Barry_, Mar 12 2005
%F Comments from _Miklos Kristof_, Mar 19 2007: (Start)
%F Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
%F L(n + m) + (-1)^m*L(n - m) = L(n)*L(m)
%F L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m)
%F 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 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 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 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 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
%F 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
%F Starting (1, 3, 4, 7, 11,...) = row sums of triangle A131774. - _Gary W. Adamson_, Jul 14 2007
%F a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n - _Gary W. Adamson_, Mar 02 2008
%F Comments from _Hieronymus Fischer_, Jan 02 2009: (Start)
%F 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 a(n) = round(1/(min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
%F 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. [From _Wolfdieter Lang_, May 15 2010]
%F L(n))/L(n - 1) - > A001622. [From _Vincenzo Librandi_, Jul 17 2010]
%F a(n) = 2*a(n - 2) + a(n - 3),n>2 [From _Gary Detlefs_, Sep 09 2010]
%F L(n) = floor(1/fract(Fib(n)*phi)), for n odd. - _Hieronymus Fischer_, Oct 20 2010
%F L(n) = ceiling(1/(1 - fract(Fib(n)*phi))), for n even. - _Hieronymus Fischer_, Oct 20 2010
%F L(n) = 2^n*(cos(Pi/5)^n + cos(3*Pi/5)^n) [From _Gary Detlefs_ Nov 29 2010]
%F L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)) [From _Gary Detlefs_ Dec 13 2010]
%F L(n) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n + 1)) [From _Gary Detlefs_ Dec 26 2010]
%F L(n) = floor(phi^n) + ((-1)^n + 1)/2, where phi = A001662. [From _Gary Detlefs_ Jan 20 2011]
%F L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2),n>2. [From _Gary Detlefs_ May 19 2011]
%F For n > = 2, a(n) = round(phi^n) where phi is the golden ratio. [Arkadiusz Wesolowski, Jul 20 2012]
%F 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
%F Contribution from Gary Detlefs, Dec 21 2012: (Start)
%F L(2*k*n) = phi^(n*k) + (2 - phi)^(n*k),
%F L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n,
%F L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k,
%F where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
%F (End)
%F L(n) = n * sum(binomial(n - k,k)/(n - k),k = 0..floor(n/2)), n>0 [H.W. Gould]. - _Gary Detlefs_, Jan 20 2013
%p with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
%p seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)
%t a[0] := 2; a[n] := Nest[{Last[ # ], First[ # ] + Last[ # ]} &, {2, 1}, n] // Last
%t Array[2 Fibonacci[ #+1] - Fibonacci[ # ] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
%t Table[LucasL[n, 1], {n, 0, 36}] (* Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 09 2009 *)
%t 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 *)
%o (MAGMA) [ Lucas(n) : n in [0..120]];
%o (PARI) a(n)=if(n<0,(-1)^n*a(-n),if(n<2,2-n,a(n-1)+a(n-2)))
%o (PARI) a(n)=if(n<0,(-1)^n*a(-n),polsym(x^2-x-1,n)[n+1])
%o (PARI) a(n)=real((2+quadgen(5))*quadgen(5)^n)
%o (PARI) a(n)=fibonacci(n+1)+fibonacci(n-1) \\ _Charles R Greathouse IV_, Jun 11 2011
%o (PARI) polsym(1+x-x^2, 50) \\ _Charles R Greathouse IV_, Jun 11 2011
%o (Sage) [lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008
%o (Haskell)
%o a000032 n = a000032_list !! n
%o a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
%o -- _Reinhard Zumkeller_, Aug 20 2011
%Y Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
%Y First row of array A103324.
%Y a(n) = A101220(2, 0, n), for n > 0.
%Y a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
%Y Cf. A131774, A001622, A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods).
%K nonn,nice,easy,core,changed
%O 0,1
%A _N. J. A. Sloane_.
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