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Cf. A000204 for Lucas numbers beginning with 1.

Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014

Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017

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) = A002390. 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 - 5*F(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 formulas 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 that have close analogs 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 formulas 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

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 nonintegers. 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

A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015

A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016

Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017

Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017

For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019

From Klaus Purath, Apr 19 2019: (Start)

While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:

First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.

Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.

Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)

L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019

If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019

From Kai Wang, Dec 18 2019: (Start)

L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).

Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)

From Peter Bala, Dec 23 2021: (Start)

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.

For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)

For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024

For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004

For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007

The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012

Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012

For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013

For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014

The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014

The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016

Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

The numerator and denominator in the definition have no common divisors >1.

Also denominators in a system of Egyptian fraction for ratios of consecutive Fibonacci numbers: 1/2 = 1/2, 3/5 = 1/2 + 1/10, 8/13 = 1/2 + 1/10 + 1/65, 21/34 = 1/2 + 1/10 + 1/65 + 1/442 etc. (Rossi and Tout). - Barry Cipra, Jun 06 2002

a(n)-1 is a square. - Sture Sjöstedt, Nov 04 2011

From Wolfdieter Lang, May 26 2020: (Start)

Partial sums of the reciprocals: Sum_{k=1..n} 1/a(k) equal 1 for n=1, and F(2*n - 1)/F(2*n - 3) for n >= 2, with F = A000045. Proof by induction. Hence a(n) = 1 for n=1, and F(2*n - 3)*F(2*n - 5) for n >= 2, with F(-1) = 1 (gcd(F(n), F(n+1)) = 1). See the comment by Barry Cipra.

Thus a(n) = 1, for n = 1, and a(n) = 1 + F(2*(n-2))^2, for n >= 2 (by Cassini-Simson for even index, e.g., Vajda, p. 178 eq.(28)). See the Sture Sjöstedt comment.

The known G.f. of {F(2*n)^2} from A049684 leds then to the conjectured formula by R. J. Mathar below, and this proves also the recurrence given there..

From the partial sums the series Sum_{k>=1} 1/a(k) converges to 1 + phi, with phi = A001622. See the formulas by Gary W. Adamson and Diego Rattaggi below. (End)

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007

Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010

Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014

Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015

The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

This is a divisibility sequence, that is, a(n) divides a(m) whenever n divides m. The sequence satisfies a linear recurrence of order 4. Cf. A273623.

More generally, for distinct integers r and s with r == s (mod 2), the sequence Lucas(r*n) - Lucas(s*n) is a fourth-order divisibility sequence. When r is even (resp. odd) the normalized sequence (Lucas(r*n) - Lucas(s*n))/(Lucas(r) - Lucas(s)), with initial term equal to 1, has the o.g.f. x*(1 - x^2)/( (1 - Lucas(r)*x + x^2)*(1 - Lucas(s)*x + x^2) ) (resp. x*(1 + x^2)/( (1 - Lucas(r)*x - x^2)*(1 - Lucas(s)*x - x^2) )) and belongs to the 3-parameter family of fourth-order divisibility sequences found by Williams and Guy, with parameter values P1 = (Lucas(r) + Lucas(s)), P2 = Lucas(r)*Lucas(s) and Q = 1 (resp. Q = -1). For particular cases see A004146 (r = 2, s = 0), A049684 (r = 4, s = 0), A215465 (r = 4, s = 2), A049683 (r = 6, s = 0), A049682 (r = 8, s = 0) and A037451 (r = 3, s = -1).

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

From William P. Orrick, Dec 20 2023: (Start)

Every term is a Markov number (see A002559) and, for n > 1, corresponds to a node of the Markov tree A368546 whose sibling and ancestors are all odd-indexed Fibonacci numbers. For n > 1, a(n) is the label of the node obtained from the root by going left n - 2 times and then right. Its Farey index, described in the comments to A368546, is 2 / (2*n - 1).

For instance, a(3) = 194 comes from going left once from the root node of the Markov tree and then right, which corresponds to the sequence of Markov numbers 5, 13, 194. The corresponding sequence of Farey indices is 1/2, 1/3, 2/5. The sibling of the final node corresponds to Markov number 34 and Farey index 1/4. (End)

Table starts

.0........1...........2...............4.................8................16

.0........9..........50.............484..............3864.............32400

.0.......64........1056...........45369...........1337130..........44315649

.0......441.......21660.........4096576.........438377904.......56606878084

.0.....3025......441720.......367565584......142283547000....71306050492681

.0....20736.....8997120.....32940435025....46087307409528.89545789526100625

.0...142129...183206016...2951293792489.14921719105582926

.0...974169..3730330800.264404315544196

.0..6677056.75953515968

.0.45765225

Even columns are perfect squares

This is a divisibility sequence.