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The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity

V(p-J(p,D)) == 2*J(p,D) (mod p) when p is prime, b=-1 and D=a^2+4.

This sequence has the odd composite integers with V(m-J(m,D)) == 2*J(m,D) (mod m).

For a=7 and b=-1, we have D=53 and V(m) recovers A086902(m).

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1)*J(p,D) (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.

The composite integers m with the property V(k*m-J(m,D)) == V(k-1)*J(m,D) (mod m) are called generalized Pell-Lucas pseudoprimes of level k- and parameter a.

Here b=-1, a=7, D=53 and k=2, while V(m) recovers A086902(m).

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1)*J(p,D) (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.

The composite integers m with the property V(k*m-J(m,D)) == V(k-1)*J(m,D) (mod m) are called generalized Pell-Lucas pseudoprimes of level k- and parameter a.

Here b=-1, a=7, D=53 and k=3, while V(m) recovers A086902(m), with V(2)=51.

If p is a prime, then A086902(p)==7 (mod p).

This sequence contains the odd composite integers for which the congruence holds.

The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.

For a=7, b=-1, V(m) recovers A086902(m).

If p is a prime, then A086902(p)==7 (mod p).

This sequence contains the even composite integers for which the congruence holds.

The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.

For a=7, b=-1, V(m) recovers A086902(m).

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

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).

a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.

As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001

a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003

Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006

Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.

The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005

Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov

1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006

For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007

Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007

{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009

Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010

For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010

The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010

n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.

a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011

Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011

a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011

Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011

If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014

The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014

a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014

a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016

Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016

For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017

Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018

a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019

a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022

From Klaus Purath, Apr 03 2025: (Start)

The odd prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.

It follows from the above that this sequence is a subsequence of A031396. (End)

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = -4. See A078356, A078357.

1/5 + 1/13 + 1/29 + ... = (Pi/8)*tanh Pi [Jolley]. - Gary W. Adamson, Dec 21 2006

Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n+1)^2 + 4), n = 1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010

(2*n + 1 + sqrt(a(n)))/2 = [2*n + 1; 2*n + 1, 2*n + 1, ...], n >= 0, with the regular continued fraction with period length 1. This is the odd case. See A087475 for the general case with the Schroeder reference and comments. For the even case see A002522.

Primes in the sequence are in A005473. - Russ Cox, Aug 26 2019

The continued fraction expansion of sqrt(a(n)) is [2n+1; {n, 1, 1, n, 4n+2}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 27 2022

Discriminant of the binary quadratic forms y^2 - x*y - A002061(n+1)*x^2. - Klaus Purath, Nov 10 2022

From Klaus Purath, Apr 08 2025: (Start)

There are no squares in this sequence. The prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*(m*y)^2 = -1 for any integer n where m = (D - 3)/2. The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*m^2 - 1), x(0) = 1, x(1) = 4*D*m^2 - 1, y(0) = 1, y(1) = 4*D*m^2 - 3. The two recurrences are of the form (4*D*m^2 - 2, -1).

It follows from the above that this sequence belongs to A031396. (End)

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019

a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]

For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

From Michael A. Allen, Feb 15 2023: (Start)

Also called the 5-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 5 kinds of squares available. (End)